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MHT CET 202624 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
24 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

MHT CET 2026 · 24 April · Evening Shift · Chemistry · Electrochemistry

Calculate the standard Gibbs energy change for the following cell...

Calculate the standard Gibbs energy change for the following cell
Cd(s) ∣ Cd++(1M) ∣∣ Sn++(1M)∣ Sn(s)\text{Cd}(s)\ |\ \text{Cd}^{++}(1\text{M})\ ||\ \text{Sn}^{++}(1\text{M})|\ \text{Sn}(s) if Ecell0E^0_{\text{cell}} is 0.27 V0.27\ \text{V}.

Options

  1. A−52.11 kJ-52.11\ \text{kJ}
  2. B52.11 kJ52.11\ \text{kJ}
  3. C−26.6 kJ-26.6\ \text{kJ}
  4. D26.6 kJ26.6\ \text{kJ}

Correct answer

A. −52.11 kJ-52.11\ \text{kJ}

Step-by-step solution

The cell reaction is Cd(s)+Sn2+→Cd2++Sn(s)\text{Cd}(s) + \text{Sn}^{2+} \rightarrow \text{Cd}^{2+} + \text{Sn}(s).

The number of electrons transferred in the reaction is n=2n = 2.

The standard Gibbs free energy change is given by the relation:

ΔG0=−nFEcell0\Delta G^0 = -nFE^0_{\text{cell}}

Substituting the given values:

ΔG0=−2×96500 C mol−1×0.27 V\Delta G^0 = -2 \times 96500\ \text{C mol}^{-1} \times 0.27\ \text{V}

ΔG0=−52110 J mol−1\Delta G^0 = -52110\ \text{J mol}^{-1}

ΔG0=−52.11 kJ mol−1\Delta G^0 = -52.11\ \text{kJ mol}^{-1}

Answer: −52.11 kJ-52.11\ \text{kJ}
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