CLASS 11–12 CHEMISTRY · ELECTROCHEMISTRY

Standard cell potential and ΔG°

E°cell = E°cathode − E°anode (both as standard reduction potentials). The electrode with the higher E° is the cathode in a galvanic cell. ΔG° = −nFE°cell.

Loading practice…

Worked examples

Example 1 · level 1

A galvanic cell is made from the Cu²⁺/Cu and H⁺/H₂ electrodes under standard conditions. Standard reduction potentials: E°(Cu²⁺/Cu) = +0.34 V, E°(H⁺/H₂) = 0.00 V. What is E°cell in volts, correct to 2 decimal places?

Show solution

Cathode (higher E°): Cu²⁺/Cu; anode: H⁺/H₂.
E°cell = 0.34 − 0.00 = 0.34 V.

Answer: 0.34 V

Example 2 · level 2

For the cell Ag | Ag⁺(1 M) || Cu²⁺(1 M) | Cu. Standard reduction potentials: E°(Cu²⁺/Cu) = +0.34 V, E°(Ag⁺/Ag) = +0.80 V. What is E°cell for the cell as written, and is the cell reaction spontaneous?

−0.46 V; non-spontaneous as written
−0.46 V; spontaneous as written
+1.14 V; spontaneous as written
+0.46 V; spontaneous as written

Show solution

Left (anode) = Ag⁺/Ag, right (cathode) = Cu²⁺/Cu.
E°cell = 0.34 − 0.80 = −0.46 V.
E°cell < 0, so ΔG° > 0: non-spontaneous as written (the reverse reaction is spontaneous).

Answer: −0.46 V; non-spontaneous as written

Example 3 · level 3

A galvanic cell is made from the Mg²⁺/Mg and H⁺/H₂ electrodes. Standard reduction potentials: E°(Mg²⁺/Mg) = −2.37 V, E°(H⁺/H₂) = 0.00 V. Calculate ΔG° for the cell reaction in kJ/mol, correct to 2 decimal places (include the sign). Take F = 96500 C mol⁻¹.

Show solution

Cathode: H⁺/H₂; anode: Mg²⁺/Mg. E°cell = 0.00 − (−2.37) = 2.37 V.
Anode: Mg → Mg²⁺ + 2e⁻. Cathode: 2H⁺ + 2e⁻ → H₂. So n = 2.
ΔG° = −nFE° = −2 × 96500 × 2.37 = −4,57,410 J/mol = −457.41 kJ/mol.

Answer: -457.41