This document discusses thermochemistry and energy changes that occur during chemical reactions. It defines exothermic and endothermic reactions, and how to construct energy level diagrams to represent them. Specific heats of reaction like combustion, precipitation, displacement, and neutralization are also explained. Experiments to determine various heats of reaction are described. The relationships between the heat of reaction and type of reactants, as well as the number of carbons in alcohols are also summarized.
This document discusses electrochemistry and voltaic cells. It begins by defining electrochemistry as the interconversion of chemical and electrical energy. It then discusses electrolysis and voltaic cells. Electrolysis involves using electricity to break down substances, while voltaic cells convert chemical energy to electrical energy. The document goes on to describe the components and reactions of voltaic cells, including simple voltaic cells and Daniell cells. It also discusses applications of electrolysis in industries such as metal extraction and electroplating.
The document discusses rate of reaction and factors that affect it. It defines rate of reaction as the change in amount of reactants or products per unit time. Rate of reaction is affected by several factors including surface area, concentration, temperature, catalysts and pressure (for gas reactions). The collision theory is also explained, stating that reactions only occur during effective collisions where particles attain sufficient kinetic energy to overcome the activation energy barrier. Examples of how scientific understanding of rate of reaction enhances quality of life through applications like food storage, cooking and petroleum processing are provided.
The document describes several experiments involving qualitative analysis of salts and acids. Salt X is identified as lead(II) carbonate from its reaction to produce solid Y and carbon dioxide gas when heated. Solution W is found to contain nitrate ions, identified through a test producing a brown ring with sulfuric acid and iron(II) sulfate. A precipitation reaction between solutions of lead nitrate and potassium iodide is used to calculate the mass of yellow precipitate formed. Strong acids such as hydrochloric acid and sulfuric acid are described as reacting with bases to produce salts and water, with metals to produce salts and hydrogen gas, and with carbonates to produce salts, carbon dioxide, and water. The role of water in
This document discusses the preparation and classification of salts. Salts are formed through the replacement of hydrogen ions in acids by metal ions or ammonium ions. There are two main methods for preparing salts - neutralization and precipitation. Neutralization involves reacting an acid with a metal, alkali, oxide or carbonate to form a soluble salt. Precipitation involves mixing two aqueous solutions of soluble salts to form an insoluble salt precipitate. The document provides examples of preparing various salts such as potassium chloride and lead chloride. It also discusses classifying salts as soluble or insoluble and purifying soluble salts through recrystallization.
This document provides a summary of Chapter 5 on Indices and Logarithms from an Additional Mathematics textbook. It includes examples and explanations of:
1. Laws of indices such as addition, subtraction, multiplication and division of indices.
2. Converting expressions between index form and logarithmic form using common logarithms and other bases.
3. Applying the laws of logarithms including addition, subtraction, and change of base.
4. Solving equations involving indices and logarithms through appropriate applications of index laws and logarithmic properties.
This document discusses thermochemistry and energy changes that occur during chemical reactions. It defines exothermic and endothermic reactions, and how to construct energy level diagrams to represent them. Specific heats of reaction like combustion, precipitation, displacement, and neutralization are also explained. Experiments to determine various heats of reaction are described. The relationships between the heat of reaction and type of reactants, as well as the number of carbons in alcohols are also summarized.
This document discusses electrochemistry and voltaic cells. It begins by defining electrochemistry as the interconversion of chemical and electrical energy. It then discusses electrolysis and voltaic cells. Electrolysis involves using electricity to break down substances, while voltaic cells convert chemical energy to electrical energy. The document goes on to describe the components and reactions of voltaic cells, including simple voltaic cells and Daniell cells. It also discusses applications of electrolysis in industries such as metal extraction and electroplating.
The document discusses rate of reaction and factors that affect it. It defines rate of reaction as the change in amount of reactants or products per unit time. Rate of reaction is affected by several factors including surface area, concentration, temperature, catalysts and pressure (for gas reactions). The collision theory is also explained, stating that reactions only occur during effective collisions where particles attain sufficient kinetic energy to overcome the activation energy barrier. Examples of how scientific understanding of rate of reaction enhances quality of life through applications like food storage, cooking and petroleum processing are provided.
The document describes several experiments involving qualitative analysis of salts and acids. Salt X is identified as lead(II) carbonate from its reaction to produce solid Y and carbon dioxide gas when heated. Solution W is found to contain nitrate ions, identified through a test producing a brown ring with sulfuric acid and iron(II) sulfate. A precipitation reaction between solutions of lead nitrate and potassium iodide is used to calculate the mass of yellow precipitate formed. Strong acids such as hydrochloric acid and sulfuric acid are described as reacting with bases to produce salts and water, with metals to produce salts and hydrogen gas, and with carbonates to produce salts, carbon dioxide, and water. The role of water in
This document discusses the preparation and classification of salts. Salts are formed through the replacement of hydrogen ions in acids by metal ions or ammonium ions. There are two main methods for preparing salts - neutralization and precipitation. Neutralization involves reacting an acid with a metal, alkali, oxide or carbonate to form a soluble salt. Precipitation involves mixing two aqueous solutions of soluble salts to form an insoluble salt precipitate. The document provides examples of preparing various salts such as potassium chloride and lead chloride. It also discusses classifying salts as soluble or insoluble and purifying soluble salts through recrystallization.
This document provides a summary of Chapter 5 on Indices and Logarithms from an Additional Mathematics textbook. It includes examples and explanations of:
1. Laws of indices such as addition, subtraction, multiplication and division of indices.
2. Converting expressions between index form and logarithmic form using common logarithms and other bases.
3. Applying the laws of logarithms including addition, subtraction, and change of base.
4. Solving equations involving indices and logarithms through appropriate applications of index laws and logarithmic properties.
It's very good for SPM students . You have to learn the ionic bond thoroughly. If you understand well you can explain it vividly. For other chemistry notes can email me puterizamrud@gmail.com or facebook Pusat Tuisyen Zamrud .
Acids and bases are defined based on their ability to produce hydrogen or hydroxide ions in water. Acids produce hydrogen ions and bases produce hydroxide ions. Examples of common acids include hydrochloric acid, sulfuric acid, and citric acid. Common bases include sodium hydroxide and calcium hydroxide. Acids and bases have many uses from manufacturing to agriculture to medicine. They require water to show their acidic or alkaline properties by dissociating into ions.
This document provides an introduction to thermochemistry and the key concepts of enthalpy, enthalpy change, and standard enthalpy of formation. It defines system and surroundings, and the three types of systems - open, closed, and isolated. The key points are:
- Enthalpy change (ΔH) is the difference in enthalpies between products and reactants and indicates whether a reaction is endothermic or exothermic.
- Standard enthalpy of formation (H°f) is the enthalpy change when 1 mole of a substance is formed from its elements under standard conditions.
- Enthalpy of combustion (H°c) is the enthalpy change when 1 mole
The document provides information on several chemistry concepts and experiments. It includes:
1) A chapter on matter that discusses states of matter, kinetic theory, and heating curves.
2) Chapters on chemical formulas, periodic table, chemical bonds, and electrochemistry.
3) An experiment on determining the end point of a neutralization reaction between potassium hydroxide and hydrochloric acid.
The document defines oxidation number and provides rules for determining oxidation numbers of elements in compounds and polyatomic ions. The rules state that the oxidation number of atoms is 0, ions take the charge, and the sum of oxidation numbers in compounds and polyatomic ions equals the overall charge. Examples are provided to demonstrate applying the rules to calculate the oxidation number of underlined elements in various compounds and polyatomic ions.
Includes a discussion of Voltaic and electrolytic cells, the Nernst equation and the relationship between electrochemical processes, chemical equilibrium and free energy.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
The document provides a 5-step process for balancing redox reactions using the half-reaction method. It explains that redox reactions are split into oxidation and reduction half-reactions, which are then balanced individually before being combined to give the overall balanced redox equation. It also describes an alternative oxidation-number change method for balancing redox reactions.
Chemical formulae, equations, calculations, and reactions are summarized. Molar mass, moles, volume, and molarity calculations are explained for gases, solids, liquids, and solutions. Common cationic and anionic symbols are listed. Formulae for molecules and ions are provided. Periodic trends and reactions of Groups 1 and 17 are summarized. Electrochemistry principles of electrolytes, discharge reactions, and test observations are condensed. Characteristics of acids, bases, and ionization are highlighted. Solubility, preparation, color, and effects of heating for various salts are summarized concisely.
This document discusses the uses of electrolysis in industries, including the purification of metals, electroplating of metals, and extraction of metals. It provides examples of how electrolysis is used to purify copper, electroplate metals like tin onto cans, and extract reactive metals like aluminum from ores. The document also notes some of the potential pollution problems caused by electrolysis in industry, such as releasing heavy metals and altering the pH of water resources.
Dokumen tersebut membahas tentang pengoksidaan dan penurunan. Pengoksidaan adalah tindak balas yang melibatkan penambahan oksigen, penurunan elektron, atau peningkatan nomor pengoksidaan, sedangkan penurunan adalah tindak balas yang melibatkan kehilangan oksigen, penerimaan elektron, atau penurunan nomor pengoksidaan. Dokumen tersebut juga menjelaskan konsep reaksi redoks yang melibatkan peng
This document provides information about carbon compounds and their properties. It discusses organic compounds such as hydrocarbons, alcohols, carboxylic acids, and esters. For hydrocarbons, it describes the properties of alkanes such as their electrical conductivity, density, and how their melting/boiling points increase with more carbon atoms. It also discusses chemical tests and reactions that can be used to differentiate alkanes from alkenes, such as their reactions with bromine water and acidified potassium manganate(VII) solution. Industrial production of ethanol is outlined involving fermentation of sugars or hydration of ethene. Key reactions of alcohols like combustion, oxidation, and dehydration are
The document discusses acids and bases. It defines acids as compounds that ionize in water to produce hydrogen ions, and bases as compounds that react with acids to produce salts and water. Alkalis are bases that ionize in water to produce hydroxide ions. Water is necessary for acids and alkalis to exhibit their properties, as it allows them to dissociate into ions. The document also outlines the chemical properties of acids and bases, such as their reactions with each other, metals, and carbonates to produce salts, water, hydrogen gas or carbon dioxide. Common uses of acids and bases in daily life are also mentioned.
The document summarizes key aspects of the periodic table including:
1) Elements are arranged in order of increasing proton number and elements with similar properties are in the same group, with groups 1-18 and periods 1-7.
2) Group 18 elements are noble gases with full outer shells making them chemically inert monoatomic gases.
3) Group 1 elements are alkali metals that react vigorously with water and oxygen and reactivity increases down the group as atomic size increases.
4) Group 17 elements are halogens that exist as diatomic molecules and reactivity decreases down the group as atomic size increases.
5) Period 3 elements include metals, nonmetals and semi-metals that show trends
Here are some potential solutions to optimize industrial processes to produce higher yields at lower cost:
1. Use a catalyst. Adding a catalyst can lower the activation energy of the reaction, allowing it to proceed at a faster rate even at lower temperatures and pressures. This reduces energy costs while maintaining or increasing product yields.
2. Improve reactor design. Advanced reactor designs that improve heat and mass transfer can allow reactions to reach optimum conditions more efficiently. Examples include continuous flow reactors, microreactors, and reactive distillation columns.
3. Employ process intensification techniques. Methods like ultrasound, microwave irradiation, and supercritical fluid processing can accelerate reaction kinetics, reducing processing time. Some also allow operation at lower temperatures and pressures.
The document reports on an experiment to determine the enthalpy change of the neutralization reaction between sodium hydroxide and hydrochloric acid. 150 mL of 1M HCl and 50 mL of 1M NaOH were mixed in a polystyrene cup calorimeter. The temperature increase of 0.75°C was used to calculate the energy transferred of 156.75 J. Thermochemistry principles are discussed including definition of enthalpy change, methods to determine it including calorimetry, and equations used to calculate energy from temperature change measurements in solution calorimetry.
The document describes the process and tests used in qualitative analysis to identify salts based on their physical properties, reaction to heat, and tests to detect specific cations and anions. It provides details on observing the color and solubility of salts, conducting gas tests, and using confirmatory tests to identify ions like Fe2+, Fe3+, Pb2+, and NH4+. The qualitative analysis plan involves examining the salt's physical properties, heating it, testing for cations and anions, and then confirming the identities of ions present.
This document provides information about circular measure including radians, conversion between radians and degrees, length of arc, and area of sectors. It defines a radian as the angle subtended by an arc equal in length to the radius. Formulas are given for converting between radians and degrees, finding the length of an arc given the radian measure of its central angle, and finding the area of a sector given its radian measure and the radius. Several examples demonstrate applying these formulas to solve problems involving radians. Exercises provide additional practice problems for students to work through.
The document discusses rate of reaction and factors that affect it. It defines rate of reaction as the change in amount of reactants or products per unit time. It describes several factors that affect rate based on collision theory, including surface area, concentration, temperature, catalysts, and pressure. It gives examples of how scientific understanding of rate of reaction enhances quality of life, such as refrigeration, pressure cooking, cutting food into smaller pieces, making margarine, and burning coal.
The document discusses balancing redox reactions using the half-reaction method. It provides several examples of writing and balancing half-reactions and using them to derive the overall balanced redox equation. Key steps include separating the reaction into oxidation and reduction half-reactions, balancing all elements except H and O, adding H2O to balance O, adding H+ or OH- to balance H, and adding electrons to balance charge.
Redox reactions involve the transfer of electrons between reactants. In an electrochemical cell, a spontaneous redox reaction occurs between two half-cells separated by a salt bridge. In the oxidation half-reaction, electrons are lost at the anode. In the reduction half-reaction, electrons are gained at the cathode. The standard electrode potential (E0) of a half-reaction indicates its tendency to be reduced or oxidized relative to the standard hydrogen electrode. The cell potential (Ecell) is equal to the cathode potential minus the anode potential and determines if the cell reaction is spontaneous.
It's very good for SPM students . You have to learn the ionic bond thoroughly. If you understand well you can explain it vividly. For other chemistry notes can email me puterizamrud@gmail.com or facebook Pusat Tuisyen Zamrud .
Acids and bases are defined based on their ability to produce hydrogen or hydroxide ions in water. Acids produce hydrogen ions and bases produce hydroxide ions. Examples of common acids include hydrochloric acid, sulfuric acid, and citric acid. Common bases include sodium hydroxide and calcium hydroxide. Acids and bases have many uses from manufacturing to agriculture to medicine. They require water to show their acidic or alkaline properties by dissociating into ions.
This document provides an introduction to thermochemistry and the key concepts of enthalpy, enthalpy change, and standard enthalpy of formation. It defines system and surroundings, and the three types of systems - open, closed, and isolated. The key points are:
- Enthalpy change (ΔH) is the difference in enthalpies between products and reactants and indicates whether a reaction is endothermic or exothermic.
- Standard enthalpy of formation (H°f) is the enthalpy change when 1 mole of a substance is formed from its elements under standard conditions.
- Enthalpy of combustion (H°c) is the enthalpy change when 1 mole
The document provides information on several chemistry concepts and experiments. It includes:
1) A chapter on matter that discusses states of matter, kinetic theory, and heating curves.
2) Chapters on chemical formulas, periodic table, chemical bonds, and electrochemistry.
3) An experiment on determining the end point of a neutralization reaction between potassium hydroxide and hydrochloric acid.
The document defines oxidation number and provides rules for determining oxidation numbers of elements in compounds and polyatomic ions. The rules state that the oxidation number of atoms is 0, ions take the charge, and the sum of oxidation numbers in compounds and polyatomic ions equals the overall charge. Examples are provided to demonstrate applying the rules to calculate the oxidation number of underlined elements in various compounds and polyatomic ions.
