Question
(JEE Main pattern) Calculate the EMF of the cell at 25°C:
Zn∣Zn2+(0.1 M)∣∣Cu2+(1 M)∣Cu
Standard electrode potentials: E°Cu2+/Cu=+0.34 V, E°Zn2+/Zn=−0.76 V.
Solution — Step by Step
E°cell=E°cathode−E°anode=0.34−(−0.76)=1.10 V
(Cu is the cathode — higher reduction potential. Zn is the anode.)
For the cell reaction Zn+Cu2+→Zn2++Cu, n=2 electrons transferred. The reaction quotient:
Q=[Cu2+][Zn2+]=10.1=0.1
Nernst equation (at 298 K):
Ecell=E°cell−n0.0591logQ=1.10−20.0591log0.1
log0.1=−1. So:
Ecell=1.10−20.0591⋅(−1)=1.10+0.0296=1.1296 V
Cell EMF ≈ 1.13 V.
Why This Works
Standard EMF assumes all species are at 1 M and 1 atm. Real cells deviate from this, so the Nernst equation corrects for non-standard concentrations.
A higher concentration of products vs reactants makes Q large, logQ>0, and Ecell smaller than E°. Lower products → smaller Q → larger Ecell. Here, [Zn2+]<[Cu2+], so Q<1, logQ<0, and Ecell>E° — exactly what we got.
Sign conventions: Ecathode−Eanode for cell EMF. Both potentials taken as reduction potentials. Don’t flip signs of the anode reduction potential and add — that’s a common mistake.
Alternative Method
Use Ecell=−(ΔG/nF) and compute ΔG from the cell reaction’s free energy. Equivalent but more cumbersome. Nernst is the direct shortcut.
Common Mistake
Students set Q=[Cu2+]/[Zn2+] (inverted). The reaction quotient uses products/reactants as written in the cell reaction. For Zn → Zn²⁺ and Cu²⁺ → Cu, products are Zn²⁺, reactants are Cu²⁺. So Q=[Zn2+]/[Cu2+].