Standard electrode potentials E ⦵ ; standard cell potentials E ⦵ cell and the Nernst equation

A standard electrode potential (E°) is the potential of a half-cell measured relative to the standard hydrogen electrode (SHE), under standard conditions: 298 K, all solutions 1 mol dm⁻³, gases at 100 kPa.

The SHE (H⁺ 1 mol dm⁻³, H₂ at 100 kPa over a platinum electrode, 298 K) is the reference, defined as exactly 0.00 V. A half-cell is connected to it through a high-resistance voltmeter and a salt bridge; the reading (with its sign) is that half-cell's E°.

The standard cell potential is found by subtracting the more negative electrode potential from the more positive one:

$E^{\circ}_{cell} = E^{\circ}_{\text{(more positive)}} - E^{\circ}_{\text{(more negative)}}$

A reaction is feasible when E°cell is positive. The species with the more positive E° is reduced (it is the oxidising agent); the more negative one is oxidised (the reducing agent).

Worked. Zn²⁺/Zn (E° = −0.76 V) and Cu²⁺/Cu (E° = +0.34 V): $E^{\circ}_{cell} = (+0.34) - (-0.76) = +1.10\,\text{V}$ Positive → feasible. Cu²⁺ is reduced; Zn is oxidised: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

To check whether one species oxidises another (e.g. will Cl₂ oxidise Fe²⁺?), combine the two relevant E° values; a positive E°cell confirms it happens.

Standard potentials assume 1 mol dm⁻³. At other concentrations, the Nernst equation gives the electrode potential (at 298 K):

$E = E^{\circ} + \frac{0.059}{z}\lg\!\left(\frac{[\text{oxidised species}]}{[\text{reduced species}]}\right)$

where z is the number of electrons transferred. For a metal/metal-ion electrode (the reduced form is the solid, taken as 1), this simplifies to $E = E^{\circ} + \dfrac{0.059}{z}\lg[\text{M}^{n+}]$.

Worked (Cu²⁺/Cu, E° = +0.34 V, [Cu²⁺] = 0.010 mol dm⁻³, z = 2): $E = 0.34 + \frac{0.059}{2}\lg(0.010) = 0.34 + (0.0295)(-2) = +0.28\,\text{V}$

So lowering the oxidised-species concentration makes E less positive; raising it makes E more positive. You can reach the same conclusion with Le Chatelier on the electrode equilibrium Cu²⁺ + 2e⁻ ⇌ Cu.