Includes a discussion of Voltaic and electrolytic cells, the Nernst equation and the relationship between electrochemical processes, chemical equilibrium and free energy.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
The document provides a 5-step process for balancing redox reactions using the half-reaction method. It explains that redox reactions are split into oxidation and reduction half-reactions, which are then balanced individually before being combined to give the overall balanced redox equation. It also describes an alternative oxidation-number change method for balancing redox reactions.
Chemical formulae, equations, calculations, and reactions are summarized. Molar mass, moles, volume, and molarity calculations are explained for gases, solids, liquids, and solutions. Common cationic and anionic symbols are listed. Formulae for molecules and ions are provided. Periodic trends and reactions of Groups 1 and 17 are summarized. Electrochemistry principles of electrolytes, discharge reactions, and test observations are condensed. Characteristics of acids, bases, and ionization are highlighted. Solubility, preparation, color, and effects of heating for various salts are summarized concisely.
This document discusses the uses of electrolysis in industries, including the purification of metals, electroplating of metals, and extraction of metals. It provides examples of how electrolysis is used to purify copper, electroplate metals like tin onto cans, and extract reactive metals like aluminum from ores. The document also notes some of the potential pollution problems caused by electrolysis in industry, such as releasing heavy metals and altering the pH of water resources.
Dokumen tersebut membahas tentang pengoksidaan dan penurunan. Pengoksidaan adalah tindak balas yang melibatkan penambahan oksigen, penurunan elektron, atau peningkatan nomor pengoksidaan, sedangkan penurunan adalah tindak balas yang melibatkan kehilangan oksigen, penerimaan elektron, atau penurunan nomor pengoksidaan. Dokumen tersebut juga menjelaskan konsep reaksi redoks yang melibatkan peng
This document provides information about carbon compounds and their properties. It discusses organic compounds such as hydrocarbons, alcohols, carboxylic acids, and esters. For hydrocarbons, it describes the properties of alkanes such as their electrical conductivity, density, and how their melting/boiling points increase with more carbon atoms. It also discusses chemical tests and reactions that can be used to differentiate alkanes from alkenes, such as their reactions with bromine water and acidified potassium manganate(VII) solution. Industrial production of ethanol is outlined involving fermentation of sugars or hydration of ethene. Key reactions of alcohols like combustion, oxidation, and dehydration are
The document discusses acids and bases. It defines acids as compounds that ionize in water to produce hydrogen ions, and bases as compounds that react with acids to produce salts and water. Alkalis are bases that ionize in water to produce hydroxide ions. Water is necessary for acids and alkalis to exhibit their properties, as it allows them to dissociate into ions. The document also outlines the chemical properties of acids and bases, such as their reactions with each other, metals, and carbonates to produce salts, water, hydrogen gas or carbon dioxide. Common uses of acids and bases in daily life are also mentioned.
The document summarizes key aspects of the periodic table including:
1) Elements are arranged in order of increasing proton number and elements with similar properties are in the same group, with groups 1-18 and periods 1-7.
2) Group 18 elements are noble gases with full outer shells making them chemically inert monoatomic gases.
3) Group 1 elements are alkali metals that react vigorously with water and oxygen and reactivity increases down the group as atomic size increases.
4) Group 17 elements are halogens that exist as diatomic molecules and reactivity decreases down the group as atomic size increases.
5) Period 3 elements include metals, nonmetals and semi-metals that show trends
Here are some potential solutions to optimize industrial processes to produce higher yields at lower cost:
1. Use a catalyst. Adding a catalyst can lower the activation energy of the reaction, allowing it to proceed at a faster rate even at lower temperatures and pressures. This reduces energy costs while maintaining or increasing product yields.
2. Improve reactor design. Advanced reactor designs that improve heat and mass transfer can allow reactions to reach optimum conditions more efficiently. Examples include continuous flow reactors, microreactors, and reactive distillation columns.
3. Employ process intensification techniques. Methods like ultrasound, microwave irradiation, and supercritical fluid processing can accelerate reaction kinetics, reducing processing time. Some also allow operation at lower temperatures and pressures.
The document reports on an experiment to determine the enthalpy change of the neutralization reaction between sodium hydroxide and hydrochloric acid. 150 mL of 1M HCl and 50 mL of 1M NaOH were mixed in a polystyrene cup calorimeter. The temperature increase of 0.75°C was used to calculate the energy transferred of 156.75 J. Thermochemistry principles are discussed including definition of enthalpy change, methods to determine it including calorimetry, and equations used to calculate energy from temperature change measurements in solution calorimetry.
The document describes the process and tests used in qualitative analysis to identify salts based on their physical properties, reaction to heat, and tests to detect specific cations and anions. It provides details on observing the color and solubility of salts, conducting gas tests, and using confirmatory tests to identify ions like Fe2+, Fe3+, Pb2+, and NH4+. The qualitative analysis plan involves examining the salt's physical properties, heating it, testing for cations and anions, and then confirming the identities of ions present.
This document provides information about circular measure including radians, conversion between radians and degrees, length of arc, and area of sectors. It defines a radian as the angle subtended by an arc equal in length to the radius. Formulas are given for converting between radians and degrees, finding the length of an arc given the radian measure of its central angle, and finding the area of a sector given its radian measure and the radius. Several examples demonstrate applying these formulas to solve problems involving radians. Exercises provide additional practice problems for students to work through.
The document discusses rate of reaction and factors that affect it. It defines rate of reaction as the change in amount of reactants or products per unit time. It describes several factors that affect rate based on collision theory, including surface area, concentration, temperature, catalysts, and pressure. It gives examples of how scientific understanding of rate of reaction enhances quality of life, such as refrigeration, pressure cooking, cutting food into smaller pieces, making margarine, and burning coal.
The document discusses balancing redox reactions using the half-reaction method. It provides several examples of writing and balancing half-reactions and using them to derive the overall balanced redox equation. Key steps include separating the reaction into oxidation and reduction half-reactions, balancing all elements except H and O, adding H2O to balance O, adding H+ or OH- to balance H, and adding electrons to balance charge.
Redox reactions involve the transfer of electrons between reactants. In an electrochemical cell, a spontaneous redox reaction occurs between two half-cells separated by a salt bridge. In the oxidation half-reaction, electrons are lost at the anode. In the reduction half-reaction, electrons are gained at the cathode. The standard electrode potential (E0) of a half-reaction indicates its tendency to be reduced or oxidized relative to the standard hydrogen electrode. The cell potential (Ecell) is equal to the cathode potential minus the anode potential and determines if the cell reaction is spontaneous.
The document discusses oxidation-reduction (redox) reactions and oxidation numbers. It provides examples of half reactions and full redox reactions formed by combining half reactions. Oxidation involves an increase in oxidation state through loss of electrons, while reduction involves a decrease in oxidation state through gain of electrons. The species donating electrons is the reducing agent, while the species gaining electrons is the oxidizing agent.
This document discusses oxidation and reduction reactions. It begins by defining oxidation as a reaction where substances combine with oxygen and reduction as a reaction where a substance "gave up" oxygen. It then explains that oxidation and reduction actually refer to the gain or loss of electrons in a chemical reaction, regardless of whether oxygen is present. Oxidation involves the loss of electrons, while reduction involves the gain of electrons. Redox reactions always involve both oxidation and reduction occurring together through the transfer of electrons. The document provides examples of how to identify the oxidizing agent, reducing agent, and what is being oxidized and reduced in redox reactions. It also discusses how to balance redox reactions through half-reactions and the role of acid and
Redox reactions involve the simultaneous oxidation and reduction of reactants. Oxidation is defined as an increase in oxidation state or loss of electrons, while reduction is a decrease in oxidation state or gain of electrons. There are two main methods for balancing redox reactions - the ion-electron method which balances the ions and electrons, and the oxidation state method which balances the changes in oxidation states of elements between reactants and products. Balancing redox reactions involves identifying changes in oxidation states, cross multiplying these changes, and then balancing other elements such as hydrogen and oxygen.
Redox reactions involve the transfer of electrons between species. There are two types of agents involved - oxidizing agents that reduce other species by accepting electrons, and reducing agents that oxidize other species by donating electrons. Identification of redox reactions involves looking for a change in oxidation state between reactants and products. Balancing redox reactions uses the ion-electron method of writing and balancing half reactions for oxidation and reduction and combining them. Organic redox reactions use a similar process by writing oxidation and reduction half reactions and balancing mass, charge, and electrons.
This document defines key terms related to redox reactions and oxidation states. It discusses oxidation and reduction in terms of electron transfer and changes in oxidation state. Oxidation involves the loss of electrons and an increase in oxidation state, while reduction involves the gain of electrons and a decrease in oxidation state. The document also covers how to calculate oxidation states and provides examples of balancing redox reactions.
This document provides information on acid-base reactions and oxidation-reduction (redox) reactions. It defines acids and bases, and explains that in acid-base reactions, acids donate protons to bases. Neutralization reactions between acids and bases produce water and a salt. The document also discusses how to determine oxidation states of elements in compounds and identify the oxidized and reduced substances in redox reactions. It provides steps for balancing redox equations, including dividing the reaction into partial equations and adding electrons to balance charges. Examples of assigning oxidation states and balancing redox reactions are included.
This document provides information on calculating oxidation numbers and discusses redox reactions. It contains:
1) Five rules for determining oxidation numbers of elements in compounds and ions. These rules state that oxidation numbers of elements in their elemental forms are zero, ions take the charge of the ion, oxygen is -2 except in peroxides, hydrogen is +1 except in hydrides, and the sum of oxidation numbers equals the net charge.
2) Examples of using the rules to calculate oxidation numbers in various compounds and ions, including transition metal compounds that use Roman numerals.
3) A statement that oxidation is an increase in oxidation number and reduction is a decrease in oxidation number during a redox reaction.
The document provides information about a chemistry problem involving the molecular weight of a compound. A 3.41 x 10-6 g sample of the compound contains 4.67 x 10^16 molecules. Calculating the molecular weight based on the moles of molecules and mass of the sample gives a value of 44.0 g/mole. Of the compounds listed, CO2 has a molecular weight that matches this value.
The document discusses two methods for balancing redox reactions:
1) Using oxidation number changes, which involves assigning oxidation numbers, identifying oxidized and reduced elements, and using coefficients to equalize oxidation number changes.
2) Using half-reactions, which involves writing separate half-reactions for oxidation and reduction, balancing atoms, adding electrons, and combining half-reactions to give the overall balanced equation. The half-reaction method is best for reactions in acidic or alkaline solutions.
The document discusses two methods for balancing redox reactions:
1) Using oxidation number changes, which involves assigning oxidation numbers, identifying oxidized and reduced elements, and using coefficients to equalize oxidation number changes.
2) Using half-reactions, which involves writing separate half-reactions for oxidation and reduction, balancing atoms, adding electrons, and combining half-reactions to give the overall balanced equation. The half-reaction method is best for reactions in acidic or alkaline solutions.
Reduction - Oxidation Titrations Theory and Application.pptAbdelrhman abooda
This document discusses redox (oxidation-reduction) titrations. It begins by defining redox reactions and key terms like oxidizing agent and reducing agent. It then discusses oxidation numbers and how to calculate them. Several examples of oxidation numbers are provided. The document also covers balancing redox reactions using the half-reaction method. Factors that affect oxidation potential are explained, including common ions, pH, complexing agents, and precipitating agents. Standard reduction potentials of various redox couples are presented. Finally, the shape of a typical redox titration curve is briefly described.
Okay, here are the steps to balance this reaction:
Step 1) Identify oxidizing and reducing agents:
MnO4- is reduced, so it is the oxidizing agent.
MnO4- + 5e- → Mn2+
SO2 is oxidized, so it is the reducing agent.
SO2 → SO42- + 4e-
Step 2) Balance other elements: No need here.
Step 3) Balance O by adding H2O:
MnO4- + 5e- → Mn2+ + 4H2O
SO2 → SO42- + 4e-
Step 4) Balance H by adding H+:
M
IB Chemistry on Redox, Oxidizing, Reducing Agents and writing half redox equa...Lawrence kok
The document discusses oxidation numbers (also called oxidation states), which are used to keep track of electrons in chemical reactions. Some key points:
- Oxidation numbers are assigned to each atom in a chemical species by assuming ionic bonding and counting electrons.
- Common rules are outlined for assigning oxidation numbers to elements, such as metals in Group 1 have a +1 oxidation state and nonmetals in Group 7 have a -1 oxidation state.
- Oxidation numbers can be used to determine if a reaction is a redox reaction by looking for changes in oxidation numbers between reactants and products.
- Transition metals can have multiple common oxidation states. Roman numerals are used to distinguish, such as
1) A chemical reaction can be differentiated from a phase change or nuclear change based on whether it involves a change in chemical composition, state of matter, or type of element.
2) In a chemical reaction, reactants yield products through the rearrangement of atoms. Chemical equations use symbols to represent the reactants and products.
3) Evidence that a chemical reaction occurred includes the release or absorption of energy, production of a gas, formation of a precipitate, or a color change.
When the following skeletal equation is balanced under basic condition.docxSUKHI5
When the following skeletal equation is balanced under basic conditions, what are the coefficients of the species shown?
SO 3 2- + Cl - --> Cl 2 + S 2 O 3 2-
Water appears in the balanced equation as a (reactant, product, neither) with a coefficient of ____. (Enter 0 for neither.)
Which species is the oxidizing agent?
Solution
Oxidation number of each element in SO3-2 is
O=-2
S=+4
Oxidation number of each element in Cl-1 is
Cl=-1
Oxidation number of each element in S2O3-2 is
O=-2
S=+2
Oxidation number of each element in Cl2 is
Cl=0
Oxidation number of each element in reactant is
O=-2
S=+4
Cl=-1
Oxidation number of each element in product is
O=-2
S=+2
Cl=0
S in SO3-2 has oxidation state of +4
S in S2O3-2 has oxidation state of +2
So, S in SO3-2 is reduced to S2O3-2
This make SO3-2 as oxidising agent
Cl in Cl- has oxidation state of -1
Cl in Cl2 has oxidation state of 0
So, Cl in Cl- is oxidised to Cl2
This make Cl- as reducing agent
Reduction half cell:
2 SO3-2 + 4e- --> S2O3-2
Oxidation half cell:
2 Cl- --> Cl2 + 2e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
2 SO3-2 + 4e- --> S2O3-2
Oxidation half cell:
4 Cl- --> 2 Cl2 + 4e-
Lets combine both the reactions.
2 SO3-2 + 4 Cl- --> S2O3-2 + 2 Cl2
Balance Oxygen by adding water
2 SO3-2 + 4 Cl- --> S2O3-2 + 2 Cl2 + 3 H2O
Balance Hydrogen by adding H+
2 SO3-2 + 4 Cl- + 6 H+ --> S2O3-2 + 2 Cl2 + 3 H2O
Add equal number of OH- on both sides as the number of H+
2 SO3-2 + 4 Cl- + 6 H+ + 6 OH- --> S2O3-2 + 2 Cl2 + 3 H2O + 6 OH-
Combine H+ and OH- to form water
2 SO3-2 + 4 Cl- + 6 H2O --> S2O3-2 + 2 Cl2 + 3 H2O + 6 OH-
Remove common H2O from both sides
Balanced Eqn is
2 SO3-2 + 4 Cl- + 3 H2O --> S2O3-2 + 2 Cl2 + 6 OH-
This is balanced chemical equation in basic medium
Water appear as reactant with a coefficient of 3
SO 3 2- is oxidizing agent
.
IB Chemistry on Redox, Oxidizing, Reducing Agents and writing half redox equa...Lawrence kok
This document provides a tutorial on redox reactions, oxidation states, and half reactions. It begins by defining redox as involving both oxidation and reduction, which is the loss or gain of electrons. It then discusses oxidation states (also called oxidation numbers), which are assigned to atoms in compounds to keep track of electrons. Rules are provided for assigning oxidation states based on electronegativity. Redox reactions involve a change in oxidation states between reactants and products. Examples of assigning oxidation states to elements in compounds are also given.
