Standard electrode potentials, cell potentials and the Nernst equation

Chemistry A2 · Electrochemistry. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

Test this topic free All Chemistry A2 topics

The idea

A standard electrode potential, E^⦵, measures how easily a species gains electrons. It is the voltage of a half-cell compared with the standard hydrogen electrode under standard conditions. Half-equations are always written as reductions.

The more positive the E^⦵, the more easily the species on the left is reduced, so it is a stronger oxidising agent. The more negative the E^⦵, the more easily the species on the right is oxidised, so it is a stronger reducing agent.

Joining two half-cells gives a cell. The half-cell with the more positive E^⦵ is the positive electrode, where reduction happens. Electrons flow to it through the external circuit.

Rules to remember

Standard conditions: 298 K, solutions of 1.00 mol dm⁻³, gases at 101 kPa.
Standard hydrogen electrode: H₂(g) at 101 kPa over platinum in 1.00 mol dm⁻³ H⁺(aq); its E^⦵ is 0.00 V.
Two ions of the same element (e.g. Fe³⁺/Fe²⁺): use a platinum electrode in a solution with both ions at 1.00 mol dm⁻³.
E^⦵_cell = E^⦵(more positive half-cell) − E^⦵(more negative half-cell)
A reaction is feasible if E^⦵_cell is positive: the oxidising agent must come from the half-equation with the more positive E^⦵.
Nernst equation at 298 K: E = E^⦵ + (0.059 ÷ z) × log([oxidised species] ÷ [reduced species]); a solid metal is left out of the ratio.
ΔG^⦵ = −nE^⦵_cellF, giving joules per mole; n = moles of electrons transferred.

Common mistake

Do not multiply an E^⦵ value when you multiply a half-equation to balance the electrons. E^⦵ is a voltage and does not depend on the number of moles.

Worked example

A cell is made from Ni²⁺/Ni (E^⦵ = −0.25 V) and Ag⁺/Ag (E^⦵ = +0.80 V) half-cells. Find E^⦵_cell and ΔG^⦵ for the cell reaction. (F = 96 500 C mol⁻¹)

Silver has the more positive E^⦵, so it is the positive electrode: E^⦵_cell = +0.80 − (−0.25) = +1.05 V.
Cell reaction: Ni + 2Ag⁺ → Ni²⁺ + 2Ag, so n = 2.
ΔG^⦵ = −nE^⦵_cellF = −2 × 1.05 × 96 500 = −202 650 J mol⁻¹.
Divide by 1000: ΔG^⦵ = −203 kJ mol⁻¹.

Answer: E^⦵_cell = +1.05 V; ΔG^⦵ = −203 kJ mol⁻¹ (feasible)

Ask about this

Type what you are stuck on. The helper finds the explanation, worked examples and the notes to read, free and instant.

Read the full notes

Watch: 3 | Standard Electrode Potentials

Keep going

← Electrolysis Acids and bases →Take a Chemistry A2 test