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MCQEasyJEE 2025Electrochemical Cells

JEE Chemistry 2025 Question with Solution

Standard electrode potentials for a few half-cells are mentioned below:

Half-cellStandard Electrode PotentialCu2+/Cu+0.34 VZn2+/Zn−0.76 VAg+/Ag+0.80 VMg2+/Mg−2.37 V\begin{array}{|c|c|} \hline \text{Half-cell} & \text{Standard Electrode Potential}\\ \hline \text{Cu}^{2+} / \text{Cu} & +0.34 \, \text{V}\\ \text{Zn}^{2+} / \text{Zn} & -0.76 \, \text{V}\\ \text{Ag}^+ / \text{Ag} & +0.80 \, \text{V}\\ \text{Mg}^{2+} / \text{Mg} & -2.37 \, \text{V}\\ \hline \end{array}

Which one of the following cells gives the most negative value of ΔG∘\Delta G^\circ?

  • A

    Zn ∣ Zn2+(1 M) ∣∣ Ag+(1 M) ∣ Ag\text{Zn} \,|\, \text{Zn}^{2+} (1\,\text{M}) \,||\, \text{Ag}^+ (1\,\text{M}) \,|\, \text{Ag}

  • B

    Zn ∣ Zn2+(1 M) ∣∣ Mg2+(1 M) ∣ Mg\text{Zn} \,|\, \text{Zn}^{2+} (1\,\text{M}) \,||\, \text{Mg}^{2+} (1\,\text{M}) \,|\, \text{Mg}

  • C

    Ag ∣ Ag+(1 M) ∣∣ Mg2+(1 M) ∣ Mg\text{Ag} \,|\, \text{Ag}^+ (1\,\text{M}) \,||\, \text{Mg}^{2+} (1\,\text{M}) \,|\, \text{Mg}

  • D

    Cu ∣ Cu2+(1 M) ∣∣ Ag+(1 M) ∣ Ag\text{Cu} \,|\, \text{Cu}^{2+} (1\,\text{M}) \,||\, \text{Ag}^+ (1\,\text{M}) \,|\, \text{Ag}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Standard electrode potentials of Cu2+/Cu\text{Cu}^{2+}/\text{Cu}, Zn2+/Zn\text{Zn}^{2+}/\text{Zn}, Ag+/Ag\text{Ag}^+/\text{Ag}, and Mg2+/Mg\text{Mg}^{2+}/\text{Mg} are provided.

Find: Which cell gives the most negative value of ΔG∘\Delta G^\circ.

Use the relation

Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

and for a galvanic cell, more positive Ecell∘E^\circ_{\text{cell}} means more negative ΔG∘\Delta G^\circ because

ΔG∘=−nFEcell∘\Delta G^\circ = -nF E^\circ_{\text{cell}}

Now evaluate each option:

  1. For Zn ∣ Zn2+∣∣Ag+ ∣ Ag\text{Zn} \,|\, \text{Zn}^{2+} || \text{Ag}^+ \,|\, \text{Ag},
Ecell∘=EAg+/Ag∘−EZn2+/Zn∘E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/\text{Ag}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} Ecell∘=(+0.80)−(−0.76)=+1.56 VE^\circ_{\text{cell}} = (+0.80) - (-0.76) = +1.56 \, \text{V}
  1. For Zn ∣ Zn2+∣∣Mg2+ ∣ Mg\text{Zn} \,|\, \text{Zn}^{2+} || \text{Mg}^{2+} \,|\, \text{Mg},
Ecell∘=EMg2+/Mg∘−EZn2+/Zn∘E^\circ_{\text{cell}} = E^\circ_{\text{Mg}^{2+}/\text{Mg}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} Ecell∘=(−2.37)−(−0.76)=−1.61 VE^\circ_{\text{cell}} = (-2.37) - (-0.76) = -1.61 \, \text{V}
  1. For Ag ∣ Ag+∣∣Mg2+ ∣ Mg\text{Ag} \,|\, \text{Ag}^+ || \text{Mg}^{2+} \,|\, \text{Mg},
Ecell∘=EMg2+/Mg∘−EAg+/Ag∘E^\circ_{\text{cell}} = E^\circ_{\text{Mg}^{2+}/\text{Mg}} - E^\circ_{\text{Ag}^+/\text{Ag}} Ecell∘=(−2.37)−(+0.80)=−3.17 VE^\circ_{\text{cell}} = (-2.37) - (+0.80) = -3.17 \, \text{V}
  1. For Cu ∣ Cu2+∣∣Ag+ ∣ Ag\text{Cu} \,|\, \text{Cu}^{2+} || \text{Ag}^+ \,|\, \text{Ag},
Ecell∘=EAg+/Ag∘−ECu2+/Cu∘E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/\text{Ag}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}} Ecell∘=(+0.80)−(+0.34)=+0.46 VE^\circ_{\text{cell}} = (+0.80) - (+0.34) = +0.46 \, \text{V}

The highest positive standard cell potential is +1.56 V+1.56 \, \text{V} for Zn ∣ Zn2+(1 M) ∣∣ Ag+(1 M) ∣ Ag\text{Zn} \,|\, \text{Zn}^{2+} (1\,\text{M}) \,||\, \text{Ag}^+ (1\,\text{M}) \,|\, \text{Ag}.

Therefore, the cell with the most negative ΔG∘\Delta G^\circ is option A.

Potential Difference Insight

Given: The cell with the most negative ΔG∘\Delta G^\circ must have the largest positive Ecell∘E^\circ_{\text{cell}}.

Find: Which option has the maximum separation between cathode and anode reduction potentials.

Choose the electrode with the most negative reduction potential as the anode and the one with the most positive reduction potential as the cathode among the pair in each option. For option A,

Ecell∘=(+0.80)−(−0.76)=+1.56 VE^\circ_{\text{cell}} = (+0.80) - (-0.76) = +1.56 \, \text{V}

This is larger than the positive value for option D, while options B and C give negative cell potentials as written in the solution.

Therefore, the correct option is A.

Common mistakes

  • Using the most negative Ecell∘E^\circ_{\text{cell}} instead of the most negative ΔG∘\Delta G^\circ. This is wrong because ΔG∘=−nFEcell∘\Delta G^\circ = -nF E^\circ_{\text{cell}}. A more negative ΔG∘\Delta G^\circ corresponds to a more positive Ecell∘E^\circ_{\text{cell}}. Compare the largest positive cell potential instead.

  • Reversing anode and cathode while applying Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}. This gives the wrong sign of Ecell∘E^\circ_{\text{cell}}. Identify oxidation at the anode and reduction at the cathode before substitution.

  • Comparing individual electrode potentials directly without taking their difference. This is wrong because cell spontaneity depends on the potential difference between cathode and anode, not on a single half-cell potential. Always calculate Ecell∘E^\circ_{\text{cell}} for each complete cell.

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