Standard conditions (E°):

• 25°C (298 K)
• 1 M concentrations
• 1 atm pressures

Real conditions often differ!

• Different concentrations
• Different temperatures
• Cell potential changes

The Nernst Equation

General form:

$E = E° - \frac{RT}{nF}\ln Q$

At 25°C (298 K):

$E = E° - \frac{0.0592}{n}\log Q$

Where:

• E = cell potential (V)
• E° = standard cell potential (V)
• R = 8.314 J/(mol·K)
• T = temperature (K)
• n = moles of electrons transferred
• F = 96,485 C/mol
• Q = reaction quotient

0.0592 V comes from (RT ln 10)/F at 298 K

Reaction Quotient (Q)

For redox reaction:

$aA + bB \rightarrow cC + dD$

Q expression:

$Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

Same as equilibrium K, but use current concentrations

Rules:

• Include aqueous and gaseous species
• Omit pure solids and liquids
• Use partial pressures (atm) for gases

Using Nernst Equation

Step-by-step:

1. Calculate E° from standard potentials
2. Write balanced equation
3. Determine n (electrons transferred)
4. Write Q expression
5. Calculate Q from concentrations
6. Substitute into Nernst equation
7. Solve for E

Effect of Concentration

Le Chatelier applies to electrochemistry!

Increase [products]:

• Q increases
• E decreases (more negative)
• Forward reaction less favorable

Increase [reactants]:

• Q decreases
• E increases (more positive)
• Forward reaction more favorable

Dilution effects:

• Cell potential changes
• Can shift from spontaneous to non-spontaneous

Concentration Cells

Same species at both electrodes, different concentrations

Example: Cu | Cu²⁺(dilute) || Cu²⁺(concentrated) | Cu

E° = 0 (same half-reactions)

But E ≠ 0! (different concentrations)

Nernst equation:

$E = 0 - \frac{0.0592}{n}\log\frac{[Cu^{2+}]_{\text{anode}}}{[Cu^{2+}]_{\text{cathode}}}$

Electrons flow from dilute to concentrated

• Dilute side: Cu → Cu²⁺ (oxidation)
• Concentrated side: Cu²⁺ → Cu (reduction)
• Equalizes concentrations

Relationship to Equilibrium

At equilibrium:

• E = 0 (no driving force)
• Q = K

Substitute into Nernst:

$0 = E° - \frac{0.0592}{n}\log K$

Rearrange:

$E° = \frac{0.0592}{n}\log K$

Or:

$\log K = \frac{nE°}{0.0592}$

Can calculate K from E°!

Relationship:

• Large positive E° → Large K (products favored)
• Large negative E° → Small K (reactants favored)

Temperature Effects

Full Nernst equation needed if T ≠ 298 K:

$E = E° - \frac{RT}{nF}\ln Q$

Temperature affects:

• Both E° and the coefficient
• Generally small effect for moderate T changes

Dead Battery

Battery dies when E → 0:

• Approaches equilibrium
• Reactants and products balanced
• Q → K
• No more driving force

Recharging:

• Apply external voltage
• Reverse reaction
• Restore reactants

pH and Electrochemistry

Many half-reactions involve H⁺:

Example: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

E depends on pH!

Lower pH (higher [H⁺]):

• Shifts reduction forward
• Increases E
• Stronger oxidizing agent

Higher pH (lower [H⁺]):

• Shifts reduction backward
• Decreases E
• Weaker oxidizing agent

Sign of E and Spontaneity

Under any conditions:

General relationship:

$\Delta G = -nFE$

At standard conditions:

$\Delta G° = -nFE°$

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

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 ✓

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