Electrochemistry

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Includes a discussion of Voltaic and electrolytic cells, the Nernst equation and the relationship between electrochemical processes, chemical equilibrium and free energy.
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Electrochemistry

  1. 1. Copyright Sautter 2015
  2. 2. ELECTROCHEMISTRY • All electrochemical reactions involve oxidation and reduction. • Oxidation means the loss of electrons (it does not always involve oxygen). • Reduction means the gain of electrons (gaining of negatives, that is electrons, reduces the oxidation number of an atom. We will discuss oxidation number latter in this program). • Whenever electrons are lost by one substance they must be gained by another. • The substance that loses electrons is referred to as a reducing agent (it lets another substance be reduced, that is, gain the electrons). • The substance that gains electrons is referred to as an oxidizing agent (it lets another substance be oxidized, that is, lose the electrons). • These terms are important in electrochemistry!! 2
  3. 3. OXIDATION & REDUCTION • OXIDATION: Zn(s)  Zn+2 (aq) + 2 e- Metallic zinc is oxidized to zinc ion. Metallic zinc is serving as a reducing agent.(electron loser) • REDUCTION: Cu+2 (aq) + 2 e- Cu(s) Copper ion is reduced to copper metal. Copper ion is serving as an oxidizing agent (electron gainer) • In the overall reaction two electrons are transferred from the zinc metal to the copper ion. Zn(s) + Cu+2 (aq) Zn+2 (aq) + Cu(s) 3
  4. 4. OXIDATION NUMBERS • In the previous reactions the loss and gain of electrons was obvious. Zn(s) must lose two negative electrons in order to form a Zn+2 (aq) ion. And Cu+2 (aq) must gain two negative electrons to form a Cu(s) atom. • What about more complicated reactions? How do we tell which substance loses and which gains electrons and how many? • For example: H+ + MnO4 - + Cl-  Mn+2 + Cl2 + H2O Here we need oxidation numbers! 4
  5. 5. OXIDATION NUMBERS • Before discussing the mechanics of oxidation numbers it is important to realize that: • (1) oxidation numbers have no physical significance. They are merely a way to tell which substances gain or lose electrons and how many. • (2) oxidation numbers are assigned to atoms never molecules. Molecules contain atoms with oxidation numbers but they themselves cannot be assigned oxidation numbers. • (3) there a several rules used to assign oxidation numbers. They must be observed carefully. 5
  6. 6. OXIDATION NUMBERS RULES • (1) Elements have oxidation numbers of zero. For example Cu has an oxidation number of zero. In Cl2 each atom of chlorine has an oxidation number of zero. • (2) The oxidation number of a monatomic ion (consisting of one atom) is the charge on the ion. In Cu+2 the oxidation number is +2. In Cl- the oxidation number is –1. • (3) The oxidation number of combined oxygen is always –2 except in peroxides such as hydrogen peroxide, H2O2. In CO2 each oxygen atom has an oxidation number of –2. In SO3 each oxygen is –2. 6
  7. 7. OXIDATION NUMBERS RULES • (4) Combined hydrogen is always +1 except in hydrides (metal atoms and hydrogen like NaH). In water H2O each hydrogen has an oxidation number of +1. In methane, CH4, each hydrogen is +1. • (5) The sum of all of the oxidation numbers of the atoms in a molecule equal its charge. If no charge is shown the sum must equal zero. This allows us to determine the oxidation numbers of elements not discussed in the first four oxidation number rules! • Oxidation numbers of other elemental groups: Elements of Column I in a compound = +1 Elements of Column II in a compound = +2 It is important to realize that the oxidation number of an atom can change when it combines differently during reaction. This is the basis for deciding how electrons are transferred during a chemical reaction! 7
  8. 8. FINDING UNKNOWN OXIDATION NUMBERS • Problem: What is the oxidation number of Mn in MnO4 -, the permanganate ion? • Solution: Mn is unknown Each O is –2 (rule 3) All of the oxidation numbers must add up to –1 (rule 5) Mn + 4(-2) = -1, Mn = +7 in permanganate • Problem: What is the oxidation number of Cr in K2Cr2O7, potassium dichromate? • Solution: Cr is unknown Each K from Column I is +1 Each O is –2 (rule 3) All the oxidation numbers add up to 0 (no charge is shown for K2Cr2O7) 2(+1) + 2 Cr + 7(-2) = 0, 2 Cr = 12, Cr = 12 / 2 = +6 Cr = +6 in potassium dichromate 8
