Redox reactions
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Redox reactions

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Redox lecture slides, conducted in Yishun Junior College (Singapore) in 2011.

Redox lecture slides, conducted in Yishun Junior College (Singapore) in 2011.

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Redox reactions Redox reactions Presentation Transcript

  • REDOX REACTIONS A.k.a. A study on why things explode
  • But I Know Every Rock and Tree and Creature…
    • Has a life, has a spirit, has a name.
  • Oxidation of Food: What a Waste!
    • Fruits and Vegetables oxidised when left in open air
      • Solution: Seal in plastic wrap
      • More radical: Add lemon juice to the cut fruit
  • Oxidation of…
    • Oxidation of nutrients causes increased activity of cells, leading to aging skin
      • Solution: Beauty products?
    People!
  • What will I learn?
    • What is a redox reaction?
    • What is oxidation and reduction? (and how to identify them)
    • What are oxidising and reducing agents?
    • What is oxidation state?
    • What is oxidation and reduction in terms of oxidation state?
    • How to assign the oxidation state of an atom?
    • How to use oxidation state to determine which species is oxidised / reduced
  • What is a redox reaction?
    • Redox – red uction + ox idation
    • Both processes occur simultaneously
    • Hence, one species is oxidised, another is reduced
    • So, what is oxidation, and what is reduction?
    • 3 different versions of the definition:
  • Redox gain of electrons loss of electrons gain in hydrogen loss of hydrogen loss of oxygen gain in oxygen Reduction Oxidation
  • Oxidation and Reduction
    • In terms of Oxygen:
      • Oxidation: Gain of oxygen in a species
        • E.g. Mg is oxidized to MgO
      • Reduction: Loss of oxygen in a species
        • E.g. H 2 O is reduced to H 2
      • Note: It’s the gain or loss of O, not O 2-
  • Oxidation and Reduction
    • In terms of Hydrogen:
      • Oxidation: Loss of hydrogen in a species
        • E.g. H 2 O is oxidised to O 2
      • Reduction: Gain of hydrogen in a species
        • E.g. O 2 is reduced to H 2 O 2
      • Note: It’s the gain or loss of H, not H +
  • Oxidation and Reduction
    • In terms of Electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain):
      • Oxidation: Loss of electrons in a species
        • E.g. Mg is oxidized to MgO (Mg from 12 electrons to 10 electrons in Mg 2+ )
      • Reduction: Gain of electrons in a species
        • E.g. O 2 is reduced to H 2 O 2 (O from 8 electrons to 9 electrons per O in O 2 2- )
  • Oxidising and Reducing agent
    • An oxidising agent is a chemical species that causes the other reactant in a redox reaction to be oxidised, and it is always reduced in the process.
    • A reducing agent is a chemical species that causes the other reactant in a redox reaction to be reduced, and it is always oxidised in the process.
  • Oxidising and Reducing Agents
    • Remember:
      • An oxidising agent is itself REDUCED when it oxidises something
      • A reducing agent is itself OXIDISED when it reduces something
      • 2Mg + O 2 -> 2MgO
      • Mg is oxidised, and thus is the reducing agent
      • O 2 is reduced, and thus is the oxidising agent
  • List of common Oxidising and Reducing Agents
    • Realise something?
      • H 2 O 2 is both an oxidising and a reducing agent!
      • If a stronger oxidising agent is present, H 2 O 2 is reducing
  • Oxidation and Reduction in ‘A’ Level
    • In terms of Oxidation States:
      • Oxidation: Gain in oxidation state in a species
        • E.g. Mg is oxidized to MgO (Mg from 0 to +2 in Mg 2+ )
      • Reduction: Gain of electrons in a species
        • E.g. O 2 is reduced to H 2 O 2 (O from 0 to -1 in O 2 2- )
    • Note: Oxidation states are always written in + x or – x , never just x or x - (e.g. Oxidation State of Mg in MgO is +2 , not 2 or -2 )
  • Common Oxidation States Chemical species Oxidation state and remarks Any element eg Fe, O 2 , S 8 zero Oxygen in any compound -2 except in peroxides example H 2 O 2 or Na 2 O 2 then oxygen atom has oxidation state of -1 or in F 2 O , then oxygen atom has oxidation state of +2 Fluorine in any compound -1 being most electronegative Hydrogen in any compound +1 except in metal hydrides example NaH then hydrogen atom has oxidation state of -1 as metals have a greater tendency to lose electrons Chlorine, bromine, iodine -ve oxidation state if bonded to less electronegative element eg NaCl; then Cl = -1. +ve oxidation state if bonded to more electronegative element eg ClO - , then Cl = +1; ClO 3 - , then Cl = +5
  • Determine the oxidation state of…
    • H in H 2 O
    • N in NH 4 +
    • S in S 2 O 3 2-
    • Cr in Cr 2 O 7 2-
  • Determine the oxidation state of…
    • * Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom
    • 1) H in H 2 O
    • Let the oxidation state of H be x .
    • Thus, in H 2 O, 2 x + (-2) = 0
    • x = 1
  • Determine the oxidation state of…
    • * Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom
    • 2) N in NH 4 +
    • Let the oxidation state of N be x .
    • Thus, in NH 4 + , x + 4(+1) = +1
    • x = -3
  • Determine the oxidation state of…
    • * Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom
    • 3) S in S 2 O 3 2-
    • Let the oxidation state of S be x .
    • Thus, in S 2 O 3 2- , 2 x + 3(-2) = -2
    • x = +2
  • Determine the oxidation state of…
    • * Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom
    • 4) Cr in Cr 2 O 7 2-
    • Let the oxidation state of Cr be x .
    • Thus, in Cr 2 O 7 2- , 2 x + 7(-2) = -2
    • x = +6
  • Example 1
    • Let the oxidation state of Mn be x .
    • Thus, in MnO 4 - , x + 4(-2) = -1
    • x = +7
    • Manganese is reduced from oxidation state of +7 in MnO 4 - to +2 in Mn 2+ , while iron is oxidised from oxidation state of +2 in Fe 2+ to +3 in Fe 3+ .
