Given:

• Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
• [Zn²⁺] = 0.10 M
• [Cu²⁺] = 2.0 M
• E°_cell = 1.10 V
• T = 25°C

Determine n:

Half-reactions:

• Zn → Zn²⁺ + 2e⁻
• Cu²⁺ + 2e⁻ → Cu

n = 2 electrons transferred

Write Q expression:

$Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}$

Solids (Zn, Cu) not included

Calculate Q:

$Q = \frac{0.10}{2.0} = 0.050$

Nernst equation (25°C):

$E = E° - \frac{0.0592}{n}\log Q$

Substitute:

$E = 1.10 - \frac{0.0592}{2}\log(0.050)$

$E = 1.10 - 0.0296 \times \log(0.050)$

$\log(0.050) = -1.30$

$E = 1.10 - 0.0296 \times (-1.30)$

$E = 1.10 - (-0.038)$

$E = 1.10 + 0.038$

$E = 1.14 \text{ V}$

Answer: E = 1.14 V

Interpretation:

E (1.14 V) > E° (1.10 V)

Why?

• [Cu²⁺] is high (2.0 M > 1.0 M)
• [Zn²⁺] is low (0.10 M < 1.0 M)
• Q < 1 makes forward reaction MORE favorable
• Cell potential increases

Q = 0.050 < 1:

• More reactants than at standard
• Le Chatelier: favors products
• Higher voltage

Still spontaneous (E > 0)!

Given:

• Reaction: Ni(s) + 2Ag⁺(aq) ⇌ Ni²⁺(aq) + 2Ag(s)
• E°(Ag⁺/Ag) = +0.80 V
• E°(Ni²⁺/Ni) = -0.23 V

Calculate E°_cell:

Oxidation: Ni → Ni²⁺ + 2e⁻ (E°_anode = -0.23 V) Reduction: Ag⁺ + e⁻ → Ag (×2) (E°_cathode = +0.80 V)

$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$

$E°_{\text{cell}} = 0.80 - (-0.23) = 1.03 \text{ V}$

Determine n:

2 electrons transferred (2e⁻ in each half-reaction)

n = 2

Use relationship between E° and K:

At equilibrium: E = 0, Q = K

$E° = \frac{0.0592}{n}\log K$

Rearrange:

$\log K = \frac{nE°}{0.0592}$

Substitute:

$\log K = \frac{(2)(1.03)}{0.0592}$

$\log K = \frac{2.06}{0.0592}$

$\log K = 34.8$

Take antilog:

$K = 10^{34.8}$

$K = 6.3 \times 10^{34}$

Answer: K = 6.3 × 10³⁴

Interpretation:

Extremely large K!

• Reaction goes essentially to completion
• Products heavily favored
• Consistent with large positive E° (1.03 V)

General pattern:

• E° = 0.06 V → K ≈ 10
• E° = 0.12 V → K ≈ 100
• E° = 0.60 V → K ≈ 10²⁰
• E° = 1.0 V → K ≈ 10³⁴

Each +0.06 V adds ~1 to log K (for n=1)

Verification:

Large positive E° (1.03 V) means:

• ΔG° very negative
• Reaction very spontaneous
• Products extremely favored
• K >> 1 ✓

Given:

• Concentration cell: Ag | Ag⁺(0.010 M) || Ag⁺(1.0 M) | Ag
• Same metal, different [Ag⁺]
• T = 25°C

(a) Calculate E_cell

Half-reactions:

Both sides: Ag⁺ + e⁻ ⇌ Ag

E° for both = +0.80 V

Standard cell potential:

$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.80 - 0.80 = 0$

Concentration cell always has E° = 0!

Overall reaction:

Dilute side oxidizes, concentrated side reduces:

Ag⁺(1.0 M) + e⁻ → Ag (cathode - concentrated) Ag → Ag⁺(0.010 M) + e⁻ (anode - dilute)

Net: Ag⁺(1.0 M) → Ag⁺(0.010 M)

Tries to equalize concentrations

Q expression:

$Q = \frac{[Ag^+]_{\text{anode}}}{[Ag^+]_{\text{cathode}}} = \frac{0.010}{1.0} = 0.010$

n = 1 (one electron)

Nernst equation:

$E = E° - \frac{0.0592}{n}\log Q$

$E = 0 - \frac{0.0592}{1}\log(0.010)$

$\log(0.010) = -2.0$

$E = 0 - 0.0592 \times (-2.0)$

$E = 0 + 0.118$

$E = 0.12 \text{ V}$

Answer (a): E_cell = 0.12 V

(b) Which is anode?

Anode: Dilute side (0.010 M)

• Oxidation occurs: Ag → Ag⁺
• Increases [Ag⁺] in dilute solution
• Electrons leave this electrode

Cathode: Concentrated side (1.0 M)

• Reduction occurs: Ag⁺ → Ag
• Decreases [Ag⁺] in concentrated solution
• Electrons enter this electrode

Answer (b): Dilute side (0.010 M) is anode

General rule for concentration cells:

Dilute → Anode (oxidation increases concentration) Concentrated → Cathode (reduction decreases concentration)

System tries to equalize concentrations!

Alternative Nernst form for concentration cells:

$E = \frac{0.0592}{n}\log\frac{[\text{concentrated}]}{[\text{dilute}]}$

$E = \frac{0.0592}{1}\log\frac{1.0}{0.010}$

$E = 0.0592 \times \log(100) = 0.0592 \times 2 = 0.12 \text{ V}$ ✓

Concentration ratio of 10 → E = 0.059 V Concentration ratio of 100 → E = 0.118 V