Electrochemistry

Oxidation Reduction Reactions (1) Oxidation: Loss e- Increase in Oxidation Number Zn (s)  Zn e - (2) Reduction: Acceptance of e - Decrease in Oxidation Number Cl 2(g) + 2e -  2Cl - (1) Oxidation: Loss e- Increase in Oxidation Number Zn (s)  Zn e - (2) Reduction: Acceptance of e - Decrease in Oxidation Number Cl 2(g) + 2e -  2Cl -

Balancing Oxidation-Reduction Equations: Use Half-Reaction Method Use Half-Reaction Method The half-reactions for Sn 2+ (aq) + 2Fe 3+ (aq)  Sn 4+ (aq) + 2Fe 3+ (aq) are Sn 2+ (aq)  Sn 4+ (aq) +2e - 2Fe 3+ (aq) + 2e -  2Fe 2+ (aq) Oxidation Half-Reaction: electrons are products. Reduction Half-Reaction: electrons are reactants.

Half-Reaction Method for Balancing Oxidation-Reduction Equations 1. Separate the equation into the two half-reactions. Write down the two half reactions. 2. Balance each half reaction: a. First balance all elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H + if have acidic solution. d. Finish by balancing charge by adding electrons.

3. Multiply each half reaction to make the number of electrons equal. 4. Add the two half-reactions and simplify. To simplify, remove components common to both reactant and product sides. 5. Check!

Example Balance: (acidic) MnO 4 - (aq) + Fe 2+ (aq)  Mn 2+ (aq) + Fe 3+ (aq) 1.The two incomplete half reactions are MnO 4 - (aq)  Mn 2+ (aq) Fe 2+ (aq)  Fe 3+ (aq)

MnO 4 -  Mn H2O 2. Balance each half reaction: Fe 2+ (aq)  Fe 3+ (aq)

MnO e -  Mn H2O Fe 2+ (aq)  Fe 3+ (aq) + e -

8H + + MnO e -  Mn H2O Fe 2+ (aq)  Fe 3+ (aq) + e -

8H + + MnO e -  Mn H2O 5 (Fe 2+ (aq)  Fe 3+ (aq) + e - )

8H + + MnO e -  Mn H2O 5 (Fe 2+ (aq)  Fe 3+ (aq) + e - ) 8H + + MnO Fe 2+ (aq)  Mn H2O + 5Fe 3+ (aq)

Example Balance: (basic) MnO 4 - (aq) + I - (aq)  MnO 2 + I 2 1.The two incomplete half reactions are MnO 4 - (aq)  MnO 2 I - (aq)  I 2

2. Balance each half reaction: MnO 4 - (aq) + 3e -  MnO 2 2I - (aq)  I 2 + 2e -

MnO 4 - (aq) + 3e -  MnO 2 + 4OH - 2I - (aq)  I 2 + 2e -

2H 2 O + MnO 4 - (aq) + 3e -  MnO 2 + 4OH - 2I - (aq)  I 2 + 2e -

2 (2H 2 O + MnO 4 - (aq) + 3e -  MnO 2 + 4OH - ) 3 (2I - (aq)  I 2 + 2e - ) 4H 2 O + 6I - (aq) + MnO4-(aq)  2MnO2 + 8OH- + 3I 2

Problem Complete and balance the following equations, and identify the oxidizing and reducing agents Cr2O72- (aq) + I-(aq)  Cr3+(aq) + IO3-(aq) (acidic solution) Pb(OH)42- (aq) + ClO- (aq)  PbO2(s) + Cl-(aq) (basic solution)

ZnCu

salt bridge (K + ) Cation flow → (Cl - ) flow ←

Brief Activity Series

Strong Reducing Agent.

Strong Oxidizing Agent

How many moles of Cu can be plated out of a solution containing Cu 2+ ions if a 1.50 amp current is passed through for 300 s? Cu e - → Cu (s) Notes: 1 amp = C/s F = 9.65 x 10 4 c/ mole e - Find: c → moles e - s → moles Cu = moles Cu 1.50 c s 300 s mole e x 10 4 C 1 mole Cu 2 mole e - s 2.33 x 10 -3

Effect of Concentration on EMF A Battery going dead. ∆G = ∆G o + RT lnQ -nFE = -nFE o + RT ln Q Dividing by –nF gives Rise to the Nerst Equation.

The Nerst Equation: Q: aA + bB ↔ cC + dD n = # of e - s transferred

The Nerst Equation allows us to find voltage under nonstandard conditions Example: Determine E (voltage) for: Fe(s) + Cd 2+ (aq) → Fe 2+ (aq) + Cd(s) When [Fe 2+ ] = 0.10 M and [Cd 2+ ] = K Half RXNs: Fe(s) → Fe e - E o = V Cd e - → Cd(s) E o = V E o = V + Value means spontaneous

Now try: [Fe 2+ ] = 1.0 M and [Cd 2+ ] = K - Value means non spontaneous and will go in opposite direction

What does an E = 0 value mean? For: Fe(s) + Cd 2+ (aq) → Fe 2+ (aq) + Cd(s) E o = +0.04V K = 22

The pray graet Dariush in Takht-e-jamshid My god preservation this country ( Iran ) from ENEMY, FALSE and ARID YEAR