Similar to Inorganic Chemistry : Electrochemistry (20)
Forestland soil was the most permeable to water, allowing water to pass through in just a few minutes with 0% porosity. Clay soil was the least permeable, not allowing any water to pass through and having 100% porosity. Riverbank soil and beach soil had intermediate permeability, with riverbank soil having lower permeability than beach soil as indicated by the longer time for water to pass through. Porosity and permeability were found to be related, with soils having more pore space (higher porosity) exhibiting lower permeability.
To study the properties, nomenclature and the physical as well chemical reactions of aliphatic and alkyl benzene. Might as well as the usage of benzene in our daily life routine
This document discusses periodicity and trends in properties across and down periods of the periodic table. It explains that atomic radius generally decreases across periods as nuclear charge increases, outweighing constant screening effects. Atomic radius increases down groups as nuclear charge rises but screening effects also increase. Ionic radius follows similar trends as atomic radius but is smaller for cations and larger for anions. Melting and boiling points are influenced by type and strength of bonding. Metallic bonding results in higher melting points for metals with more delocalized electrons. Network covalent bonding in nonmetals produces high melting points due to needing to overcome many bonds. Molecular nonmetals have weaker van der Waals forces between molecules. First ionization energies also follow trends
Inorganic Chemistry : Transition Elements (Chemical properties of first row i...Thivyaapriya Sambamoorthy
This document discusses the chemical properties of transition elements in the first row of the periodic table. It explains their variable oxidation states in terms of the relative energies of the 3d and 4s orbitals. Transition metals can exist in multiple oxidation states from +1 to +7 due to the small energy difference between these orbitals. Higher oxidation states form covalent oxo ions bonded to oxygen or fluorine. The stability of the +2 and +3 oxidation states varies across the period based on standard electrode potentials, with the +3 state being more stable for early transition metals and the +2 state becoming more stable later in the period.
In this topic , I have classified the classifications of silicates as well as its uses and functions in this modern age . Same goes to silicon and silicone . I also have discussed also the structure of silicone itself . Other than silicon , silicone and silicate , I have also discussed about Zeolites and Tin & Alloys . Enjoy .
Allowing students to rate their teachers could have benefits. Students have first-hand experience in the classroom and can provide feedback on a teacher's teaching ability, subject knowledge, and ability to engage and connect with students. Their feedback could help identify both good teachers and those needing improvement, leading to better quality of education. However, there are also concerns such as students grading teachers poorly due to personal dislikes rather than objective evaluation of teaching. Overall, student ratings may help improve teacher quality if implemented with safeguards against purely subjective ratings.
The document summarizes trends in atomic properties across periods 2 and 3 of the periodic table. Atomic radius decreases across periods due to increasing nuclear charge, while it increases down groups due to greater screening effect. Ionic radius, melting/boiling points, and enthalpy of vaporization follow similar trends. Electrical conductivity increases with more delocalized electrons. Electronegativity increases across periods but decreases down groups.
Some of you might having trouble on understanding the concept of the protein carrier in Co-Transport, well same goes to me. And i do hope this will help you in understanding it better.
Good luck
The pressure-flow hypothesis explains translocation in phloem plants. It states that solutes like sucrose move through phloem via a pressure gradient between source and sink regions. At sources like leaves, photosynthesis produces sucrose which is actively loaded into sieve tubes. This increases pressure, pushing sap toward sinks like roots. At sinks, sucrose is unloaded, decreasing pressure and drawing more sap. The pressure difference drives mass flow through phloem from high to low pressure. Various experiments support this hypothesis, though some questions remain.
The binding of cosmological structures by massless topological defectsSérgio Sacani
Assuming spherical symmetry and weak field, it is shown that if one solves the Poisson equation or the Einstein field
equations sourced by a topological defect, i.e. a singularity of a very specific form, the result is a localized gravitational
field capable of driving flat rotation (i.e. Keplerian circular orbits at a constant speed for all radii) of test masses on a thin
spherical shell without any underlying mass. Moreover, a large-scale structure which exploits this solution by assembling
concentrically a number of such topological defects can establish a flat stellar or galactic rotation curve, and can also deflect
light in the same manner as an equipotential (isothermal) sphere. Thus, the need for dark matter or modified gravity theory is
mitigated, at least in part.
Immersive Learning That Works: Research Grounding and Paths ForwardLeonel Morgado
We will metaverse into the essence of immersive learning, into its three dimensions and conceptual models. This approach encompasses elements from teaching methodologies to social involvement, through organizational concerns and technologies. Challenging the perception of learning as knowledge transfer, we introduce a 'Uses, Practices & Strategies' model operationalized by the 'Immersive Learning Brain' and ‘Immersion Cube’ frameworks. This approach offers a comprehensive guide through the intricacies of immersive educational experiences and spotlighting research frontiers, along the immersion dimensions of system, narrative, and agency. Our discourse extends to stakeholders beyond the academic sphere, addressing the interests of technologists, instructional designers, and policymakers. We span various contexts, from formal education to organizational transformation to the new horizon of an AI-pervasive society. This keynote aims to unite the iLRN community in a collaborative journey towards a future where immersive learning research and practice coalesce, paving the way for innovative educational research and practice landscapes.
The debris of the ‘last major merger’ is dynamically youngSérgio Sacani
The Milky Way’s (MW) inner stellar halo contains an [Fe/H]-rich component with highly eccentric orbits, often referred to as the
‘last major merger.’ Hypotheses for the origin of this component include Gaia-Sausage/Enceladus (GSE), where the progenitor
collided with the MW proto-disc 8–11 Gyr ago, and the Virgo Radial Merger (VRM), where the progenitor collided with the
MW disc within the last 3 Gyr. These two scenarios make different predictions about observable structure in local phase space,
because the morphology of debris depends on how long it has had to phase mix. The recently identified phase-space folds in Gaia
DR3 have positive caustic velocities, making them fundamentally different than the phase-mixed chevrons found in simulations
at late times. Roughly 20 per cent of the stars in the prograde local stellar halo are associated with the observed caustics. Based
on a simple phase-mixing model, the observed number of caustics are consistent with a merger that occurred 1–2 Gyr ago.
We also compare the observed phase-space distribution to FIRE-2 Latte simulations of GSE-like mergers, using a quantitative
measurement of phase mixing (2D causticality). The observed local phase-space distribution best matches the simulated data
1–2 Gyr after collision, and certainly not later than 3 Gyr. This is further evidence that the progenitor of the ‘last major merger’
did not collide with the MW proto-disc at early times, as is thought for the GSE, but instead collided with the MW disc within
the last few Gyr, consistent with the body of work surrounding the VRM.
Remote Sensing and Computational, Evolutionary, Supercomputing, and Intellige...University of Maribor
Slides from talk:
Aleš Zamuda: Remote Sensing and Computational, Evolutionary, Supercomputing, and Intelligent Systems.
11th International Conference on Electrical, Electronics and Computer Engineering (IcETRAN), Niš, 3-6 June 2024
Inter-Society Networking Panel GRSS/MTT-S/CIS Panel Session: Promoting Connection and Cooperation
https://www.etran.rs/2024/en/home-english/
Phenomics assisted breeding in crop improvementIshaGoswami9
As the population is increasing and will reach about 9 billion upto 2050. Also due to climate change, it is difficult to meet the food requirement of such a large population. Facing the challenges presented by resource shortages, climate
change, and increasing global population, crop yield and quality need to be improved in a sustainable way over the coming decades. Genetic improvement by breeding is the best way to increase crop productivity. With the rapid progression of functional
genomics, an increasing number of crop genomes have been sequenced and dozens of genes influencing key agronomic traits have been identified. However, current genome sequence information has not been adequately exploited for understanding
the complex characteristics of multiple gene, owing to a lack of crop phenotypic data. Efficient, automatic, and accurate technologies and platforms that can capture phenotypic data that can
be linked to genomics information for crop improvement at all growth stages have become as important as genotyping. Thus,
high-throughput phenotyping has become the major bottleneck restricting crop breeding. Plant phenomics has been defined as the high-throughput, accurate acquisition and analysis of multi-dimensional phenotypes
during crop growing stages at the organism level, including the cell, tissue, organ, individual plant, plot, and field levels. With the rapid development of novel sensors, imaging technology,
and analysis methods, numerous infrastructure platforms have been developed for phenotyping.
Authoring a personal GPT for your research and practice: How we created the Q...Leonel Morgado
Thematic analysis in qualitative research is a time-consuming and systematic task, typically done using teams. Team members must ground their activities on common understandings of the major concepts underlying the thematic analysis, and define criteria for its development. However, conceptual misunderstandings, equivocations, and lack of adherence to criteria are challenges to the quality and speed of this process. Given the distributed and uncertain nature of this process, we wondered if the tasks in thematic analysis could be supported by readily available artificial intelligence chatbots. Our early efforts point to potential benefits: not just saving time in the coding process but better adherence to criteria and grounding, by increasing triangulation between humans and artificial intelligence. This tutorial will provide a description and demonstration of the process we followed, as two academic researchers, to develop a custom ChatGPT to assist with qualitative coding in the thematic data analysis process of immersive learning accounts in a survey of the academic literature: QUAL-E Immersive Learning Thematic Analysis Helper. In the hands-on time, participants will try out QUAL-E and develop their ideas for their own qualitative coding ChatGPT. Participants that have the paid ChatGPT Plus subscription can create a draft of their assistants. The organizers will provide course materials and slide deck that participants will be able to utilize to continue development of their custom GPT. The paid subscription to ChatGPT Plus is not required to participate in this workshop, just for trying out personal GPTs during it.
When I was asked to give a companion lecture in support of ‘The Philosophy of Science’ (https://shorturl.at/4pUXz) I decided not to walk through the detail of the many methodologies in order of use. Instead, I chose to employ a long standing, and ongoing, scientific development as an exemplar. And so, I chose the ever evolving story of Thermodynamics as a scientific investigation at its best.
Conducted over a period of >200 years, Thermodynamics R&D, and application, benefitted from the highest levels of professionalism, collaboration, and technical thoroughness. New layers of application, methodology, and practice were made possible by the progressive advance of technology. In turn, this has seen measurement and modelling accuracy continually improved at a micro and macro level.
Perhaps most importantly, Thermodynamics rapidly became a primary tool in the advance of applied science/engineering/technology, spanning micro-tech, to aerospace and cosmology. I can think of no better a story to illustrate the breadth of scientific methodologies and applications at their best.
EWOCS-I: The catalog of X-ray sources in Westerlund 1 from the Extended Weste...Sérgio Sacani
Context. With a mass exceeding several 104 M⊙ and a rich and dense population of massive stars, supermassive young star clusters
represent the most massive star-forming environment that is dominated by the feedback from massive stars and gravitational interactions
among stars.
Aims. In this paper we present the Extended Westerlund 1 and 2 Open Clusters Survey (EWOCS) project, which aims to investigate
the influence of the starburst environment on the formation of stars and planets, and on the evolution of both low and high mass stars.
The primary targets of this project are Westerlund 1 and 2, the closest supermassive star clusters to the Sun.
Methods. The project is based primarily on recent observations conducted with the Chandra and JWST observatories. Specifically,
the Chandra survey of Westerlund 1 consists of 36 new ACIS-I observations, nearly co-pointed, for a total exposure time of 1 Msec.
Additionally, we included 8 archival Chandra/ACIS-S observations. This paper presents the resulting catalog of X-ray sources within
and around Westerlund 1. Sources were detected by combining various existing methods, and photon extraction and source validation
were carried out using the ACIS-Extract software.
Results. The EWOCS X-ray catalog comprises 5963 validated sources out of the 9420 initially provided to ACIS-Extract, reaching a
photon flux threshold of approximately 2 × 10−8 photons cm−2
s
−1
. The X-ray sources exhibit a highly concentrated spatial distribution,
with 1075 sources located within the central 1 arcmin. We have successfully detected X-ray emissions from 126 out of the 166 known
massive stars of the cluster, and we have collected over 71 000 photons from the magnetar CXO J164710.20-455217.
The use of Nauplii and metanauplii artemia in aquaculture (brine shrimp).pptxMAGOTI ERNEST
Although Artemia has been known to man for centuries, its use as a food for the culture of larval organisms apparently began only in the 1930s, when several investigators found that it made an excellent food for newly hatched fish larvae (Litvinenko et al., 2023). As aquaculture developed in the 1960s and ‘70s, the use of Artemia also became more widespread, due both to its convenience and to its nutritional value for larval organisms (Arenas-Pardo et al., 2024). The fact that Artemia dormant cysts can be stored for long periods in cans, and then used as an off-the-shelf food requiring only 24 h of incubation makes them the most convenient, least labor-intensive, live food available for aquaculture (Sorgeloos & Roubach, 2021). The nutritional value of Artemia, especially for marine organisms, is not constant, but varies both geographically and temporally. During the last decade, however, both the causes of Artemia nutritional variability and methods to improve poorquality Artemia have been identified (Loufi et al., 2024).
Brine shrimp (Artemia spp.) are used in marine aquaculture worldwide. Annually, more than 2,000 metric tons of dry cysts are used for cultivation of fish, crustacean, and shellfish larva. Brine shrimp are important to aquaculture because newly hatched brine shrimp nauplii (larvae) provide a food source for many fish fry (Mozanzadeh et al., 2021). Culture and harvesting of brine shrimp eggs represents another aspect of the aquaculture industry. Nauplii and metanauplii of Artemia, commonly known as brine shrimp, play a crucial role in aquaculture due to their nutritional value and suitability as live feed for many aquatic species, particularly in larval stages (Sorgeloos & Roubach, 2021).
2. 2.1 Oxidation number
Oxidation numbers are a convenient way of determining if a
substance has been oxidised or reduced.These numbers are
assigned arbitrarily to atoms and are equal to the charge the
atom would have if its bonds were purely ionic.
1. All free atoms in element have an oxidation number of zero
2. For simple ions (and ionic compounds), the oxidation number is
the same as the charge of ion
Na = Mg = H2 = Cl2 = P4 =
K+ = Ca2+ = B3+ = P3- = O2- = F- =
0 0 0 0 0
+1 +2 +3 – 3 – 2 – 1
3. 3. For covalent compounds, the covalent bonds are changed into
“ionic bonds” by assuming that the bonded electrons are on the
more electronegative atom.Table below shows some elements
oxidation number
Element
Oxidation
number
Notes
Group I +1 --
Group 2 +2 --
Group 17 –1 True only to halogen without O in it
Oxygen –2
Exception : –1 for peroxide and
+2 for F2O
Hydrogen +1 Except : metal hydride MH (H = -1)
4. Na2O B2O3 CO2
SO3 Cl2O7 HF
H2S NH3 CH4
H2CO3 H2SO4 HBrO3
3. In a neutral molecule, the sum of the oxidation numbers of all
atom are equals
2 Na + O = 0
2(Na) + (-2) = 0
Na = +1
2 B + 3 O = 0
2(B) + 3(-2) = 0
B = +3
1 C + 2 O = 0
1(C) + 2(-2) = 0
C = +4
1S + 3O = 0
1(S) + 3(-2) = 0
S = +6
2Cl + 7O = 0
2(Cl) + 7(-2) = 0
Cl = +7
1H + 1F = 0
1(F) + 1(+1) = 0
F = -1
1S + 2H = 0
1(S) + 2(+1) = 0
S = -2
1N + 3H = 0
1(N) + 3(+1) = 0
N = -3
1C + 4H = 0
1(C) + 4(+1) = 0
N = -4
2H + 1C + 3O = 0
2(+1) +1C + 3(-2) = 0
C = +4
2H + 1S + 4O = 0
2(+1) +1S + 4(-2) = 0
S = +6
1H + 1Br + 3O = 0
1(+1) +1Br + 3(-2) = 0
Br = +5
5. 4. In a molecular ion, the sum of the oxidation numbers of all
atoms in the formula unit equals to the charge on the ion.
