Redox Reactions and Electrochemistry

The Nernst equation gives the EMF of a half-cell (or full cell) at any temperature, concentration, and partial pressure:

E = E° − (RT / nF) ln Q

At 25°C (298 K), converting ln to log₁₀:

E = E° − (0.0591 / n) log Q

where:

• E° = standard electrode/cell potential
• n = number of electrons transferred in the balanced equation
• F = Faraday's constant (96,485 C/mol)
• Q = reaction quotient ([products]/[reactants] in same form as K_eq)

For a half-cell: Q = [reduced form] / [oxidized form] (for the half-reaction written as reduction).

Worked example. For Zn²⁺ + 2e⁻ → Zn (E° = −0.76 V), at [Zn²⁺] = 0.01 M:

Q = 1 / 0.01 = 100. n = 2.
E = −0.76 − (0.0591/2) log 100 = −0.76 − 0.0591 = −0.819 V.

Lower [Zn²⁺] → reduction is less favorable → more negative E.

Cell potential and Q:

E_cell = E°_cell − (0.0591/n) log Q_cell

For Daniell cell: Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺] / [Cu²⁺].

At equilibrium, E_cell = 0 and Q = K (equilibrium constant). So:

E°_cell = (0.0591/n) log K

This connects thermodynamics (K), electrochemistry (E°), and kinetics in one equation.

Concentration cells. Same electrode in both half-cells, different concentrations. E° = 0; EMF comes purely from concentration difference:

E_cell = (0.0591/n) log ([conc at cathode] / [conc at anode])

ΔG° and E°: ΔG° = −nFE°. Spontaneous reaction (ΔG° < 0) means E°_cell > 0.

Common JEE pitfalls:

• Sign of E_cell: if E_cell > 0, reaction as written is spontaneous; if < 0, the reverse is spontaneous.
• "Reduction potential" vs "oxidation potential" — JEE uses reduction (Nernst as written above). Anode oxidation potential is the negative of its reduction potential.
• n in Nernst = electrons in the cell reaction, not the half-reaction (unless you're computing a half-cell).