Electrochemistry: Edge Cases and Subtle Traps (9)

Question

Calculate the EMF of the cell at $25°C$:

$\text{Zn} | \text{Zn}^{2+}(0.1 \text{ M}) \, || \, \text{Cu}^{2+}(0.01 \text{ M}) | \text{Cu}$

Standard EMF $E°_{\text{cell}} = 1.10$ V.

Solution — Step by Step

Why This Works

The Nernst equation accounts for non-standard concentrations. When $[\text{Cu}^{2+}]$ is lower than $[\text{Zn}^{2+}]$, the reaction is “less favourable” thermodynamically, so EMF drops below the standard value. The $\log$ term is positive here, lowering $E_{\text{cell}}$.

The $0.0591/n$ factor incorporates $RT/F$ at $25°C$. At other temperatures, this constant changes — but for JEE/NEET, $25°C$ is standard.

Alternative Method

Use the natural-log form: $E = E° - (RT/nF) \ln Q$ with $RT/F = 0.0257$ V at $25°C$. Both give the same answer; the base-10 version is more common in JEE.

Common Mistake

A subtler trap: forgetting that solid electrodes don’t appear in the reaction quotient $Q$. Only ionic concentrations enter. Students sometimes write $[\text{Zn}][\text{Cu}^{2+}]/...$ — wrong. Solids and pure liquids have activity $= 1$ and drop out.