Question
For the cell Zn∣Zn2+(0.1M)∣∣Cu2+(1.0M)∣Cu, given EZn2+/Zn∘=−0.76 V and ECu2+/Cu∘=+0.34 V, find the cell EMF at 298 K. A student computes E=1.10+(0.0591/2)log(1/0.1)=1.13 V. Is this correct?
Solution — Step by Step
Anode (oxidation): Zn→Zn2++2e−.
Cathode (reduction): Cu2++2e−→Cu.
Overall: Zn+Cu2+→Zn2++Cu. n=2.
Ecell∘=Ecathode∘−Eanode∘=0.34−(−0.76)=1.10 V
Nernst: E=E∘−n0.0591logQ where Q is the reaction quotient with products over reactants.
Q=[Cu2+][Zn2+]=1.00.1=0.1
E=1.10−20.0591log(0.1)=1.10−0.02955×(−1)=1.10+0.02955≈1.13 V
The student wrote log(1/0.1)=log(10)=+1, while we wrote log(0.1)=−1. Both lead to +0.02955 added to E∘, so the answer is numerically correct: E≈1.13 V.
But the student’s setup with 1/0.1 inside the log corresponds to using −logQ as log(1/Q). That’s algebraically equivalent here because Q=[Zn2+]/[Cu2+] — but if they had reversed the cell or set up Q incorrectly, the shortcut would fail. The robust approach is: compute Q as products over reactants, take its log directly, and plug into the standard Nernst form.
Why This Works
The Nernst equation governs the dependence of EMF on concentration. At 298 K, the prefactor RT/F⋅ln10=0.0591 V. The reaction quotient Q is products over reactants, just like K, except for non-equilibrium conditions.
Why does decreasing [Zn2+] increase E? The forward reaction is favoured when products are scarce (Le Chatelier in disguise), so the cell drives harder — higher EMF.
Alternative Method
Plug into E=E∘+n0.0591log([products][reactants]). For this reaction: E=1.10+(0.0591/2)log(1/0.1)=1.13 V. This is the form many books use — sign-flipped, products in denominator. Either form works if you’re consistent.
Common Mistake
Three traps. (1) Forgetting that pure solid metals (Zn, Cu) have activity = 1 and don’t appear in Q. (2) Sign error on Ecell∘ — always reduction-cathode minus reduction-anode. (3) Using log vs ln — at 298 K, the constant is 0.0591 for log10, 0.0257 for ln. Mismatch loses the entire 4 marks in JEE Main.