Electrochemistry: Conceptual Doubts Cleared (2)

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Tags Electrochemistry

Question

For the cell Zn∣Zn2+(0.1 M)∣∣Cu2+(1.0 M)∣CuZn|Zn^{2+}(0.1\,M)||Cu^{2+}(1.0\,M)|Cu at 25°C25°C, calculate the cell EMF. Given E°Zn2+/Zn=−0.76E°_{Zn^{2+}/Zn} = -0.76 V and E°Cu2+/Cu=+0.34E°_{Cu^{2+}/Cu} = +0.34 V. CBSE 2024 boards and NEET 2023 pattern.

Solution — Step by Step

The Daniell cell uses ZnZn as anode (oxidation) and CuCu as cathode (reduction):

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

Anode: Zn→Zn2++2e−Zn \to Zn^{2+} + 2e^-

Cathode: Cu2++2e−→CuCu^{2+} + 2e^- \to Cu

Overall: Zn+Cu2+→Zn2++CuZn + Cu^{2+} \to Zn^{2+} + Cu. Number of electrons transferred: n=2n = 2.

At 25°C:

Ecell=E°cell−0.059nlog⁡QE_{\text{cell}} = E°_{\text{cell}} - \frac{0.059}{n}\log Q

The reaction quotient Q=[Zn2+][Cu2+]=0.11.0=0.1Q = \dfrac{[Zn^{2+}]}{[Cu^{2+}]} = \dfrac{0.1}{1.0} = 0.1.

Ecell=1.10−0.0592log⁡(0.1)=1.10−0.0592(−1)=1.10+0.0295E_{\text{cell}} = 1.10 - \frac{0.059}{2}\log(0.1) = 1.10 - \frac{0.059}{2}(-1) = 1.10 + 0.0295

Ecell≈1.13 VE_{\text{cell}} \approx 1.13 \text{ V}

Final answer: Ecell≈1.13E_{\text{cell}} \approx 1.13 V.

Why This Works

The Nernst equation extends standard cell potentials (measured at 1 M) to any concentration. The logarithmic dependence on concentrations comes from the connection between Gibbs free energy and the reaction quotient: ΔG=ΔG°+RTln⁡Q\Delta G = \Delta G° + RT \ln Q, divided by −nF-nF to convert to potential.

When [Zn2+][Zn^{2+}] is small (here 0.1 M), the forward reaction is favoured, increasing the EMF slightly above the standard 1.10 V. Conversely, high [Zn2+][Zn^{2+}] would lower the EMF.

Alternative Method

Some textbooks use RTnFln⁡Q\dfrac{RT}{nF}\ln Q instead of 0.059nlog⁡Q\dfrac{0.059}{n}\log Q. Both are equivalent because RT/F⋅ln⁡10≈0.059RT/F \cdot \ln 10 \approx 0.059 V at 25°C. The simplified version is faster for exam computations.

Reaction quotient QQ for a Daniell cell uses anode ion concentration in the numerator and cathode ion in the denominator. The opposite of what students sometimes write. Always: Q=products of net cell reactionreactantsQ = \dfrac{\text{products of net cell reaction}}{\text{reactants}}.

For 2-mark NEET MCQs on Nernst equation, memorise: at 25°C, every factor of 10 difference in concentrations changes the EMF by 0.059/n0.059/n V. With n=2n = 2, that’s about 30 mV per decade.

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