Question
For the cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate ΔG° and predict whether the reaction is spontaneous.
Solution — Step by Step
We calculate the standard cell potential using:
Here, Cu²⁺ is reduced (cathode) and Zn is oxidised (anode):
A positive E°cell already tells us the reaction is spontaneous — but we still need the number.
Write the half-reactions:
- Anode:
Zn → Zn²⁺ + 2e⁻ - Cathode:
Cu²⁺ + 2e⁻ → Cu
Two electrons are transferred, so n = 2.
The relationship between Gibbs free energy and cell potential is:
where F = 96500 C mol⁻¹ (Faraday’s constant).
Since ΔG° is negative, the reaction is spontaneous under standard conditions. This is consistent with E°cell being positive — the two criteria always agree.
Why This Works
The equation ΔG° = −nFE°cell is the bridge between electrochemistry and thermodynamics. Gibbs free energy measures the maximum useful work a system can do; in an electrochemical cell, that useful work is electrical work.
When electrons flow spontaneously from anode to cathode, the cell does work on the surroundings — meaning the system loses free energy (ΔG° < 0). The more positive E°cell is, the greater the driving force, and the more negative ΔG° becomes.
This is why the Daniel cell — Zn-Cu — was historically the first reliable battery. With E°cell = 1.10 V, it delivers a solid potential difference, which translates to a large negative ΔG°.
Alternative Method — Using Equilibrium Constant
We can also connect ΔG° to the equilibrium constant K:
Combined with ΔG° = −nFE°cell, we get:
This enormous K value confirms the reaction goes nearly to completion — essentially irreversible in the forward direction. This approach appears occasionally in JEE Main as a two-part question where they ask for both ΔG° and K.
The shortcut formula at 298 K: log K = nE°cell / 0.0592. No need to use ln and 0.0257 separately — this saves 30 seconds in the exam.
Common Mistake
Swapping cathode and anode in the E°cell formula.
Many students write E°cell = E°anode − E°cathode instead of E°cathode − E°anode. For this problem, that gives E°cell = −0.76 − 0.34 = −1.10 V, and ΔG° comes out positive — wrong sign, wrong conclusion.
The memory trick: “Reduction at cathode, so cathode comes first.” Always write it as cathode minus anode.
Another frequent slip — forgetting to convert J to kJ in the final answer. CBSE marking schemes specifically ask for ΔG° in kJ mol⁻¹. Divide your answer by 1000 before writing it down.