CrO4
- Cr2O7
2- MnO4
-
C2O4
2- ClO2
- HSO4
-
1 Cr + 4 O = -1
1 Cr + 4(-2) = -1
Cr = +7
2 Cr + 7 O = -2
2 Cr + 7 (-2) = -2
Cr = +6
1 Mn + 4 O = -1
1 Mn + 4(-2) = -1
Mn = +7
2 C + 4 O = -2
2 C + 4(-2) = -2
C = +3
1 Cl + 2 O = -1
1 Cl + 2 (-2) = -1
Cl = +3
1 H + 1 S + 4 O = -1
1(+1) + 1S + 4(-2) = -1
S = +6
6. 2.2 Half equation and redox reaction.
Half equation ~ equation which shows how electrons are
accept / donate in a chemical reaction
When a substance is oxidise, electron is ………………. ; e- is
written at the ……….. side equation.
When a substance is reduce, electron is ………………. ; e- is
written as the ……… side equation.
Simple half equation : State the changes of oxidation number and
write the half equation.
Reaction
Oxidation
no change
Reaction Half equation
Na Na+
Mg Mg2+
Al Al3+
Cu2+ Cu
donated
right
received
left
0 +1 oxidation Na Na+ + e-
0 +2 oxidation Mg Mg2+ + 2 e-
0 +3 oxidation Al Al3+ + 3 e-
+2 0 reduction Cu2+ + 2 e- Cu
8. When it comes to the reaction involving molecular ion, the overall
charge has to be balanced in such order.
1. Write a skeleton half equation. Determine the reaction (oxidation
or reduction) using oxidation number
2. Balance the charge by adding electrons at the appropriate side
3. Balance the number of atoms other than oxygen.
4. Based on the changes in number of oxygen, write the number of
water molecule formed/used.
5. From the number of water molecule formed/used, write the
number of hydrogen ion (H+) required.
9. a) ClO3
- Cl-
Half equation : …………………………………………………………………
b) CrO4
2- Cr3+
Half equation : …………………………………………………………………
c) Cr2O7
2- Cr3+
Half equation : …………………………………………………………………
Changes in OS : +5 -1 ; reduction
Different in OS = 6, so 6 e- at the LHS of equation
ClO3
- + 6 e- Cl-
ClO3
- + 6 e- Cl- + 3 H2O
6 H+ + ClO3
- + 6 e- Cl- + 3 H2O
6 H+ + ClO3
- + 6 e- Cl- + 3 H2O
Changes in OS : +6 +3 ; reduction
Different in OS = 3, so 3 e- at the LHS of equation
CrO4
2- + 3 e- Cr3+
CrO4
2- + 3 e- Cr3+ + 4 H2O
8 H+ + CrO4
2- + 3 e- Cr3+ + 4 H2O
8 H+ + CrO4
2- + 3 e- Cr3+ + 4 H2O
Changes in OS : +6 +3 ; reduction
Different in OS = 3, so 3 e- at the LHS of equation
Since there are 2 Cr , so total e- = 6 ; Cr2O7
2- + 6 e- 2 Cr3+
Cr2O7
2- + 6 e- 2 Cr3+ + 7 H2O
14 H+ + Cr2O7
2- + 6 e- 2 Cr3+ + 7 H2O
14 H+ + Cr2O7
2- + 6 e- 2 Cr3+ + 7 H2O
10. d) MnO4
- Mn2+
Half equation : ……………………………………………………………………..
e) NO2
- NO3
-
Half equation : ……………………………………………………………………..
f) CrO2
- CrO4
2-
Half equation : ……………………………………………………………………..
Changes in OS : +7 +2 ; reduction
Different in OS = 5, so 5 e- at the LHS of equation
MnO4
- + 5 e- Mn2+
MnO4
- + 5 e- Mn2+ + 4 H2O
8 H+ + MnO4
- + 5 e- Mn2+ + 4 H2O
8 H+ + MnO4
- + 5 e- Mn2+ + 4 H2O
Changes in OS : +3 +5 ; oxidation
Different in OS = 2, so 2 e- at the RHS of equation
NO2
- NO3
- + 2 e-
NO2
- + H2O NO3
- + 2 e-
NO2
- + H2O 2 H+ + NO3
- + 2 e-
NO2
- + H2O 2 H+ + NO3
- + 2 e-
Changes in OS : +3 +6 ; oxidation
Different in OS = 3, so 3 e- at the RHS of equation
CrO2
- CrO4
2- + 3 e-
2 H2O + CrO2
- CrO4
2- + 3 e-
2 H2O + CrO2
- 4 H+ + CrO4
2- + 3 e-
2 H2O + CrO2
- 4 H+ + CrO4
2- + 3 e-
11. g) As2O3 As2O5
Half equation : ………………………………………………………………………
When half equation of both oxidation and reduction reaction are
written, a redox reaction can be balanced.
Example 10 : Cu2+ (aq) + Na (s) Cu (s) + Na+ (aq)
Oxidation half equation : ……………………………………….………………
Reduction half equation : ………………………………………………………
Overall equation : …………………………………………………………..
Example 11 : Fe2+ (aq) + MnO4
- (aq) Fe3+ (aq) + Mn2+ (aq)
Oxidation half equation : ………………………………………………………
Reduction half equation : ………………………………………………………
Overall equation : …………………………………………………………..
Changes in OS : +3 +5 ; oxidation
Different in OS = 2, so 2 e- at the RHS of equation
Since there are 2 As , so total e- = 4 ; As2O3 As2O5 + 4 e-
As2O3 + 2 H2O As2O5 + 4 e-
As2O3 + 2 H2O As2O5 + 4 e- + 4 H+
As2O3 + 2 H2O As2O5 + 4 e- + 4 H+
Na Na+ + e-
Cu2+ + 2 e- Cu
X 2
Cu2+ + 2 Na Cu + 2 Na+
Fe2+ Fe3+ + e-
8 H+ + MnO4
- + 5 e- Mn2+ + 4 H2O
X 5
5 Fe2+ 8 H+ + MnO4
- Mn2+ + 4 H2O + 5 Fe3+
13. Other than using half equation, a redox reaction can also be
balanced using the change of oxidation number.
Supposed we have a reaction : x A + y B products
If the oxidation of reactant A increased by m while B reduced by n ;
Then x (+ m) + y (– n ) = 0
Using a simple reaction :
ySn4+ (aq) + xFe2+ (aq) Sn2+ (aq) + Fe3+ (aq)
For Sn ; O.N changed from …… to …… ; so the difference is ……..
For Fe ; O.N changed from …… to ..…. ; so the difference is ……..
This will makes the equation become :
Balanced the number of atoms on both side of the equation
+4 +2 – 2
+2 +3 + 1
x (+ 1) + y (– 2) = 0
So, x = 2 ; y = 1
1 Sn4+ (aq) + 2 Fe2+ (aq) Sn2+ (aq) + Fe3+ (aq)
Sn4+ (aq) + 2 Fe2+ (aq) Sn2+ (aq) + 2 Fe3+ (aq)
14. Example 14 : Br – (aq) + SO4
2- (aq) SO2 (g) + Br2 (l)
For Br ; O.N changed from …… to ……. ; so the difference is ………
For S ; O.N changed from …… to ……. ; so the difference is ………
This will make the equation become :
- 1 0 + 1
+6 +4 - 2
x (+ 1) + y (– 2) = 0
So, x = 2 ; y = 1
2 Br – (aq) + 1 SO4
2- (aq) Br2 (l) + SO2 (aq)
2 Br – (aq) + SO4
2- (aq) + 4 H+ Br2 (l) + SO2 (aq) + 2 H2O
15. Balanced the number of atoms on both side of the equation
Example 15 : CrO4
2- + Cl- Cr3+ + Cl2
Example 16 : Cr2O7
2- + NO2
- Cr3+ + NO3
-
For Cr ; O.N changed from +6 to +3 ; so the difference is –3
For Cl ; O.N changed from –1 to 0 ; so the difference is +1
x (+ 1) + y (– 3) = 0
So, x = 3 ; y = 1
8 H+ + CrO4
2- + 3 Cl- Cr3+ + 3/2 Cl2 + 4 H2O
16 H+ + 2 CrO4
2- + 6 Cl- 2 Cr3+ + 3 Cl2 + 8 H2O
For Cr ; O.N changed from +6 to +3 ; so the difference is –3
Since there are 2 Cr involved, diff. = – 6
For N ; O.N changed from +3 to +5 ; so the difference is +2
x (+ 2) + y (– 6) = 0
So, x = 3 ; y = 1
Cr2O7
2- + 3 NO2
- Cr3+ + NO3
-
8 H+ + Cr2O7
2- + 3 NO2
- 2 Cr3+ + 3 NO3
- + 4 H2O
16. If the redox reaction occur in a basic solution, the number of H+
shall be neutralise by the number of OH-.
Example 17 : MnO4
- + SO3
2- MnO2 + SO4
2-
Example 18 : Fe(OH)2 + CrO4
2− Fe(OH)3 + Cr(OH)3
Oxidation ½ eq : H2O + SO3
2- 2 H+ + SO4
2- + 2 e-
Reduction ½ eq : 4 H+ + MnO4
- + 3 e- MnO2 + 2 H2O
Overall : 2 H+ + 2 MnO4
- + 3 SO3
2- 2 MnO2 + 3 SO4
2- + H2O
H2O + 2 MnO4
- + 3 SO3
2- 2 MnO2 + 3 SO4
2- + 2 OH-
X 3
X 2
Oxidation ½ eq : H2O + Fe(OH)2 → Fe(OH)3 + e− + H+
Reduction ½ eq : 5 H+ + 3 e− + CrO4
2− → Cr(OH)3 + H2O
Overall : 2 H2O + 3 Fe(OH)2 + 2 H+ + CrO4
2− → 3 Fe(OH)3 + Cr(OH)3
In basic : 4 H2O + 3Fe(OH)2 + CrO4
2− → 3Fe(OH)3 + Cr(OH)3 + 2OH−
X 3
17. Disproportionation reactions
~ Substances which are able to undergo self oxidation – reduction
are called disproportionation
~ Examples of disproportionation reaction.
18. Cu+ (aq) + Cu+ (aq) Cu2+ (aq) + Cu (s)
19. NaOH (aq) + Cl2 (aq) NaCl (aq) + NaOCl (aq) + H2O
20. NaOBr (aq) NaBrO3 (aq) + NaBr (aq)
+1 +2 0
0 +1-1
+1 +5 -1
18. 2.3 Electrode Potential
When a strip of metal, M (s) (known as electrode) is placed in a
solution of its aqueous solution, Mn+ (aq), the following
equilibrium is established : Mn+ (aq) + n e- → M (s)
At equilibrium, there is a separation of charge between metal
(M) and ions (Mn+) in the solution. as a result, there is a
potential difference between the metal and the solution.This
potential difference is known as electrode potential and is
written as Eo.
Electrode potential can be measure under these circumstances
where
19. Metal
Cu2+ + 2e- ↔ Cu
The positive value of E0 indicates the equilibrium favours to
the ……… position. Copper (II) ions (Cu2+), have a greater
tendency to …………...… at copper electrode.
Zn2+ + 2e- ↔ Zn
The negative value of E0 indicates the equilibrium favours to
the ….. position. Zinc ion (Zn2+) have a greater tendency to
…………….. at zinc electrode.
V34.0E Cu/Cu2
V76.0E Zn/Zn2
M
M+
M+
M+ M+
M+
M+
right
be reduced
left
be oxidised
20. Non – Metal
F2 + 2e- ↔ 2 F-
Cl2 + 2e- ↔ 2 Cl-
Positive value of ECl2/Cl- and EF2/F- indicates the equilibrium
favours to the…………. position. ……………………… has a
greater tendency to …………… under platinum electrode
The more positive the value, higher the tendency of non-
metal to …………….
In another words, fluorine is a stronger ………….. agent than
chlorine.
V87.2E F/F2
V36.1E Cl/Cl2
Cl2 (g)
Cl- (aq) [1.0 M]
be reduced
Fluorine and chlorineright
be reduced
oxidising
21. Mixture of aqueous ion
A potential difference also exists between ions in an aqueous
solution. Example :
Cr3+ + e- ↔ Cr2+
Fe3+ + e- ↔ Fe2+
Base on the Eo value, Cr3+ is ……… stable than Cr2+ as
equilibrium favour to ………….. (Eo is negative)
Base on the Eo value, Fe3+ is …….. stable than Fe2+ as
equilibrium favour to ………….. (Eo is positive)
V41.0E 23
Cr/Cr
V77.0E 23
Fe/Fe
Ma+ [1.0 M] / Mb+ [1.0 M]
more
backward
less
forward
V
22. 2.3.1 Standard Electrode Potential
Definition : The standard electrode potential, Eo
M
n+
/ M of a metal M
is the ………… difference between the metal M and the …………
solution of the metal ions of concentration ……………… at …… K
and …. atm, measured relatively to ……………………………….…
Standard Hydrogen Electrode ( S.H.E.)
It is impossible to measure the electrode potential for an
…………….. half-cell. It can only be measured for a complete
circuit with 2 ………. , i.e. only differences in electrode potentials
are measurable.
The standard chosen for electrode potentials is the standard
hydrogen electrode (SHE). The standard electrode potentials of
other half-cells are measured relative to the SHE’s electrode
potential.
By convention, the standard electrode potential for this
reference hydrogen half-cell is taken to be …………...
2 H+ (aq) + 2 e- H2 (g)
Condition : ….... oC ; H2 (g) at …… atm ; [H+] = 1.00 M
potential aqueous
1.0 mol dm-3 298
1.0
Standard Hydrogen Electrode
incomplete
half cell
standard
25
1.0
23. Measuring standard electrode potential of a
metal / metal aqueous solution
The set-up of the apparatus to measure the standard
potential electrode, Eo. is described as below :
Standard hydrogen electrode Zinc half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Zn (s)
Zn2+ (aq)
[1.0 M]
0.76 V
24. The chemical cell is set-up by connecting a standard
………….. half-cell to a standard …………. electrode.
The e.m.f. for the cell is ………. V. The potentiometer point
to the direction of …….. electrode in the external circuit,
indicating electrons flow from ………….. to …………. half-
cell.
Eq. Zn half-cell :
Eq. H half-cell :
Overall reaction :
The cell notation can be written as :
At zinc electrode ; electrons are ………..... ; ….……..…
reaction occur
At platinum electrode ; electrons are ………….. ; ………….
reaction occur
Since zinc is oxidised in a SHE, the standard e.m.f value is
……………
H+/H2Zn2+/Zn
0.76
H2
H+ / H2Zn2+ / Zn
Zn (s) Zn2+ (aq) + 2 e-
2 H+ (aq) + 2 e- H2 (g)
Zn (s) + 2 H+ (aq) H2 (g) + Zn2+ (aq)
Zn (s) I Zn2+ (aq) II H+ (aq) , H2 (g) I Pt (s)
donated oxidation
received reduction
– 0.76 V
25. Another example : silver / silver aqueous solution (Ag / Ag+)
The set-up of the apparatus to measure the standard potential
electrode, Eo. is described as below :
Standard hydrogen electrode Silver half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Ag (s)
Ag+ (aq)
[1.0 M]
0.80 V
26. The chemical cell set-up by connecting a standard
……………… half-cell to a standard …………….. electrode.
The e.m.f. for the cell is ……. V. The galvanometer point to
the direction of ………….. electrode in the external circuit,
indicating electrons flow from ………….. to ………… half-cell.