  9. 9. USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS • Using oxidation numbers we will find the oxidizing agent, the reducing agent, the number of electrons lost and gained, the oxidation half-reaction, the reduction half-reaction and the final balanced equation for: H+ + MnO4 - + Cl-  Mn+2 + Cl2 + H2O • Step I – Find the oxidation number of each atom. H+ + MnO4 - + Cl-  Mn+2 + Cl2 + H2O +1 (+7, -2) -1 +2 0 (+1, -2) • Step II – Determine which oxidation numbers change MnO4 -  Mn+2 2 Cl-  Cl2 +7  +2 2(-1)  0 5 electrons gained 2 electrons lost (MnO4 - is the oxidizing agent. It gained electrons from Cl- ) (Cl- is the reducing agent. It lost electrons to MnO4 - ) If oxidization # s decrease, Reduction occurs If oxidization # s increase, Oxidation occurs 9
  10. 10. USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS • Step III - Electrons lost by a reducing agent must always equal electrons gained by the oxidizing agent! • Therefore: 2 (MnO4 - + 5e-  Mn+2 ) 2MnO4 - + 10e-  2Mn+2 (reduction half-reaction*) • And 5(2Cl-  Cl2 + 2e-) 10 Cl-  5Cl2 + 10e- (oxidation half-reaction) • Step IV -Completing the first half-reaction with H+ ions and H2O molecules: 16 H+ + 2MnO4 - + 10e-  2Mn+2 + 8 H2O • Step V - Adding the oxidation half-reaction and reduction half- reaction the 10 electrons gained and lost cancel to give the overall reaction: 16 H+ + 2MnO4 - + 10 Cl-  2 Mn+2 + 5 Cl2 + 8 H2O • * Half-reaction refers to the reaction showing either the electron gain (reduction) or the electron loss (oxidation) step of the reaction. 10
  11. 11. ELECTRICAL POTENTIAL, CHARGE, ELECTRON FLOW • For electrons to transfer from one substance to another the oxidizing agent (electron gainer) must have a greater attraction for those electrons than the reducing agent (electron loser). This where electrical potentials come into play. • Electrical potential is measured in volts. Electrical charge is measured in coulombs. The flow of electrons from one substance to another is measured in amperes. • 1 coulomb is the charge on 6.25 x 1018 electrons.(The charge on 1 electron is –1.6 x 10-19 coulombs). 1 ampere (amp) is the flow rate of 1 coulomb per second 1 volt is 1 joule of energy moving 1 coulomb of charge between two substances. 11
  12. 12. 6.25 x 10 18electrons 1 coulomb 1 second 1 amp = 1 coul / sec 1 joule 1 coulomb of charge 1 volt = 1 joule / coul ELECTRICAL QUANTITIES 12
  13. 13. ELECTRICAL POTENTIAL • Electrical potential is measured in volts and indicates the tendency of electrons to move from one substance to another. Potential depends on a variety of factors such as the concentration of reactant materials, temperature, gas pressures and the nature of the materials involved. • Standard Reduction Potentials (measuring the tendency of a substance to gain electrons) are determined at 25 0C, 1 molar concentrations, 1 atm pressures and are compared to hydrogen ion’s tendency to be reduced (gain electrons) to hydrogen gas (2H+ (aq) + 2e-  H2(g)) which is assigned a potential of 0.00 volts. • Substances which gain electrons better than H+ ion are assigned positive potentials. Those which gain electrons more poorly than H+ ion are assigned negative potentials.13
  14. 14. ELECTRICAL POTENTIAL Some Standard Reduction Potentials E0 (volts) 14
  15. 15. ELECTRICAL POTENTIALS • The very best oxidizing agent (electron gainer) available is flourine gas, F2. Its reduction potential is +2.87 volts. • The very best reducing agent (electron loser) is lithium metal, Li. Its oxidization potential is +3.05 volts. • When Li is combined with F2, the reaction is highly spontaneous. Electrons move readily from the Li to the F2. The standard potential of this cell (redox reaction) is +5.92 volts. • By contrast, combining Li+ ion with F- ion results in a nonspontaneous process.(Li+ has already lost its electrons and F- has already gained its electrons). The standard potential for this cell is –5.92 volts 15
  16. 16. ELECTRICAL POTENTIALS • From the Standard Reduction Potential Table: (1) F2(g) + 2e-  2 F- (aq) 2.87 volts (2) Li+ (aq) + 1 e-  Li(s) -3.05 volts • Since Li metal serves as a reducing agent, it must be shown to lose electrons, and the half-reaction must be reversed. This requires that the E0 sign be changed. (3) Li(s)  Li+ (aq) + 1e- +3.05 volts • and since 2 electrons must be transferred to F2 (4) 2 Li(s)  2 Li+ (aq) + 2e- +3.05 volts (note that the E0 is not multiplied by two.) • Combining half-reaction (1) and (4) we get Li(s) + F2(g)  Li+ (aq) + 2 F- (aq) + 5.92 volts For the reverse reaction which is nonspontaneous, the potential is –5.92 volts. 16
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