1. Electrochemistry AP Chemistry

2. During oxidation-reduction (redox) reactions, the oxidation states of two substances change. oxidation = reduction = loss of e – gain of e – both occur in redox LEO: “GER…” “OIL RIG.” Oxidation is loss; reduction is gain. Losing electrons: oxidation. Gaining electrons: reduction.

3. Oxidation-Reduction Reactions oxidizing agent (oxidant): is reduced (or has a component that is reduced) reducing agent (reductant): is oxidized (or has a component that is oxidized) e.g., Zn(s) + 2 H + (aq)  H 2 (g) + Zn 2+ (aq) is oxidized is the reducing agent is reduced is the oxidizing agent (charge goes from 0 to 2+) (charge goes from 1+ to 0) 001+2+

4. Identify the oxidant and the reductant. 2 H 2 O(l) + Al(s) + MnO 4 – (aq)  Al(OH) 4 – (aq) + MnO 2 (s) 07+3+4+ oxidant: reductant: MnO 4 – Al Permanganate ion is a strong oxidizer. Powdered aluminum will speed up the reaction rate.

5. Balancing Oxidation-Reduction Reactions -- conserve mass AND conserve charge half-reaction: oxidation by itself, or reduction by itself When iron rusts, one half-reaction is: Fe Fe 3+ + 3 e – (oxidation) The reduction half-reaction might be: O 2 + 4 e – 2 O 2–

6. The sum of the two half-rxns should be the overall rxn. Write half-reactions for… Sn 2+ (aq) + 2 Fe 3+ (aq)  Sn 4+ (aq) + 2 Fe 2+ (aq) Sn 2+ (aq) Sn 4+ (aq) (oxidation) Fe 3+ (aq) Fe 2+ (aq) (reduction) + 2 e – e – +222 In line notation, this reaction would be written: Sn 2+ Sn 4+ Fe 3+ Fe 2+ anode (oxidation) to L of, cathode (reduction) on R

7. Steps in Balancing Equations by the Method of Half-Reactions 1. Break overall equation into two half-reactions. 2. a. Balance everything but H and O. b. Balance O by adding H 2 O as needed. c. Balance H by adding H + as needed (assuming acidic solution). d. Add e – as needed. *4. BASIC SOLN ONLY: Add enough OH – to cancel any H +. Simplify again. e. Multiply each half-reaction by integers to cancel e –. 3. Add the two half-reactions and simplify.

8. Balance this reaction, which takes place in acidic solution. Cr 2 O 7 2– + Cl –  Cr 3+ + Cl 2 Cr 2 O 7 2–  Cr 3+ Cl –  Cl 2 2 2 + 7 H 2 O 14 H + + + 2 e – 6 e – + 66 3 14 H + +Cr 2 O 7 2– +6 Cl – 2 Cr 3+ +7 H 2 O +3 Cl 2

9. CN – + MnO 4 –  CNO – + MnO 2 Balance this reaction, which takes place in basic solution. CN –  CNO – MnO 4 –  MnO 2 4 H + + + 2 H + H 2 O + 3 e – + + 2 e – 6 + 2 H 2 O 3 CN – + 2 MnO 4 – + H2OH2O + 2 OH –  2 H + 8 22 4 6 6 333 + 3 CNO – + 2 MnO 2 + 2 OH – 3 CN – + 2 MnO 4 – + H 2 O  3 CNO – + 2 MnO 2 + 2 OH – 2 H 2 O