Ag half-cell :
H2 half-cell :
Overall :
The cell notation can be written as :
At silver electrode ; electrons are …………. ; ……………
reaction occur
At platinum electrode ; electrons are …………… ; ……………
reaction occur
Since silver is reduced in a SHE, the standard value is
………………
donated oxidation
received reduction
+ 0.80 V
H+ / H2Ag+ / Ag
0.80
H+ / H2
silver
Ag+ / Ag
Ag+ (aq) + e- Ag (s)
H2 (g) 2 H+ (aq) + 2 e-
H2 (g) + 2 Ag+ (aq) 2 Ag (s) + 2 H+ (aq)
Pt (s) I H2 (g) , H+ (aq) II Ag+ (aq) I Ag (s)
27. Measuring a standard electrode potential of a gaseous
substance
The chemical cell set-up by connecting a standard …………
half-cell to a standard …………… electrode. Note that the
set-up of the half-cells are the same for gaseous substances
Cl2 / Cl–
H+ / H2
Standard hydrogen electrode Chlorine half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Cl2 (g)
1 atm
Cl- (aq)
[1.0 M]
1.36 V
28. The e.m.f. for the cell is …….V. The galvanometer point to
the direction of ……….. electrode in the external circuit,
indicating electrons flow from ………….. to …………. half-
cell.
Chlorine half-cell :
Hydrogen half-cell :
Overall : :
The cell notation can be written as :
At platinum electrode in the half-cell of hydrogen ; electrons
are ………… ; ………… reaction occur
At platinum electrode in the half-cell of chlorine ; electrons
are ………….. ; …………… reaction occur
Since chlorine is ………… by SHE, the standard value is
……….
1.36
Pt (Cl2)
H+ / H2 Cl2 / Cl–
Cl2 (g) + 2 e- 2 Cl– (aq)
H2 (g) 2 H+ (aq) + 2 e-
H2 (g) + Cl2 (g) 2 Cl– (aq) + 2 H+ (aq)
Pt (s) I H2 (g) , H+ (aq) II Cl2 (g), Cl– (aq) I Pt (s)
donated oxidation
received reduction
reduced
+ 1.36 V
29. Measuring a standard electrode potential of a mixture of metal
ions.
The electrode potential of a mixture of ions can be measured
in the similar way, using standard hydrogen electrode (SHE)
as the other half-cell of the chemical cell
For example, in a mixture of iron (II) and iron (III) ion
Standard hydrogen electrode Fe2+ / Fe3+ half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Fe2+ (aq)
Fe3+ (aq)
[1.0 M]
0.77 V
30. The chemical cell set-up by connecting a standard …………
half-cell to a standard …………… electrode. Note that the set-
up of the half-cells are a mixture of iron (II) and iron (III) ion
under standard condition with …………… as electrode.
The e.m.f. for the cell is ……. V. The galvanometer point to the
direction of …………... half cell in the external circuit, indicating
electrons flow from …………. to …………. half-cell.
Fe3+ / Fe2+ half-cell :
Hydrogen half-cell :
Overall reaction :
The cell notation can be written as :
At half-cell of hydrogen ; electrons are ………… ; ……………
reaction occur
At half-cell of Fe3+/Fe2+ ; electrons are …………. ; …………...
reaction occur
Since the mixture is ………... by SHE, the value of ……………
Fe3+ / Fe2+
H+ / H2
platinum
0.77
Fe3+ / Fe2+
Fe3+ / Fe2+
H+ / H2
Fe3+ (aq) + e- Fe2+ (aq)
H2 (g) 2 H+ (aq) + 2 e-
H2 (g) + 2 Fe3+(aq) 2 Fe2+ (aq) + 2 H+ (aq
Pt (s) I H2 (g) , H+(aq) II Fe3+(aq), Fe2+(aq) I Pt (s)
donated oxidation
received reduction
reduced + 0.77 V
31. The calomel electrode
Platinum electrode is known as the …………… reference electrode.
However, it is relatively difficult to set up and operate under
standard condition. It is more easier and safer to use a calomel
electrode as a ……………… electrode. [calomel = ……………………..].
Diagram of a typical calomel electrode
primary
secondary Mercury base alloy
32. 2.4 Factors Affecting Electrode Potential
By convention, the half equation is written with ………….… as the
forward reaction.
The magnitude of the electrode potential depends on the position
of the above equilibrium
When value is positive ; a ……………… reaction is favoured
When value is negative ; a ………………….. reaction is favoured
Factors which affect the position of equilibrium would therefore
affect the value of electrode potential
1. Nature of metal
When a metal is highly …………………….., the metal atoms have a
greater tendency to become positive ions, leaving the ……………
behind on the metal electrode.The electrode potential therefore
become more ………………. and the position of equilibrium shift
more to ……… (……….……… is favoured)
reduction
forward
backward
electropositive
electron
negative
left oxidation
33. Metal Half equation E (V)
Silver + 0.80
Lead – 0.13
Zinc – 0.76
Magnesium – 2.38
2. Concentration of metal
• If the concentration of the hydrated metal ions is increased in the
equilibrium, the position of equilibrium will shift to the …………,
favouring …………. ; electrode potential become more ……………
Pb2+ (aq) + 2 e- Pb (s) E = – 0.13 V [ Conc = 1.0 M ]
• If concentration Pb2+ changed to 0.001 M ; equilibrium shift to
………………. ; E – 0.13 V
• If concentration Pb2+ changed to 10.0 M ; equilibrium shift to
………………. ; E – 0.13 V
Ag+ + e- ↔ Ag
Pb2+ + 2 e- ↔ Pb
Zn2+ + 2 e- ↔ Zn
Mg2+ + 2 e- ↔ Mg
right
forward positive
backward <
forward >
34. 3 Temperature
Most of the reduction processes are exothermic process.
Increasing the temperature will cause the equilibrium to
shift to the position of ……………….. process ; which is to
the ……Thus, the electrode potential becomes more
…………………………….
4. Pressure for gaseous species
From what we’ve learned from chemical equilibria, when
pressure increased, equilibrium will shift to the position
with …….. mole of gas ; while decreasing pressure will
cause equilibrium to shift to position with ……….mole of
gas.
Eg : Cl2 (g) + 2 e- 2 Cl- (aq) E = + 1.36V
Increasing pressure will cause equilibrium shift to
…………….. ; E + 1.36V
Decreasing pressure will cause equilibrium shift to
…………….. ; E + 1.36V
endothermic
left
negative / less positive
less
more
right side
left side
>
<
35. 2.5 The electrochemical series (ECS)
When a series of standard reduction potential of different
substances are determined and are arranged in order, a
electrochemical series is obtained.
36. Below are some important facts about electrochemical series.
Half cell of the standard electrode potential is always written
as ………… processes. Due to this reason, sometimes it is
also known as ……………………………………
The positive / negative sign shows how substances favour to
each of the reaction.
If the EO is positive, substances favour a ……..… reaction.
In another words, it serve well as …………. agent. The
more positive the value ; the stronger the ………….. agent
If the EO is negative, substances favour ………..… reaction.
In another words, it serve well as …………… agent. The
more negative the value ; the stronger the ………… agent
The number of electron involve does not affect the standard
electrode potential value. If
Cl2 (g) + 2 e- 2 Cl- (aq) Eo = + 1.36 V; then
½ Cl2 (g) + e- Cl- (aq) Eo = ………. V
reduction
Standard reduction potential (SRP)
forward
oxidising
oxidising
backward
reducing
reducing
+ 1.36
37. Some substances have more than one Eo value. For example
Fe2+ ; H2O2 ; NO2
- ; Cu+. These substances can act as an
oxidising or reducing agent. Examples
In Fe2+
Fe3+ (aq) + e- Fe2+ (aq) (Fe2+ act as reducing agent)
Fe2+ (aq) + 2 e- Fe (s) (Fe2+ act as oxidising agent)
38. 2.6 Redox reaction and electromotive forces (e.m.f.)
In standard hydrogen electrode, we had seen on how to
measure the standard electrode potential of 3 types of half-
cell, which are metal / metal ion half-cell ; non-metal / ion half-
cell ; ion / ion half-cell
Imagine if we replace the hydrogen half-cell with other half-
cell, will we still get the same value?
The potential difference between 2 half-cells can be
measured using the same way. There are various types of the
set-up of a complete chemical cell other than the one
introduced during measuring the standard electrode potential,
such as Daniel cell (diagram below)
39. A Daniel cell is built using a copper and
zinc half-cell
A porous pot is used to……………………
………………………………………………
Substance which has a higher position in
ECS (more negative the value of Eo) is the
……… of the cell whereas the substance
which has a lower position in ECS (more
positive the value of Eo) is the ……………
of the cell.
The half equation occur at
Anode :
Cathode :
Overall :
The e.m.f. of cell is …………. V
Cell notation is written as
C
u
Z
n
V
complete the cell
And to separate between the 2 electrolytes
anode
cathode
Zn Zn2+ + 2 e- Eo =+0.76 V
Cu2+ + 2 e- Cu Eo = +0.34 V
Zn + Cu2+ Zn2+ + Cu Ecell =+1.10 V
+ 1.10
Zn (s) I Zn2+ (aq) II Cu2+ (aq) I Cu (s)
Cu2+ Zn2+
40. A U tube cell is built using iron (II) ion, Fe2+
and bromine water, Br2
H2SO4 is used to ………………………………….…
…..............…………………………………………….
The substance which has more negative / less
positive the value of Eo) is the ………… of the
cell whereas the substance which has a more
positive / less negative value of Eo is the
…………... of the cell.
The half equation occur at
Anode :
Cathode :
Overall :
The e.m.f. of cell is …………..V
Cell notation is written as
G
complete the cell
And to separate between the 2 electrolytes
anode
cathode
Fe2+ Fe3+ + e- Eo = - 0.77 V
Br2 + 2 e- 2 Br - Eo = +1.07 V
Br2 + 2 Fe2+ 2 Fe3+ + 2 Br-
+ 0.30
Pt (s) I Fe2+ (aq) , Fe3+ (aq) II Br2 (l) , Br- (aq) I Pt (s)
Pt
Br2/Br-
Fe3+/Fe2+
41. Note the following in a chemical cell :
The e.m.f. of a cell is always …………… In another words,
we must always subtract the ..……………… standard
electrode potential with a ………………. standard electrode
potential.
Electrons flow from ………… to …………… in the external
circuit
…………… reaction occur at anode while ………… reaction
occur at cathode of the cell.
POSITIVE
more positive
less positive
anode cathode
Oxidation reduction
42. SRP for both cell : Fe2+ + 2e- Fe E0 = - 0.44 V
Mg2+ + 2e- Mg E0 = - 2.38 V
Since E0 for Mg2+/Mg is more negative than Fe2+/Fe, so Mg2+/Mg
will be oxidised, so SPR of Mg2+/Mg is reversed
Oxidation ½ eq : Mg Mg2+ + 2 e- E0 = + 2.38 V
Reduction ½ eq : Fe2+ + 2e- Fe E0 = – 0.44 V
Overall eq : Fe2+ + Mg Fe + Mg2+ Ecell = + 1.94 V
Cell diagram : Mg (s) I Mg2+ (aq) II Fe2+ (aq) I Fe (s)
G
Fe / Fe2+ half cell Mg / Mg2+ half cell
43. G
MnO4
- / Mn2+ half cell Ti3+ / Ti2+ half cell
SRP : MnO4
- + 5 e- + 8 H+ Mn2+ + 4 H2O E0 = + 1.52 V
Ti3+ + e- Ti2+ E0 = – 0.37 V
Since E0 for Ti3+/Ti2+ is more negative than MnO4
-/Mn2+, so Ti3+/Ti2+
will be oxidised, so SPR of Ti3+/Ti2+ is reversed
Oxidation ½ eq : Ti2+ Ti3+ + e- E0 = + 0.37 V
Red ½ eq : MnO4
- + 5 e- + 8 H+ Mn2+ + 4 H2O E0 = + 1.52 V
Overall: MnO4
- + 5 Ti2+ + 8 H+ Mn2+ + 4 H2O + 5 Ti3+ Ecell = + 1.89V
Cell diagram: Pt(s) I Ti2+(aq), Ti3+ (aq)II MnO4
- (aq), Mn2+(aq) I Pt(s)
44. 2.7 Feasibility of a redox reaction
If a reaction occurs on its own record when the reactants are
mixed, the reaction is a ...……………… reaction
Compare the following reaction to distinguish between a
spontaneous reaction and not spontaneous reaction.
Immerse a zinc plate into HCl 1.0 M Immerse a copper plate into HCl 1.0 M
Observation Observation
spontaneous
-Bubbling is observed
-Zinc plate is corroded by HCl
- No changes occur
45. We can use e.m.f. to predict the feasibility of the reaction.
Supposedly, in the reaction above, the 2 half equation for
the reactions can be written as
It can also be used to deduce the strength as an oxidising
agent in halogen.
Halogens are strong oxidising agent. This is supported with
the value of standard reduction potential where
½ F2 (aq) + e- F– (aq) Eo = + 2.87 V
½ Cl2 (aq) + e- Cl– (aq) Eo = + 1.36 V
½ Br2 (aq) + e- Br– (aq) Eo = + 1.07 V
½ I (aq) + e- I– (aq) Eo = + 0.54 V
-Since Zn react with H+, so the 2 half
equation can be written
Zn Zn2+ + 2e- E0 = + 0.76 V
2 H+ + 2e- H2 E0 = + 0.00 V
Zn + 2 H+ Zn2+ + H2 E = + 0.76 V
Since Ecell is positive, the reaction is
spontaneous
-Since Cu react with H+, so the 2 half
equation can be written
Cu Cu2+ + 2e- E0 = - 0.34 V
2 H+ + 2e- H2 E0 = + 0.00 V
Cu + 2 H+ Cu2+ + H2 E = - 0.34 V
Since Ecell is negative, the reaction is
non–spontaneous
Stronger
oxidisingagent
46. Observation Half equation & overall equation
Chlorine in
Tetrachloromethane
is added to aqueous
potassium bromide
(KBr).
Bromine in
tetrachloromethane
is added to aqueous
potassium iodide (KI)
Iodine is
tetrachloromethane
is added to aqueous
potassium chloride
(KCl)
Pale yellow solution in
CCl4 turned brown when
shaken with KBr.