    +7 +2 +2 +3
  • A Special Redox: Disproportionation
    • Definition: A disproportionation reaction is a redox reaction in which one species is simultaneously oxidised and reduced .
    • Chlorine is simultaneously reduced from oxidation state of 0 in Cl 2 to -1 in Cl - , and oxidised from oxidation state of 0 in Cl 2 to +1 in ClO - .
    0 +1 -1
  • Disproportionation Reaction
    • Example: Is this a disproportionation reaction?
    • This is NOT a disproportionation reaction
    • Disproportionation requires that the same atom is both oxidised and reduced simultaneously.
    • In this case, different atoms (of nitrogen) are oxidised and reduced.
    -3 +5 +1
  • Non-redox reactions
    • The oxidation states of the elements remained unchanged in the following reactions:
    • Neutralisation reactions:
  • Non-redox reactions
    • The oxidation states of the elements remained unchanged in the following reactions:
    • Precipitation reactions:
  • Non-redox reactions
    • The oxidation states of the elements remained unchanged in the following reactions:
    • Complex formation:
    Tetraammine copper(II) complex (deep blue solution) ligand
  • Non-redox reactions
    • The oxidation states of the elements remained unchanged in the following reactions:
    • Another reaction:
  • Interlude: More real-life redox!
  • Balancing redox reactions
    • To make calculations in redox titrations, you need a balanced equation
  • Balancing redox reactions
    • Example: Try to balance the following reaction by trial and error.
    • Possible answer:
  • Balancing redox reactions
    • Example: Try to balance the following reaction by trial and error.
    • Actual answer:
    • Note: You might not even be told at the beginning that H + is reactant, H 2 O is product.
  • The half-equation method
    • Write down the given reactants and products of the reaction
    • Identify the atoms in the given species that are undergoing oxidation / reduction and construct the unbalanced oxidation / reduction half-equations
    • Balance both the half-equations using the following steps:
      • Balance the “odd” atoms (“odd” atoms refer to atoms other than oxygen and hydrogen)
      • Balance oxygen atoms by adding H 2 O molecules
      • Balance hydrogen atoms by adding H + ions
      • Balance charges by adding electrons
    • Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal
    • Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.
  • Oh yeah, the blue bottle…
    • “ Angry bottle”
    • The blue colour is methylene blue in the oxidised form
    • Before the bottle was shaken, the methylene blue is in the reduced state (colourless)
    • After the bottle was shaken, the oxygen in the air oxidised the methylene blue to the oxidised state (blue)
    • After a while, the glucose (reducing sugar) in the solution reduces the methylene blue back to the reduced state (colourless)
  • End of Lecture 1 of Redox “ I have never let my schooling interfere with my education.” Mark Twain
  • The half-equation method
    • Example: Balance the following reaction:
  • The half-equation method
    • Step 1: Write down the given reactants and products of the reaction
  • The half-equation method
    • Step 2: Identify the atoms in the given species that are undergoing oxidation / reduction and write the unbalanced oxidation / reduction half-equations
    • Reduction half-equation:
    • Oxidation half-equation:
    Reduced Oxidised
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance the atoms undergoing oxidation / reduction
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance oxygen atoms by adding H 2 O molecules
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance oxygen atoms by adding H 2 O molecules
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance hydrogen atoms by adding H + ions
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance hydrogen atoms by adding H + ions
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance charges by adding electrons
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance charges by adding electrons
    • Reduction half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance the atoms undergoing oxidation / reduction
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance oxygen atoms by adding H 2 O molecules
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance hydrogen atoms by adding H + ions
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance hydrogen atoms by adding H + ions
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance charges by adding electrons
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 3: Balance both the half-equations using the following steps:
      • Balance charges by adding electrons
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal
    • Reduction half-equation:
    • Oxidation half-equation:
  • The half-equation method
    • Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal
    • Reduction half-equation:
    • Oxidation half-equation:
    x 2 x 5
  • The half-equation method
    • Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal
    • Reduction half-equation:
    • Oxidation half-equation:
    x 2 x 5
  • The half-equation method
    • Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal
    • Reduction half-equation:
    • Oxidation half-equation:
    x 2 x 5
  • The half-equation method
    • Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.
    • Reduction half-equation:
    • Oxidation half-equation:
    x 2 x 5
  • The half-equation method
    • Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.
    +6H +
  • The half-equation method
    • Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.
    • Balanced Equation:
    +6H +
  • The half-equation method
    • Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.
    • Balanced Equation: