1. Electrochemistry Use of spontaneous chemical reactions to produce electricity; use of electricity to drive non-spontaneous reactions. Zn(s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu(s) Redox reactions involve the transfer of electrons from the species being oxidized (the reducing agent) to the species being reduced (the oxidizing agent) Examples of redox reactions include: reactions in photosynthesis, metabolism, combustion, extraction of metals from ores, reactions in batteries.

2. The transfer of electrons in the reaction Zn(s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu(s) is thermodynamically favorable; the reaction proceeds spontaneously.  G f o Zn 2+ (aq) = -147.06 kJ/mol  G f o Cu 2+ (aq) = +65.49 kJ/mol  G r o = -212.55 kJ/mol The transfer of electrons can used to produce energy in the form of electricity - chemical energy converted to electrical energy.

3. Balancing Oxidation-Reduction Reactions Sn 2+ (aq) + Fe 3+ --> Sn 4+ (aq) + Fe 2+ (aq) This reaction is balanced in terms of mass but not charge. As written, the oxidation process (Sn 2+ to Sn 4+ ) involves two electrons and the reduction process (Fe 3+ to Fe 2+ ) one electron To balance redox reactions it is easier to consider the oxidation and reduction reactions separately, even though one cannot occur without the other.

4. Oxidation: Sn 2+ (aq) --> Sn 4+ (aq) + 2e - Reduction: Fe 3+ (aq) + e - --> Fe 2+ (aq) Each of the above reactions is called a half reaction. In writing half reactions, the number of electrons lost in the oxidation half reaction must equal the number of electrons gained in the reduction half reaction Oxidation: Sn 2+ (aq) --> Sn 4+ (aq) + 2e - Reduction: 2Fe 3+ (aq) + 2e - --> 2Fe 2+ (aq)

5. The overall reaction is then the sum of the two half reactions Sn 2+ (aq) + 2Fe 3+ --> Sn 4+ (aq) + 2Fe 2+ (aq) Writing the half reactions so that the number of electrons lost in the oxidation half reaction equals the number of electrons gained in the reduction half reaction ensures that the overall reaction is balanced in terms of charge.

6. Systematic procedure for balancing redox reactions The unbalanced redox reaction between MnO 4 - and C 2 O 4 2- reaction in an acidic aqueous solution is: MnO 4 - (aq) + C 2 O 4 2- (aq) --> Mn 2+ (aq) + CO 2 (g) Step 1) Divide the equation into two incomplete half- reactions, one for oxidation and one for reduction MnO 4 - (aq) --> Mn 2+ (aq) C 2 O 4 2- (aq) --> CO 2 (g) +7 +4+3 +2 Reduction Oxidation

7. Step 2) Balance each half reaction a) First balance the elements other than H and O MnO 4 - (aq) --> Mn 2+ (aq) C 2 O 4 2- (aq) --> 2CO 2 (g) b) Next balance the O atoms by adding H 2 O MnO 4 - (aq) --> Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2- (aq) --> 2CO 2 (g) c) Balance the H atoms: for acidic solution balance H by adding H +, for basic solutions balance H by adding H 2 O to the side deficient in H and an equal amount of OH - to the other side. MnO 4 - (aq) + 8 H + (aq) --> Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2- (aq) --> 2CO 2 (g)

8. d) Balance the charge by adding electrons (e - ) to the side with the greater overall positive charge, so that the sum of the charge on the left = sum of charge on the right MnO 4 - (aq) + 8 H + (aq) + 5e - --> Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2- (aq) --> 2CO 2 (g) + 2e - Step 3) Multiply each half reaction by an integer so that the number of electrons lost in one half reaction equals the number gained in the other. MnO 4 - (aq) + 8 H + (aq) + 5e - --> Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2- (aq) --> 2CO 2 (g) + 2e - Multiply the by MnO 4 - reaction 2 and the C 2 O 4 2- reaction by 5 to balance charge 2MnO 4 - (aq) + 16 H + (aq) + 10e - --> 2Mn 2+ (aq) + 8H 2 O(l) 5C 2 O 4 2- (aq) --> 10CO 2 (g) + 10e -

9. Step 4) Add the two half-reactions and simplify where possible by canceling species appearing on both sides of the equation. 2MnO 4 - (aq) + 16 H + (aq) + 10e - --> 2Mn 2+ (aq) + 8H 2 O(l) 5C 2 O 4 2- (aq) --> 10CO 2 (g) + 10e - 2MnO 4 - (aq) + 16 H + (aq) + 5C 2 O 4 2- (aq) + 10e - --> 2Mn 2+ (aq) + 8H 2 O(l) + 10CO 2 (g) + 10e - 2MnO 4 - (aq) + 16 H + (aq) + 5C 2 O 4 2- (aq) --> 2Mn 2+ (aq) + 8H 2 O(l) + 10CO 2 (g) Step 5) Check the equation for mass and charge balance

10. Balance the following reaction which takes place in a basic solution Step 1) the two half reactions are: Ag(s) + HS - (aq) --> Ag 2 S(s) CrO 4 2- (aq) --> Cr(OH) 3 (s) Step 2a) balance elements other than H and O 2 Ag(s) + HS - (aq) --> Ag 2 S(s) CrO 4 2- (aq) --> Cr(OH) 3 (s) Ag(s) + HS - (aq) + CrO 4 2- (aq) --> Ag 2 S(s) + Cr(OH) 3 (s) 0+1-2+2+3+6-2 +1