2Br- Br2 + 2e- Eo = - 1.07 V
Cl2 + 2e- 2Cl- Eo = + 1.36 V
Cl2 + 2Br- Br2 + 2Cl–
Ecell = + 0.29 V
Brown solution in CCl4
turned purple when
shaken with KI
2I- I2 + 2e- Eo = - 0.54 V
Br2 + 2e- 2Br- Eo = + 1.07 V
Br2 + 2I- I2 + 2Br–
Ecell = + 0.53 V
No changes occur. Purple
solution remain after
shaken with KCl
2Cl- CI2 + 2e- Eo = - 1.36 V
I2 + 2e- 2I- Eo = + 0.54 V
I2 + 2 Cl- Cl2 + 2 I–
Ecell = - 0.82 V
47. a) Iron nail are placed in zinc sulphate b) Copper is placed in concentrated nitric
acid solution (Assume NO2 (g) is produced)
c) Chlorine gas is bubbled into acidified
potassium dichromate
d) potassium iodide is added to acidified
potassium manganate (VII) solution
Reactant : Fe and Zn2+
Suitable half equation
Fe Fe2+ + 2e- Eo = + 0.44 V
Zn2+ + 2e- Zn Eo = - 0.76 V
Fe + Zn2+ Fe2+ + Zn Ecell = - 0.32V
Since Ecell is negative, reaction is not
spontaneous (cannot react)
Reactant : Cu and NO3
-
Suitable half equation
Cu Cu2+ + 2e- Eo = - 0.34 V
NO3
– + 2H+ +e– NO2 + H2O E0= +0.81 V
Cu + 2 NO3
– + 4H+ 2NO2 +2H2O + Cu2+
Ecell = + 0.47 V
Since Ecell is positive, reaction is
spontaneous (can react)
Reactant : Cl2 and Cr2O7
2–
Suitable half equation
Cl2 + 2H2O 2HOCl + 2H+ + 2e–
Eo = – 1.64 V
Cr2O7
2- + 6e- + 14H+ 2Cr3+ + 7H2O
Eo = + 1.33 V
Cr2O7
2- + 6HOCl + 2H+ 3Cl2 + H2O
Ecell = – 0.31V
Since Ecell is negative, reaction is non
spontaneous (cannot react)
Reactant : I- and MnO4
-
Suitable half equation
2I- I2 + 2 e- Eo = - 0.54 V
MnO4
- + 8 H+ + 5e- Mn2+ + 4 H2O
Eo = + 1.52 V
10 I- + 2 MnO4
- + 16H+ 2 Mn2+ + 8 H2O
+ 5 I2 Ecell = + 0.98 V
Since Ecell is positive, reaction is
spontaneous (can react)
48. e) Calcium metal is added to water f) Acidified potassium dichromate solution
is added to a solution of iron (II) sulphate
Reactant : Ca and H2O
Suitable half equation
2H2O + 2e- H2 + 2OH- Eo = - 0.83 V
Ca Ca2+ + 2e- Eo = + 2.87 V
Ca + 2H2O H2 + Ca2+ + 2 OH-
Ecell = + 2.04 V
Since Ecell is positive, reaction is
spontaneous (can react)
Reactant : Fe2+ and Cr2O7
2-
Suitable half equation
Cr2O7
2- + 6e- + 14H+ 2Cr3+ + 7H2O
Eo = + 1.33 V
Fe2+ Fe3+ + e- Eo = - 0.77 V
Cr2O7
2- + 6Fe2+ + 14H+
2Cr3+ + 7H2O + 6Fe3+
Ecell = + 0.56 V
Since Ecell is positive, reaction is
spontaneous (can react)
49. Among the oxidation states available in d-orbital, +2 and +3
oxidation states are the most common states available in the d-
block elements.
The stability of the oxidation state can be explained in terms of
electrochemistry. The standard reduction potential of a few
transition metals is given in the table below.
Half equation of
reduction
Eo (V) Stable ion
Cr3+ + e- Cr2+ – 0.41 Cr3+
Ti3+ + e- Ti2+ – 0.37 Ti3+
V3+ + e- V2+ – 0.26 V3+
Fe3+ + e- Fe2+ + 0.77 Fe2+
Mn3+ + e- Mn2+ + 1.51 Mn2+
Co3+ + e- Co2+ + 1.82 Co2+
50. The action of dilute acids on metal are usually carried out in
the presence of oxygen. We must therefore determine
whether oxygen has any effect on such reactions. For
example, in oxidation of iron (II) ion
…………….. and …………… can also react in the same way
as iron does.
For the case of cobalt and manganese, it does not react in
the same way as iron does. Consider the reaction of cobalt
(II) ion with acid in the absence / presence of oxygen
Action of acids on iron (II) ion in the absence
of air (oxygen)
Action of acids on iron (II) ion in the
presence of air
Fe2+ Fe3+ + e- E0 = - 0.77 V
2 H+ + 2 e- H2 E0 = + 0.00 V
2 Fe2+ + 2H+ 2 Fe3+ + H2
Ecell = - 0.77 V
Fe2+ Fe3+ + e- E0 = - 0.77 V
O2 + 4H+ + 4e– 2H2O E0 = +1.23 V
4Fe2+ + O2 + 4H+ 4Fe3+ + 2H2O
Ecell = + 0.46 V
Ti2+ V2+ Cr2+
51. Action of acids on cobalt in the absence of air Action of acids on cobalt in the presence of air
Co2+ Co3+ + e- E0 = - 1.82 V
2 H+ + 2 e- H2 E0 = + 0.00 V
2 Co2+ + 2H+ 2 Co3+ + H2
Ecell = - 1.82 V
Co2+ Co3+ + e- E0 = - 1.82 V
O2 + 4H+ + 4e– 2H2O E0 = +1.23 V
4Co2+ + O2 + 4H+ 4Co3+ + 2H2O
Ecell = - 0.59 V
•Graph below shows the relative stability of ions which exist in different oxidation
state
52. Other than the presence of oxygen, the presence of ligands can
also affect the stability of ions.
Consider the following electrode reactions for cobalt :
[Co(NH3)6]3+ + e- ↔ [Co(NH3)6]2+ E0 = + 0.10 V
O2 + 4 H+ + 4 e- ↔ 2 H2O E0 = + 1.23 V
[Co(H2O)6]3+ + e- ↔ [Co(H2O)6]2+ E0 = + 1.82 V
In water, Co2+ is stable toward oxidation, even in the presence of
oxygen, since the E for the reaction is -0.59 (based on the
calculation above). Therefore, +2 is more stable than +3 oxidation
state in aqueous solution (ligand is water)
When aqueous NH3 is added to the solution of Co2+, the complex
ion [Co(NH3)6]2+ is formed (since NH3 is a strong ligand, water
molecule can easily displaced by NH3 ligand).
Eq : [Co(H2O)6]2+ + 6 NH3 [Co(NH3)6]2+ + 6 H2O
When [Co(NH3)6]2+ is formed, it can react easily with acids with the
presence of air
53. When [Co(NH3)6]2+ is formed, it can react easily with acids
with the presence of air
[Co(NH3)6]2+ [Co(NH3)6]3+ + e- E0 = - 0.10 V
O2 + 4 H+ + 4 e- 2 H2O E0 = + 1.23 V
4 [Co(NH3)6]2+ + O2 + 4 H+ 2 H2O + 4 [Co(NH3)6]3+
Ecell = + 1.13 V
Therefore, although Co3+ is not stable in the presence of air,
but Co3+ is stable in ammonia aqueous solution. Hence, this
is one of the ways to prepare an ion solution which is not
stable in air.
54. 2.7 Nernst Equation and Its application
All the electrochemical cells that discussed so far are
standard Eo. [At ……..oC ; ……… atm ; ……. M]
If concentration of ions and temperature change, it will affec
the value of electrode potential. At this moment, we can use
an equation to study the changes of concentration of ions
using Nernst Equation.
25 1.00 1.00
y
x
0
]ionsproduct[
]ionsttanreac[
ln
nF
RT
EE
R = 8.31 J mol-1 K-1
T = 250C = 298 K
F (Faraday constant)
F = 96500 C mol-1
y
x
0
]ionsproduct[
]ionsttanreac[
lg
)96500(n
)303.2)(298)(31.8(
EE
y
x
0
]ionsproduct[
]ionsttanreac[
lg
n
059.0
EE
55. Ag / Ag+ (1.5 mol dm-3) Cu / Cu2+ (2.0 mol dm-3)
Cl2 / Cl- (0.50 mol dm-3) Fe3+ (0.800 mol dm-3) / Fe2+ (1.30 mol
dm-3)
1
]Ag[
lg
n
059.0
EE
1
0
Ag+ + e- Ag E0 = + 0.80 V
1
)5.1(
lg
1
059.0
80.0E
1
E = + 0.81 V
1
]Cu[
lg
n
059.0
EE
12
0
Cu2+ + 2e- Cu E0 = + 0.34 V
1
)0.2(
lg
2
059.0
34.0E
1
E = + 0.35 V
2
1
20
]Cl[
]Cl[
lg
n
059.0
EE
Cl2 + 2e- 2 Cl- E0 = + 1.36 V
2
1
)50.0(
)1(
lg
2
059.0
36.1E
E = + 1.38 V
12
13
0
]Fe[
]Fe[
lg
n
059.0
EE
Fe3+ + e- Fe2+ E0 = + 0.77 V
)30.1(
)800.0(
lg
1
059.0
77.0E
1
E = + 0.758 V
56. Ti3+ (1.20 mol dm-3) / Ti2+ (0.700 mol
dm-3)
MnO4
- (1.10 mol dm-3) ; H+ (0.800 mol
dm-3) / Mn2+ (17.0 mol dm-3)
12
13
0
]Ti[
]Ti[
lg
n
059.0
EE
Ti3+ + e- Ti2+ E0 = - 0.37 V
)700.0(
)20.1(
lg
1
059.0
37.0E
E = - 0.356 V
]Mn[
]H][MnO[
lg
n
059.0
EE 2
8
40
MnO4
- + 8 H+ + 5e- 4 H2O + Mn2+
E0 = + 1.52 V
)0.17(
)800.0)(10.1(
lg
5
059.0
52.1E
8
E = + 1.50 V
57. 13.7.1 Nernst Equation and e.m.f. of a chemical cell.
Consider the following redox reaction in an chemical cell
p A + q B ↔ r C + s D
At 25oC, Nernst Equation :
*For pure solids and liquids, it will not appear in the
equation
y
x
0
]ionsproduct[
]ionsttanreac[
ln
nF
RT
EE
sr
qp
cell
]D[]C[
]B[]A[
lg
n
059.0
EE
58. a) Cr (s) Cr3+ (0.010 mol dm-3) Ni2+ (0.20 mol dm-3) Ni (s)
Oxidation : Cr Cr3+ + 3 e– E0 = + 0.74 V
Reduction : Ni2+ + 2 e– Ni E0 = – 0.25 V
Overall : 2 Cr + 3 Ni2+ 3 Ni + 2 Cr3+ Ecell = + 0.49 V
E = + 0.51 V
23
32
cell
]Cr[
]Ni[
lg
6
059.0
EE
2
3
)010.0(
)20.0(
lg
6
059.0
49.0E
59. b) Mg (s) Mg2+ (0.500 mol dm-3) Fe3+ (1.80 mol dm-3) , Fe2+
(0.750 mol dm-3) Pt (s)
Oxidation : Mg Mg2+ + 2e– E0 = + 2.38 V
Reduction : Fe3+ + e- Fe2+ E0 = + 0.77 V
Overall : Mg + 2 Fe3+ Mg2+ + 2 Fe2+ Ecell = + 3.15 V
E = + 3.18 V
222
23
cell
]Fe][Mg[
]Fe[
lg
2
059.0
EE
2
2
)750.0)(500.0(
)80.1(
lg
2
059.0
15.3E
60. c) Pt (s) Sn2+ (0.300 mol dm-3), Sn4+ (0.500 mol dm-3)
Mn3+ (1.20 mol dm-3) , Mn2+ (0.250 mol dm-3) Pt (s)
Oxidation : Sn2+ Sn4+ + 2e– E0 = – 0.15 V
Reduction : Mn3+ + e- Mn2+ E0 = + 1.49 V
Overall : Sn2+ + 2 Mn3+ Sn4+ + 2 Mn2+ Ecell = + 1.34 V
E = + 1.37 V
224
232
]][[
]][[
lg
2
059.0
MnSn
MnSn
EE cell
2
2
)250.0)(500.0(
)20.1)(300.0(
lg
2
059.0
34.1 E
61. 2.7.1 Nernst Equation and Equilibrium Constant, KC
Consider the following reaction :
Cu2+ (aq) + Zn (s) Zn2+ (aq) + Cu (s) E0 = + 1.10 V
The Kc of the reaction can be expressed as
As time past, the concentration of [Cu2+] ……………… while
[Zn2+] ……………..
Using standard reduction potential of copper and zinc
Cu2+ (aq) + 2 e- Cu (s) Eo = + 0.34 V ; when [Cu2+]
……………… ; equilibrium shift to ……… Eo ……………
Zn2+ (aq) + 2 e- Zn (s) Eo = – 0.76 V ; when [Zn2+]
……………… ; equilibrium shift to ……… Eo …………...
]Cu[
]Zn[
Kc 2
2
decrease
increase
decrease left decrease
increase right increase
63. Applied to Nernst equation where
When system achieved equilibrium at room temperature,
Nernst equation is simplified to :
Since and at equilibrium, Ecell = 0, so
y
x
0
]ionsproduct[
]ionsttanreac[
ln
nF
RT
EE
x
y
0
]ionreactat[
]ionproduct[
lg
n
059.0
EE
]Cu[
]Zn[
Kc 2
2
C
0
Klg
n
059.0
EV0
64. Consider the following reaction :
2 Fe3+ (aq) + Cu 2 Fe2+ (aq) + Cu2+ (aq)
a) Calculate the E0 of the cell
b) Calculate the Kc for the reaction.
Consider the following reaction :
Fe3+ (aq) + Ag (s) Fe2+ (aq) + Ag+ (aq)
a) Calculate the E0 of the cell
b) Calculate the Kc for the reaction.
Fe3+ + e- Fe2+ E0 = + 0.77 V
Cu 2 e- + Cu2+ E0 = - 0.34 V
2 Fe3+ + Cu Cu2+ + 2 Fe2+
Ecell = + 0.43 V
23
222
C
]Fe[
]Cu[]Fe[
K
C
0
Klg
2
059.0
EV0
CKlg
2
059.0
43.0V0
lg KC = 14.57
KC = 3.77 x 1014 mol dm-3
Fe3+ + e- Fe2+ E0 = + 0.77 V
Ag e- + Ag+ E0 = – 0.80 V
Fe3+ + Ag Fe2+ + Ag+
Ecell = – 0.03 V
]Fe[
]Fe][Ag[
K 3
2
C
CKlg
1
059.0
03.0V0
CKlg
1
059.0
03.0V0
lg KC = – 0.5085
KC = 0.310 mol dm-3
65. 2.7.2 Nernst Equation and solubility product, Ksp
Similar to calculating equilibrium constant, solubility product of a
sparingly soluble salt can also be calculated in the same way
mentioned above. Consider the following reaction of dissociation of
AgCl (s)
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
Half-cell : AgCl (s) + e- ↔ Ag (s) + Cl- (aq) Eo = + 0.22 V
Half-cell : Ag (s) ↔ Ag+ (aq) + e- Eo = - 0.80 V
Overall Eq: AgCl (s) ↔ Ag+ (aq) + Cl- (aq) E0 = - 0.58 V
Using Nernst equation
and given Ksp = [Ag+][Cl–]
x
y
0
]ionreactat[
]ionproduct[
lg
n
059.0
EE
1
]Cl][Ag[
lg
n
059.0
EE 0
sp
0
Klg
n
059.0
EE
66. At equilibrium ; E = 0 ; so replace in the equation
lg Ksp = – 9.83 mol2 dm-6
Ksp = 1.48 x 10–10 mol2 dm-6
The question may be extend to calculate the solubility
from the solubility product calculated
Note that Eo is always negative for sparingly soluble salt
as Ksp is ………………
spKlg
1
059.0
V58.00
very small
67. Example : Given HgCl2 + 2 e- Hg + 2 Cl- Eo = + 0.27 V
and Hg2+ + 2 e- Hg Eo = + 0.85 V.
Calculate the Ksp for HgCl2.
Oxidation Hg Hg2+ + 2 e– E0 = – 0.85 V
Reduction HgCl2 + 2 e- Hg + 2 Cl- E0 = + 0.27 V
Overall HgCl2 Hg2+ + 2 Cl- Ecell = – 0.58 V
sp
0
Klg
n
059.0
EE
spKlg
2
059.0
V58.00
lg Ksp = – 19.66
Ksp = 2.18 x 10-20 mol3 dm-9
68. 13.7.3 Nernst Equation and pH of a Solution
Under SHE, if the solution of H+ is not 1.00 mol dm-3, the
Ecell can be calculated using Nernst equation.
