JEE MAIN - Chemistry (2022 - 26th June Evening Shift - No. 21)
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]
Answer
983
Explanation
$$
\begin{aligned}
\mathrm{Cu} &+\mathrm{Sn}^{+2} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Sn}(\mathrm{s}) \\\\
\mathrm{E}_{\text {cell }}^{\circ} &=\mathrm{E}_{\mathrm{OX}}^{\circ}+\mathrm{E}_{\text {Red }}^{\circ} \\\\
&=-0.34-0.14 \\\\
&=-0.48 \mathrm{~V} \\\\
\mathrm{E}=& ~\mathrm{E}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Sn}^{+2}\right]} \\\\
=&-0.48-0.0295 \log 10 \\\\
=&-0.5095 \mathrm{~V} \\\\
\Delta \mathrm{G}=&-\mathrm{nFE} \\\\
&=-2 \times 96500 \times-0.5095 \mathrm{~J} / \mathrm{mol} \\\\
&=98333.5 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \\\\
&=983.3 \times 10^{-1} \mathrm{~kJ} / \mathrm{mol} \\\\
&=983 \times 10^{-1} \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
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