11. Step 2b) Balance O 2Ag(s) + HS - (aq) --> Ag 2 S(s) CrO 4 2- (aq) --> Cr(OH) 3 (s) + H 2 O(l) Step 2c) Balance H; this is a basic solution. For basic solutions balance H by adding H 2 O to the side deficient in H and an equal amount of OH - to the other side. 2Ag(s) + HS - (aq) + OH - --> Ag 2 S(s) + H 2 O(l) CrO 4 2- (aq) + 5H 2 O(l) --> Cr(OH) 3 (s) + H 2 O(l) + 5OH - (aq) CrO 4 2- (aq) + 4H 2 O(l) --> Cr(OH) 3 (s) + 5OH - (aq) Step 2d) Balance charge in each half reaction 2Ag(s) + HS - (aq) + OH - --> Ag 2 S(s) + H 2 O(l) +2e - CrO 4 2- (aq) + 4H 2 O(l) +3e - --> Cr(OH) 3 (s) + 5OH - (aq)

12. Step 3) Multiply each half reaction by integers to balance charge between the two half reactions 6Ag(s) + 3HS - (aq) + 3OH - --> 3Ag 2 S(s) + 3H 2 O(l) + 6e - 2CrO 4 2- (aq) + 8H 2 O(l) + 6e - --> 2Cr(OH) 3 (s) + 10OH - (aq) Step 4) Add half reactions, accounting for species that appear on both sides. 6Ag(s) + 3HS - (aq) + 3OH - + 2CrO 4 2- (aq) + 8H 2 O(l) + 6e - --> 3Ag 2 S(s) + 3H 2 O(l) + 6e - + 2Cr(OH) 3 (s) + 10OH - (aq) 6Ag(s) + 3HS - (aq) + 2CrO 4 2- (aq) + 5 H 2 O(l) --> 3Ag 2 S(s) + 2Cr(OH) 3 (s) + 7OH - (aq) Step 5) Check for charge and mass balance

13. Balancing Disproportionation Reactions In a disproportionation reaction, the same species is both oxidized and reduced. To balance such reactions, realize that the same species may appear on the left of BOTH half reactions. Example: Balance the reaction below which takes place in an acidic solution Cl 2 (aq) --> ClO 3 - (aq) + Cl - (aq) Step 1 Cl 2 --> ClO 3 - Cl 2 --> Cl -

14. Balancing O and H and charges in each half reaction: Cl 2 + 6H 2 O --> 2ClO 3 - + 12H + + 10e - (oxidation) Cl 2 + 2e- --> 2Cl - (reduction) Accounting for the different charges in each half reaction - multiply reduction reaction by 5. The add the two half reactions 6Cl 2 + 6H 2 O --> 2ClO 3 - + 10Cl - + 12H + Divide through by 2 3Cl 2 (aq) + 3H 2 O(l) --> ClO 3 - (aq) + 5Cl - (aq) + 6H + (aq)

15. Electrochemical Cells The energy released in a spontaneous redox reaction can be used to perform electrical work. This is accomplished through a voltaic or galvanic cell, a device in which the transfer of electrons takes place through an external pathway rather than directly between reactants. Consider the spontaneous reaction that occurs when a strip of Zn is placed in a solution containing Cu 2+. The reaction taking place is: Zn(s) + Cu 2+ (aq) --> Cu(s) + Zn 2+ (aq)

16. As the reaction proceeds the blue color of the Cu 2+ fades and Cu(s) is deposited on the Zn strip The Zn strip is in direct contact with the Cu 2+ and the exchange of electrons between Zn and Cu 2+ occur directly between reactants.

17. The Zn and Cu 2+ are no longer in direct contact. The reduction of the Cu 2+ can occur only by a flow of electrons through an external circuit, namely the wire connecting the Zn(s) and Cu(s) strips; the Zn and Cu strips are the ELECTRODES. voltaic or galvanic cell

18. The voltaic cell can be considered to be made up of two “half-cells”. In each half cell a half-reaction takes place.

19. In the solution containing the Zn strip the following half reaction takes place: Zn(s) --> Zn 2+ (aq) + 2 e - Oxidation In the solution containing the Cu strip the following half reaction takes place: Cu 2+ (aq) + 2 e - --> Cu(s)Reduction

20. The electrode at which oxidation occurs is called the ANODE; here Zn is the anode. The anode is negative. The electrode at which reduction occurs is called the CATHODE ; here Cu is the cathode. The cathode is positive As the Zn metal is oxidized, the electrons flow from the anode (-) through the external circuit to the cathode (+) where they are taken up in the reduction of Cu 2+ to Cu(s)

21. As the oxidation-reduction reactions are proceeding, since there is charge transfer between the two solutions, one side will become more positive and the other more negative, preventing charge flow between the two cells. To prevent charge buildup, a salt bridge is setup between the two cells.

22. The salt bridge consists of a solution of an electrolyte like NaNO 3 whose ions will not react with the ions in the cell. As oxidation and reduction takes place, the ions in the salt bridge migrate in the directions required to maintain charge neutrality.