Assuming the metal is oxidise by SHE, the cell notation is
written as
M (s) I M2+ (aq) II H+ (aq) (x mol dm-3) , H2 (g) I Pt (s)
Standard hydrogen electrode Zinc half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq)
[x M]
25oC
M (s)
M2+ (aq)
[1.0 M]
69. The overall reaction can be written as
M (s) + 2 H+ (aq) M2+ (aq) + H2 (g) E0 = + a V
y
x
0
]ionsproduct[
]ionsttanreac[
lg
n
059.0
EE
12
2
0
]M[
]H[
lg
2
059.0
EE
]1[
]H[
lg2
2
059.0
EE 0
pH059.0EE 0
70. Example : The e.m.f. of of the following cell at 25oC is 0.093 V
Pb (s) I Pb2+ (1.00 mol dm-3) II H+ (test solution), H2 (g) I Pt (s)
Calculate the pH of the solution
Overall equation : Pb + 2 H+ Pb2+ + H2 E0 = + 0.13 V
E = E0 – 0.059 pH
0.059 pH = + 0.13 – 0.093
pH = 0.63
Calculate the e.m.f. of the chemical cell compared relatively to
SHE at [H+] = 0.0030 mol dm-3 for a calcium half cell.
71. 2.9 Type of cell.
A battery is a galvanic cell, or a series of combined galvanic cells,
that can be used as a source of direct electric current at a
constant voltage.
Although the operation of a battery is similar in principle to that of
the galvanic cells. A battery has the advantage of being completely
self-contained and requiring no auxiliary components such as salt
bridges.
Here we will discuss several types of batteries that are in
widespread use.
Lithium ion battery
Fuel Cell
72. 2.9.1 Lithium ion battery
Figure below shows a schematic diagram of a lithium-ion battery.
The anode is made of a conducting carbonaceous material, usually
graphite, which has tiny spaces in its structure that can hold both Li
atoms and Li+ ions.
During the discharge of the battery,
the half-cell reactions are
Anode : Li(s) → Li+ + e-
Cathode : Li+ + CoO2 + e- → LiCoO2 (s)
Overall : Li (s) + CoO2 → LiCoO2 (s)
Ecell = + 3.4 V
73. The cathode is made of a transition metal oxide such as
CoO2, which can also hold Li+ ions. Because of the high
reactivity of the metal, non - aqueous electrolyte (organic
solvent plus dissolved salt) must be used.
The advantage of the battery is that lithium has the most
negative standard reduction potential and hence the greatest
reducing strength. Furthermore, lithium is the lightest metal
so that only 6.941 g of Li (its molar mass) are needed to
produce 1 mole of electrons.
A lithium-ion battery can be recharged literally hundreds of
times without deterioration. These desirable characteristics
make it suitable for use in cellular telephones, digital
cameras, and laptop computers.
74. 2.9.2 Fuel Cell
Fossil fuels are a major source of energy, but conversion of
fossil fuel into electrical energy is a highly inefficient process.
Consider the combustion of methane:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) H = - x kJ mol-1
To generate electricity, heat produced by the reaction is first
used to convert water to steam, which then drives a turbine
that drives a generator.
An appreciable fraction of the energy released in the form of
heat is lost to the surroundings at each step; even the most
efficient power plant converts only about 40 percent of the
original chemical energy into electricity.
Combustion reactions are redox reactions, it is more desirable
to carry them out directly by electrochemical means, thereby
greatly increasing the efficiency of power production
75. This objective can be accomplished by a device known as a
fuel cell, a galvanic cell that requires a continuous supply of
reactants to keep functioning.
In its simplest form, a hydrogen-oxygen fuel cell consists of an
electrolyte solution, such as potassium hydroxide solution,
and two inert electrodes. Hydrogen and oxygen gases are
bubbled through the anode and cathode compartments where
the following reactions take place.
76. Anode : 2 H2 (g) + 4 OH- (aq) → 4 H2O (l) + 4 e- Eo = + 0.83 V
Cathode : O2 (g) + 2 H2O (l) + 4e- → 4 OH-(aq) Eo = + 0.40 V
Overall : 2 H2 (g) + O2 (g) → 2 H2O(l) Eo
cell = + 1.23 V
The electrodes used have a two-fold function. They serve as
electrical conductors, and they provide the necessary
surfaces for the initial decomposition of the molecules into
atomic species, prior to electron transfer. They are also known
as electrocatalysts. Metals such as platinum, nickel, and
rhodium are good electrocatalysts.
77. 2.9.2.1 Propane-oxygen fuel cell
In addition to the H2-O2 system, a number of other fuel cells
have been developed. Among these is the propane-oxygen
fuel cell.
The half-cell reactions are
Anode : C3H8 (g) + 6 H2O (l) → 3 CO2 (g) + 20 H+ (aq) + 20 e-
Cathode : 5 O2 (g) + 20 H+ (aq) + 20 e- → 10 H2O (l)
Overall : C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)
Unlike batteries, fuel cells do not store chemical energy.
Reactants must be constantly resupplied, and products
must be constantly removed from a fuel cell. In this respect,
a fuel cell resembles an engine more than it does a battery.
However, the fuel cell does not operate like a heat engine
and therefore is not subject to the same kind of
thermodynamic limitations in energy conversion
78. Properly designed fuel cells may be as much as 70 percent
efficient, about twice as efficient as an internal combustion
engine. In addition, fuel-cell generators are free of the noise,
vibration, heat transfer, thermal pollution, and other
problems normally associated with conventional power
plants.
79. 2.10 Electrolysis ~ decomposition of a substance by
direct current electricity.
Electrolyte – substance that can conduct electricity when in
aqueous solution or in molten state
Electrolytic cell – cell consisting of 2 electrodes immersed in
an electrolyte for carrying out electrolysis In an electrolytic
cell, the following apply
Positive terminal is called as ………… whereas the negative
terminal is called as …………
At anode, …………. process occur where
as at cathode, ………..…process occur
Cations are attracted to ………….. while
anions are attracted to ……………
Electrons flow from the ………… to
……..….. in the external circuit.
A
anode
cathode
oxidation
reduction
cathode
anode
anode
cathode
80. 2.11 Faraday’s Law
Faradays Law stated that 1 Faraday is the quantity of electricity
(9.65 x 104 C) that must be supplied to an electrolytic cell in order
to produce one mole of electrons for reactions in the cell.
The extension of Faraday Law is stated in Faraday’s First Law,
where it stated that the mass of a substance produced at an
electrode during electrolysis is …………….… to the quantity of
electricity (in Coulumb) passed. From the statement above, the
factors that influence the mass of a substance liberated during
electrolysis are
• The greater the number of electrons transferred, the greater the
mass of the product.
• The longer the time taken, the greater the electrical current
produced, the more the mass produced
Based on the statement above, the quantity of electrical current
can be calculate according to time where
Q = electric current I = current t = time in second
proportional
Q = I t
81. Example 1 : Calculate the mass of silver
formed when a current of 0.200 A is applied to
a electrolytic cell filled with aqueous silver
nitrate for 2 hour.
Example 2 : An aqueous solution of copper (II)
sulphate is electrolysed using a current of 0.50
A for 4 hours. Calculate the mass of Copper,
Cu deposited at cathode.
Example 3 : Calculate the mass of chromium
formed when 1.20 A of current is directed into
molten chromium (III) chloride for 3 hours.
Example 4 : Calculate the time taken to
produce 18.0 g of silver from silver nitrate by a
current of 0.900A
Q = It @ Q = (0.200)(2 x 60 x 60)
= 1440 C
Eq : Ag+ + e- Ag
mol of e- = Q / F @ mol = 1440 / 96500
mol of e- = 0.0149 mol
Since 1 e- = 1 Ag ; mol Ag = 0.0149 mol
Mass Ag = 0.0149 x 108
= 1.61 g
Q = It @ Q = (0.50)(4 x 60 x 60)
= 7200 C
Eq : Cu2+ + 2 e- Cu
mol of e- = Q / F @ mol = 7200 / 96500
mol of e- = 0.0746 mol
Since 2 e- = 1 Cu ; mol Cu = 0.0373 mol
Mass Cu= 0.0373 x 63.5
= 2.4 g
Q = It @ Q = (1.20)(3 x 60 x 60)
= 12960 C
Eq : Cr3+ + 3 e- Cr
mol e- = Q / F @ mol = 12960 / 96500
mol of e- = 0.134 mol
Since 3 e- = 1 Cr ; mol Cr = 0.0448 mol
Mass Cr = 0.0448 x 52.0
= 2.33 g
Mol of Ag = 18.0 / 108 = 0.167 mol
Eq : Ag+ + e- Ag
1 Ag = 1 e- ; so mol e- = 0.167 mol
Q = mol e- x F @ Q = 0.167 x 96500
Q = 16083 C
Q = It @ t =16083 / 0.900
t = 1.78 x 104 s
82. Example 5 : Calculate the time required to
form 200g of lead from molten lead (II)
bromide by a current of 1.50 A.
Example 6 : Calculate the time required to
form 10 g of aluminium from molten
aluminium oxide by a current of 10 A
Mol of Pb = 200 / 207
= 0.966 mol
Eq : Pb2+ + 2 e- Pb
1 Pb = 2 e- ; so mol e- =
1.93 mol
Q = mol e- x F @
Q = 1.93 x 96500
Q = 186473 C
Q = It @ 186473 / 1.50
t = 1.24 x 105 s
Mol of Al = 10 / 27
= 0.370 mol
Eq : Al3+ + 3 e- Al
1 Al = 3 e- ; so mol e- =
1.11 mol
Q = mol e- x F @
Q = 1.11 x 96500
Q = 107222 C
Q = It @ t = 107222 / 10
t = 1.1 x 104 s
83. 14.2.1 Faraday’s Second Law
~ stated that if the same quantity of electricity is passed
through different electrolytes, the mass of substance liberated
at electrode is inversely proportional to the charge of ions.
1 Faraday
Silver (I) nitrate Copper (II)
sulphate
Chromium (III)
chloride
Sulphuric acid
84. Half equation occur at cathode for
From the diagram above, 1 F will discharge …. mol of Ag+
ions ; …… mol of Cu2+ ion ; …… mol of Cr3+ ion ; ….… mol of
O2.
Type of
electrode
Half equation
Mol of metal
deposited
Silver
Copper
Carbon in CrCl3
Carbon in
H2SO4 (occur at
anode)
Mol of non-metal
Ag+ (aq) + e- Ag (s) 1 mol
Cu2+ (aq) + 2 e- Cu (s) 1/2 mol
Cr3+ (aq) + 3 e- Cr (s) 1/3 mol
2 H2O O2 + 4 H+ + 4 e–
1/4 mol
1
1/2 1/3 1/4
85. Example 7 : Calculate the mass of copper
deposited under the same cell if the
amount of silver formed under the same
amount of quantity charge is 1.8 g.
An electric current produced 0.56 g of
aluminium from molten aluminium
oxide. If the same current was used to
electrolysed molten lead (II) bromide,
calculate the mass of lead deposited.
Mol of Ag = 1.8 / 108
= 0.01667 mol
Eq : Ag+ + e- Ag
Since 1 Ag = 1 e- ;
mol e- = 0.01667 mol
For Cu ; Cu2+ + 2 e– Cu
Since 2 mol e- = 1 mol Cu
So mol of Cu = 0.008333 mol
Mass of Cu = 0.008333 x 63.5
= 0.53 g
Mol of A1 = 0.56 / 27
= 0.02074 mol
Eq : Al3+ + 3 e- Al
Since 1 Al = 3 e- ;
mol e- = 0.06222 mol
For Pb ; Pb2+ + 2 e– Pb
Since 2 mol of e = 1 mol of Pb
So mol of Pb = 0.03111 mol
Mass of Pb = 0.03111 x 207
= 6.44 g
86. Predicting the product for electrolysis
• In electrolysis, there may be more than one type of cation /
anion inside the electrolytes.
• Under such circumstance, since an electrode can only discharge
one cation / anion, the ion must be choose under certain
guidelines.
• The selectivity of ions are based on electrochemical series
87. No matter it is an electrolytic cell or chemical cell,
• At anode, …………. reaction occur ; electrons are ………………
• At cathode, …………. reaction occur ; electrons are ………………
Electrolytes can only discharge under 2 conditions : …..…... state or
………… solution.
• When in molten state, the electrolytes contain only the cation and
anion of the substance involve
Molten lead (II) bromide : PbBr2 (l) Pb2+ + 2 Br–
Molten aluminium oxide : Al2O3 (l) 2 Al3+ + 3 O2–
Molten barium chloride : BaCl2 (l) Ba2+ + 2 Cl–
Molten silver (I) iodide : AgI (l) Ag+ + I–
oxidation donated
reduction received
molten
aqueous
88. Electrolyte
Subst.
present
Half equation at anode
Substance at
anode
Half equation at
cathode
Substance
at cathode
PbBr2 (l)
Al2O3 (l)
NaCl (aq)
Pb2+
Br –
Br2 + 2e- 2Br-
E0 = + 1.07
V
Rev:
2Br- 2e- + Br2
bromine
Pb2+ + 2e- Pb
E0 = – 0.13 V
lead
Al3+
O2–
O2 + 4e- 2O2-
E0 = ? V
Rev:
2O2- O2 + 4e-
oxygen
Al3+ + 3e- Al
E0 = – 1.67 V
Aluminu
m
Na+
Cl–
Cl2 + 2e- 2Cl-
E0 = +1.36
V
Rev:
2Cl- Cl2 + 2e-
chlorine
Na+ + e- Na
E0 = – 2.71 V Sodium
89. No matter it is an electrolytic cell or chemical cell,
• At anode, …………. reaction occur ; electrons are ………………
• At cathode, …………. reaction occur ; electrons are ………………
• When in aqueous solution, not only it contains the cation and anion of
substance involve, but it also involves water. thus there is a selectivity
of ion occur
• In the terms of E0, a more ……….. value will be selected for
discharge at cathode, while a more …………. value will be selected
for discharge at anode.
At anode : E0 = …………. V
At cathode : E0 = …………. V
However, water is a weak electrolytes. At 250C and 1 atm, the E0 value
varies with the solution used. So in deciding which E0 value we should
used, we need to consider if the solution is different or not.
Under neutral condition, where [H+] = [OH-], using Nernst Equation, Eo
values are (At [H+] = [OH-]1.0 x 10-7 mol dm-3)
At anode : E0 = ….……. V
At cathode : E0 = ….……. V
oxidation donated
reduction received
positive
negative
4 H+ + O2 + 4 e– 2 H2O + 1.23
2 H2O + 2 e– 2 OH– + H2
– 0.83
4 H+ + O2 + 4 e– 2 H2O + 0.81
2 H2O + 2 e– 2 OH– + H2
– 0.41
90. Electrolyte
Subst.
present
Half equation at anode
Substance
at anode
Half equation at cathode
Substance
at cathode
NaCl (aq)
Na+
Cl–
H2O
Cl2 + 2e- 2Cl-
E0 = +1.36
V
4 H+ + O2 + 4 e–
2 H2O
E0 = +1.23
V
Rev: 2 H2O
4 H+ + O2 + 4 e–
oxygen
Na+ + e- Na
E0 = – 2.71 V
2 H2O + 2 e–
2 OH– + H2
E0 = – 0.83 V
Hydroge
n
92. Effect on concentration towards the selectivity of the ions to
discharge
• In general, an ion with a very high concentration is
preferentially discharged.
• For example if Pb2+ ion and Cu2+ ion are mixed under the
same concentration, ……… ion is preferred to be
discharge at cathode as it has a lower position in
electrochemical series.
• However, if the concentration of Pb2+ ion concentration is
raised much higher than Cu2+ ion, ………. ion is more
readily to be discharged.
• Another example is potassium chloride in aqueous
solution. Under dilute solution of KCl, ……….. will be
selected at anode and ………...… gas is given out, as it
has a lower position in electrochemical series.
