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MHT CET 202621 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
21 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

MHT CET 2026 · 21 April · Evening Shift · Chemistry · Electrochemistry

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

Calculate the standard Gibbs energy for the following cell reaction,
A(s)+\textB2+(aq)→\textA2+(aq)+B(s)\text{A(s)} + \textB^{2+}\text{(aq)} \rightarrow \textA^{2+}\text{(aq)} + \text{B(s)}, if E^{}^{\circ}_{cell} is 0.556 V0.556\ \text{V}.

Options

  1. A−53.65 kJ-53.65\ \text{kJ}
  2. B53.65 kJ53.65\ \text{kJ}
  3. C107.31 kJ107.31\ \text{kJ}
  4. D−107.31 kJ-107.31\ \text{kJ}

Correct answer

D. −107.31 kJ-107.31\ \text{kJ}

Step-by-step solution

The given cell reaction is A(s)+\textB2+(aq)→\textA2+(aq)+B(s)\text{A(s)} + \textB^{2+}\text{(aq)} \rightarrow \textA^{2+}\text{(aq)} + \text{B(s)}.

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

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

\Delta G^{}^{\circ} = -n F E^{}^{\circ}_{cell}

Substituting the given values (F=96500 C mol−1F = 96500\ \text{C mol}^{-1} and E^{}^{\circ}_{cell} = 0.556\ \text{V}):

\Delta G^{}^{\circ} = -2 \times 96500 \times 0.556

\Delta G^{}^{\circ} = -107308\ \text{J mol}^{-1}

\Delta G^{}^{\circ} \approx -107.31\ \text{kJ mol}^{-1}

Answer: −107.31 kJ-107.31\ \text{kJ}
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