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IIT JEE Advanced 2012 Paper 2

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The electrochemical cell shownbelow is a concentration cell.

\mathrm{M}|\mathrm{M}^{2+}

(saturated solution of a sparingly soluble salt,

\mathrm{MX}_2

) ||

\mathrm{M}^{2+}

(0.001

\mathrm{mol\,dm}^{-3}

)|M

The emf of the cell depends on the difference in concentrations of

\mathrm{M}^{2+}

ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

Section: Chemistry

Question : 31 of 60

Marks: +1, -0

The value of ΔG

(\mathrm{kJ}\,\mathrm{mol}^{-1})

\mathrm{C}\,\mathrm{mol}^{-1}

Solution:

At the anode: M →

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

At the cathode:

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

→ M.

We know that ΔG =

-nFE_{\text{cell}}

Given that

E_{\text{cell}}

= 0.059 V, 1 F = 96,500 C and n = number of electrons = 2.

Substituting, we get

ΔG = - 2 × 96500 × 0.059 = - 113873

\mathrm{J}\,\mathrm{mol}^{-1}

= - 11.387

\mathrm{kJ}\,\mathrm{mol}^{-1}

~ - 11.4

\mathrm{kJ}\,\mathrm{mol}^{-1}

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