Cu2+
Pb2+
H2O
oxygen
93. • However, if concentrated KCl is used as electrolyte, the
concentrated of Cl- increase, and …… is selected to be
discharge and ……..….… gas is given out.
• Still, if the position in electrochemical series differ too
much, like K+ and water in the example above, K+ ion
…………… be discharge as the position is much too high.
At the end, ……………. gas is given out at cathode.
• If chloride ion is replaced with fluoride ion, F– ion, and
concentration of F– ion is increased, …....... is still be
preferred as F– ion has a ……………. position in
electrochemical series
In the table below, predict the element that is expected to
form when electrolyse.
Cl–
chlorine
will not
hydrogen
water
very high
95. 2.12.3 Overvoltage
Overvoltage ~ the difference between electrode potential and
discharge potential. In another words, Overvoltage is the
voltage that must be applied to an electrolytic cell in addition
to the theoretical voltage to cause an electrode reaction to
occur.
Example of over voltage phenomenon is the electrolysis of
aqueous sodium chloride. Consider the electrolysis of sodium
chloride in aqueous solution where the substance presence
are Na+, Cl-, H2O
Half equation for substance attracted to
anode
Half equation for substance attracted to
cathode
Cl2 (g) + 2 e- → 2 Cl- (aq) E0 = + 1.36 V
4 H+ + O2 (g) + 4 e- → 2 H2O (l)
E0 = + 0.81 V
(conc. H+ = 1 x 10-7 mol dm-3)
Na+ (aq) + e- → Na (s) E0 = - 2.71 V
2 H2O (l) + 2 e- → 2 OH- + H2 (g)
E0 = - 0.41 V
(conc. OH- = 1 x 10-7 mol dm-3)
96. From the E0 values at anode, it is suggested that H2O should
be preferentially oxidized at the anode. However, by
experiment we find that the gas liberated at the anode is Cl2,
not O2.
In studying electrolytic processes, we sometimes find that the
voltage required for a reaction is considerably higher than the
electrode potential indicates.
The overvoltage is the difference between the electrode
potential and the actual voltage required to cause electrolysis.
In this case, overvoltage for O2 formation is quite high.
Therefore, under normal operating conditions Cl2 gas is
actually formed at the anode instead of O2.
As for the selectivity at cathode, H2O is selected since the E0
value for H2O is less negative than Na+.
97. Thus the half equation occur at both anode and cathode are
Anode
Cathode
Overall
As the overall reaction shows, the concentration of the Cl2
ions decreases during electrolysis and that of the OH- ions
increases.
Therefore, in addition to H2 and Cl2, the useful by-product
NaOH can be obtained by evaporating the aqueous solution
at the end of the electrolysis.
98. 2.12.4 Electrorefining and electroplating
The purification of a metal by means of electrolysis is called
electrorefining. For example, impure copper obtained from
ores is converted to pure copper in an electrolytic cell that
has impure copper as the anode and pure copper as the
cathode. The electrolyte is an aqueous solution of copper
sulphate
99. At the impure Cu anode, copper is oxidized along with more
easily oxidized metallic impurities such as zinc and iron. Less
easily oxidized impurities such as silver, gold, and platinum fall
to the bottom of the cell as anode mud, which is reprocessed
to recover the precious metals. At the pure Cu cathode, ions
are reduced to pure copper metal, but the less easily reduced
metal ions (and so forth) remain in the solution
Half equations occur for electrorefining process of copper
above are
• At anode Cu (s) → Cu2+ (aq) + 2 e-
• At cathode Cu2+ (aq) + 2 e- → Cu (s)
Thus, the net cell reaction simply involves transfer of copper
metal from the impure anode to the pure cathode, hence
purified the copper.
100. Closely related to electrorefining is electroplating, the
coating of one metal on the surface of another using
electrolysis. For example, steel automobile bumpers are
plated with chromium to protect them from corrosion, and
silver-plating is commonly used to make items of fine table
service.
The object to be plated is carefully cleaned and then set up
as the cathode of an electrolytic cell that contains a solution
of ions of the metal to be deposited (as shown in diagram
above)
Half equations occur for electroplating process of silver
above are
• At anode Ag (s) → Ag+ (aq) + e-
• At cathode Ag+ (aq) + e- → Ag (s)
101. 2.13 Industrial Electrolysis :
In this Chapter, we shall discussed the manufacturing of aluminium
and chlorine gas using the principle of electrolysis
Part 1 : Getting pure aluminium oxide (alumina) from bauxite.
1st step: Removal of impurities from the ore by dissolving
powdered bauxite in hot concentrated sodium hydroxide
solution.
2nd step: Insoluble impurities are filtered off. Filtrate contain
aluminium and silicon ions. Aluminium ion is precipitated as
aluminium hydroxide which is filtered out later as white gelatinous
precipitate.
3rd step: Aluminium hydroxide is filtered, washed, dried and
finally heated out to 12000C to produce pure aluminium oxide
(alumina), Al2O3
Al2O3 + 2 NaOH + 3 H2O 2 NaAl(OH)4
SiO2 + 2 NaOH Na2SiO3 + H2O
Use acid : 2 [Al(OH)4]- + 2 H+ 2 Al(OH)3 + 2 H2O
Or use CO2 : 2 [Al(OH)4]- + CO2 2 Al(OH)3 + H2O + CO3
2-
2 Al(OH)3 Al2O3 + 3 H2O
102. Part 2 : Extracting aluminium out from aluminium oxide
Hall-Heroult process:- A process of electrolysing aluminium oxide
(alumina) to extract out aluminium.
Aluminium metal is extracted by the cell electrolytic reduction of
alumina. Melting point of alumina is 20300C. To lower the
temperature of the electrolyte, alumina is dissolved in molten
cryolite (Na3AlF6), to maintain a temperature at about 9600C.
When alumina dissolve in molten cryolite :
Al2O3 (s) 2 Al3+ (l) + 3 O2– (l)
103. Electrolyte mixture is then placed in carbon-lined iron vat
(cathode). The heating effect of the electric current melts the
electrolyte mixture, producing Na+, Al3+, O2- and F- ions.
Half equation occur at cathode Half equation occur at anode
Na+ + e- Na E0 = – 2.71 V
Al3+ + 3e- Al E0 = – 1.66 V
F2 + 2 e- 2 F- E0 = + 2.87 V
O2 + 4e- 2O2- E0 = + 1.++V
Al3+ + 3e- Al E0 = – 1.66 V 2O2- O2 + 4e- E0 = + 1.++V
104. Aluminium alloy parts are anodized to greatly increase the
thickness of this layer for corrosion resistance. The corrosion
resistance of aluminium alloys is significantly decreased by
certain alloying elements or impurities : copper, iron,
and silicon, tend to be most susceptible.
By making an aluminium the anode of cell in which dilute
sulphuric acid is the electrolytes, it is possible to produce a
thicker and harder film of aluminium oxide on the surface of
metal.
Al C
105. 2.13.1.2 Recycling aluminium
Environmental pollution arises as cans are littered
everywhere.
The best solution of environmental pollution is recycling.
The benefits of recycling can be seen by comparing the
energy consumed in the extraction of aluminium from the
bauxite or using Hall process with that consumed when
aluminium is recycle.
106. Pure aluminium has a rather low melting print of 6600C, thus
requiring only 26.1 kJ mol-1 of energy.
On comparison between the Hall process and recycling,
Energy used in recycling = 26.1 297 100%
= 8.8%
This means that about 91% of the energy is saved for
every 1 mole of the aluminium produced through recycling
107. In industrial process, chlorine gas, together with sodium
metal, is prepared using molten sodium chloride (brine) using
mercury-cathode cell.
Mercury is specially used to attract the sodium formed in
cathode and form an alloy named amalgam
This method is not environment friendly as the mercury used
is poisonous.
Half equation occur at cathode Half equation occur at anode
Na+ + e- Na E0 = – 2.71
V
2Cl- Cl2 + 2e- E0 = – 1.36 V
108. Similar to the mercury-cathode cell, the electrolytes used in
diaphragm cell is also …………………………..
The process inside the diaphragm cell is known as the chlor-
alkali process.
When sodium chloride dissociates under the effect of an
electric current, the chloride ions are discharged. Half
equation :
brine (sodium chloride)
2 Cl- Cl2 + 2 e-
109. Titanium is chosen as the anode because it resists corrosion
by the very reactive chlorine
At cathode, since the sodium ion (Na+) is attracted to cathode
through the diaphragm, the selectivity to discharge is between
sodium ion and water molecule.
Standard reduction potential of sodium :
Standard reduction potential of water :
Since ………… has a higher E0 value, ………… is discharge
and ………………… is produced.
The level of brine (left or anode position) is always placed
higher than the water (right or cathode position) to
………………………..…………………………………………......
Na+ + e- Na E0 = – 2.71 V
2 H2O + 2 e– 2 OH– + H2 E0 = – 0.83
V water water
hydrogen gas
prevent water from crossing to brine portion. This will dilute
the solution and chlorine will not be discharged.
110. 13.8 Corrosion of metal
Corrosion is the oxidative deterioration of a metal, such as
the conversion of ………...… to ……………..
2 main important components for rusting are
………………….. and ……………………
A possible mechanism for rusting, consistent with the known
facts, is illustrated in Figure below
metal metal oxide
oxygen water
111. At anode :
Fe (s) Fe2+ (aq) + 2e- Eo = +0.44 V
At cathode :
O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)
Eo = + 0.40 V
2 Fe (s) + O2 (aq) + 2 H2O 2 Fe2+ + 4 OH– [or 2 Fe(OH)2] Ecell = + 0.84 V
Fe(OH)2 (aq) + OH– Fe(OH)3 + e –
Eo = +0.56 V
O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)
Eo = + 0.40 V
4 Fe(OH)2 (aq) + O2(aq) + 2H2O(l) 4 Fe(OH)3 (aq) Ecell = + 0.96 V
Forming rust : 2 Fe(OH)3 (s) Fe2O3.x H2O + (3 – x) H2O
In alkaline / neutral condition
112. In acidic condition
At anode :
Fe (s) Fe2+ (aq) + 2e- Eo = +0.44 V
At cathode :
O2(aq) + 4 H+(aq) + 4e− → 2H2O(l)
Eo = + 1.23 V
2 Fe (s) + O2 (aq) + 4 H+ 2 Fe2+ + 2 H2O Ecell = + 1.67 V
At anode :
Fe2+ (aq) Fe3+ (aq) + e-
Eo = -0.77 V
At cathode :
O2(aq) + 4 H+(aq) + 4e− → 2H2O(l)
Eo = + 1.23 V
4 Fe2+ (aq) + O2 (aq) + 4 H+(aq) → 4 Fe3+ (aq) + 2H2O(l) Ecell = + 0.46 V
Forming rust : 2 Fe3+ (aq) + 4 H2O (l) Fe2O3.H2O (s) + 6 H+ (aq)
113. 13.8.2 Prevention of rusting
Various methods are used to prevent / slowing down rusting.
Methods Explanation
Alloying
• Iron is alloyed with nickel and chromium to form …………………….
The chromium forms an impervious oxide layer on the surface of
iron increasing its resistance to corrosion. Chromium at the same
time ……………….. ………the steel
Barrier
Painting the iron object
Use grease / oil to coat the moving parts of machine
Coating ironwith chromium (plating) or zinc (galvanising)
Sacrificial
Also known as …………… protection
Metal with a ………………. position in electrochemical series is
‘connected’ to iron. Under such way, ………………... Will be
oxidised first before iron.
Stainless steel
decorated / coated
anodic
higher
reactive metal
115. 13.7.4 Effect of pH on Electrode Potential of a Half cell
Some reaction involve H+ ions. Examples of the are
14 H+ + Cr2O7
2- + 6 e- 2 Cr3+ + 7 H2O
8 H+ + MnO4
- + 5 e- Mn2+ + 4 H2O
6 H+ + ClO3
- + 6 e- Cl- + 3 H2O
Under standard conditions, the [H+] is 1.00 mol dm-3. varying the
concentration of [H+] and hence its pH, would change the
electrode potential of the half cell.
Consider the following half cell reaction :
MnO4
- (aq) + 8 H+ (aq) + 5 e- Mn2+ (aq) + 4 H2O (l) E = + 1.52 V
Using Nernst Equation, Ecell can be expressed as
When [MnO4
-] = [Mn2+] = 1.00 mol dm-3- ; Ecell =
]Mn[
]H][MnO[
lg
5
059.0
52.1E 2
8
4
]H[lg8
5
059.0
52.1E
pH0944.052.1E
116. [H+] / mol dm-3 1.0 0.1 0.01 0.00001
pH 0 1 2 5
Ecell (V) + 1.52 + 1.43 + 1.33 +1.05
It is then compare to the standard electrode potential of chlorine, bromine
and iodine
Cl2 + 2 e- 2 Cl- Eo = + 1.36 V @ 2 Cl- Cl2 + 2 e- Eo = - 1.36 V
Br2 + 2 e- 2 Br- Eo = + 1.07 V @ 2 Br- Br2 + 2 e- Eo = - 1.07 V
I2 + 2 e- 2 I- Eo = + 0.54 V @ 2 I- I2 + 2 e- Eo = - 0.54 V
• Under pH = 1 , Ecell of manganate (VII) ion is ................ V …………… are
able to ………… by manganate (VII) ion as it is still a strong ……………
agent.
• Under pH = 2, Ecell of manganate (VII) ion is .............. V. Only ……………
are able to ………… by manganate (VII) ion. …… cannot oxidise as the
reaction is not ……………… (Ecell = …………)
• Under pH = 5, Ecell of manganate (VII) ion is .............. V. Only …… are
able to ………… by manganate (VII) ion. ………. cannot oxidise as the
reaction is not ……………… (Ecell = ……………………)
+ 1.43 Cl- , Br- , I-
oxidise oxidising
+ 1.33 Br- , I-
oxidise Cl-
spontaneous - 0.03 V
+ 1.05 I-
oxidise Cl- , Br-
spontaneous -0.03 V for Br-
-0.31 V for Cl-
119. Fuel Cell Lithium ion cell
Diagram
Anode Lithium metal
Equation occur
at anode
Li (s) Li+ (aq) + e-
Cathode Manganese (IV) oxide
Equation occur
at cathode MnO2 (s) + Li+ + e- LiMnO2
Electrolytes Lithium chlorite (VII), LiClO4
Hydrogen
2 H2 + 4 OH– 2 H2O + 4 e–
Oxygen
O2+ 2 H2O + 4 e– 4 OH–
Hot potassium hydroxide (aq)
120. 13.9 Dental Filling
The material commonly used to fill decaying teeth is an ………………….
(a …………………… base alloy). The component in dental filling of
amalgam are …………………. , ……………… and …………..
The standard electrode potential of these electrode system are :
Hg2
2+ (aq) / Ag2Hg3 (s) Eo = + 0.85 V
Sn2+ (aq) / Ag3Sn (s) Eo = – 0.55 V
Sn2+ (aq) / Sn8Hg (s) Eo = – 0.13 V
The diagram shows the reaction take place when gold is contact with
dental amalgam, which result a electrochemical cell. ……………………
act as the anode of the cell, while ………… act as the cathode and
……….. act as the electrolyte.
Since tin is more electro………………… than gold, hence tin will corrode
to form Sn2+ and mixed with saliva. This will result an unpleasant taste in
the mouth.
If the dental amalgam is in contact with an aluminium foil, an
electrochemical cell will also produced. Unlike gold, aluminium is more
………………….. than any of the electrode above, which makes
aluminium serves as an ……………… of the cell, while …………………….
as the cathode of cell. This will result a weak current flow between the
electrode and cause an unpleasant sensation in the tooth.
amalgam
mercury
mercury silver tin
dental amalgam
gold
saliva
positive
electropositive
anode amalgam