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:

$\Delta G° = -nFE°$

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)!