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← IIT-JEE 2012 · Paper 2
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IIT-JEE 2012 · Paper 2, Chemistry question 32

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The electrochemical cell shown below is a concentration cell.

M | MX2+\ce{M^{2+}} (saturated solution of a sparingly soluble salt, MXX2\ce{MX2}) || MX2+\ce{M^{2+}} (0.001 mol dm⁻³) | M

The emf of the cell depends on the difference in concentrations of MX2+\ce{M^{2+}} ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

The value of ΔG\Delta G (kJ mol⁻¹) for the given cell is (take 1 F = 96500 C mol⁻¹)

Choose one option
Answer and worked solution

Official answer: D

Free energy. n=2n=2:
ΔG=−nFE=−2×96500×0.059 J mol−1=−11.4 kJ mol−1.\Delta G=-nFE=-2\times96500\times0.059\ \mathrm{J\,mol^{-1}}=-11.4\ \mathrm{kJ\,mol^{-1}}.
It is negative, as for any cell with positive emf.

(D)

Topics:Gibbs energy and EMF
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