The electrochemical cell shown below is a concentration cell. M | M^2+ (saturated solution of a sparingly soluble salt, MX_2 ) || M^2+ (0.001 mol dm–3) | M The emf of the cell depends on the difference in concentrations of M^2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. The value of ∆G (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1)

Electrochemistry PYQ — JEE Advanced 2012 · Shift 2 · Q2 with Solution

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Correct answer: D

Use the relation between Gibbs free energy and emf

For an electrochemical cell,

\Delta G = -n F E_{\text{cell}}

where:

$n$ = number of electrons transferred
$F = 96500\ \text{C mol}^{-1}$
$E_{\text{cell}} = 0.059\ \text{V}$

Find $n$

The electrode process is:

M^{2+} + 2e^- \rightleftharpoons M

So, the number of electrons transferred is

n=2

Calculate $\Delta G$

\Delta G = -(2)(96500)(0.059)

First compute:

96500 \times 0.059 = 5693.5

Then,

\Delta G = -2 \times 5693.5 = -11387\ \text{J mol}^{-1}

Convert to kJ mol $^{-1}$

\Delta G = -11.387\ \text{kJ mol}^{-1}

Rounding,

\Delta G \approx -11.4\ \text{kJ mol}^{-1}

Match with the options

The correct option is:

\boxed{\text{D: } -11.4}

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