QJEEM22Chfc9b: JEE Main 2022 Chemistry - Class 12 JEEM22Chfc9b Solution: Cu(s) + Sn 2+ (0.001M) Cu 2+ (0.01M) + Sn(s) The Gibbs free…

Question

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Cu(s) + Sn²⁺ (0.001M) $\to$ Cu²⁺ (0.01M) + Sn(s)

The Gibbs free energy change for the above reaction at 298 K is x $\times$ 10^{$-$ 1} kJ mol^{$-$ 1}. The value of x is __________. [nearest integer]

[Given : $E_{C{u^{2 + }}/Cu}^\Theta = 0.34\,V$ ; $E_{S{n^{2 + }}/Sn}^\Theta = - 0.14\,V$ ; F = 96500 C mol^{$-$ 1}]

Accepted Answer

\begin{aligned}\mathrm{Cu} &amp;+\mathrm{Sn}^{+2} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Sn}(\mathrm{s}) \\\\\mathrm{E}_{\text {cell }}^{\circ} &amp;=\mathrm{E}_{\mathrm{OX}}^{\circ}+\mathrm{E}_{\text {Red }}^{\circ} \\\\&amp;=-0.34-0.14 \\\\&amp;=-0.48 \mathrm{~V} \\\\\mathrm{E}=&amp; ~\mathrm{E}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Sn}^{+2}\right]} \\\\=&amp;-0.48-0.0295 \log 10 \\\\=&amp;-0.5095 \mathrm{~V} \\\\\Delta \mathrm{G}=&amp;-\mathrm{nFE} \\\\&amp;=-2 \times 96500 \times-0.5095 \mathrm{~J} / \mathrm{mol} \\\\&amp;=98333.5 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \\\\&amp;=983.3 \times 10^{-1} \mathrm{~kJ} / \mathrm{mol} \\\\&amp;=983 \times 10^{-1} \mathrm{~kJ} / \mathrm{mol}\end{aligned}

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