Electrochemistry: PYQ Walkthrough (4)

easy 2 min read

Question

Calculate the standard EMF of the cell: ZnZn2+(1M)Cu2+(1M)Cu\text{Zn} \mid \text{Zn}^{2+}(1\,\text{M}) \parallel \text{Cu}^{2+}(1\,\text{M}) \mid \text{Cu}. Given E°Zn2+/Zn=0.76 VE°_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V} and E°Cu2+/Cu=+0.34 VE°_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}. (JEE Main 2024)

Solution — Step by Step

Convention: the right-hand electrode (after the salt bridge \parallel) is the cathode. So Cu is the cathode (reduction), Zn is the anode (oxidation).

E°cell=E°cathodeE°anodeE°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

(Both potentials are written in their standard reduction form.)

E°cell=(+0.34)(0.76)=1.10 VE°_{\text{cell}} = (+0.34) - (-0.76) = 1.10\ \text{V}

Final answer: E°cell=1.10 VE°_{\text{cell}} = 1.10\ \text{V}.

Why This Works

The cell EMF formula always uses reduction potentials as listed in standard tables, with the cathode value minus the anode value. Don’t flip signs of half-reactions; just plug in the table values directly.

A positive E°cellE°_{\text{cell}} confirms the reaction is spontaneous in the direction shown — Zn is oxidised, Cu²⁺ is reduced. This is the famous Daniell cell.

Alternative Method

Add the half-reactions: oxidation ZnZn2++2e\text{Zn} \to \text{Zn}^{2+} + 2e^- (E°=+0.76 VE° = +0.76\ \text{V} when reversed), and reduction Cu2++2eCu\text{Cu}^{2+} + 2e^- \to \text{Cu} (E°=+0.34 VE° = +0.34\ \text{V}). Sum: E°cell=0.76+0.34=1.10 VE°_{\text{cell}} = 0.76 + 0.34 = 1.10\ \text{V}. Same answer.

Mnemonic: “Cathode is positive, anode is negative” — but only in galvanic cells. In electrolytic cells (where you push current with external EMF), the polarities reverse.

Common Mistake

Flipping the sign of the anode’s reduction potential and adding instead of subtracting. Both methods give the right answer, but mixing them (e.g., subtracting after flipping) gives 0.340.76=0.42 V0.34 - 0.76 = -0.42\ \text{V} — wrong sign, predicts a non-spontaneous cell.

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