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Chemistry · 2020

JEE Main · 2 September 2020, Shift 1 · Q47

The Gibbs energy change (in J) for the given reaction at [ Cu^2+]=[ Sn^2+]=1 M and 298 K is: Cu ( s )+ Sn^2+( aq.) → Cu^2+( aq.)+ Sn ( s ); ( E_Sn^2+…

The Gibbs energy change (in J) for the given reaction at $\displaystyle \left[\mathrm{Cu}^{2+}\right]=\left[\mathrm{Sn}^{2+}\right]=1 \mathrm{M}$ and $\displaystyle 298$ $\displaystyle K$ is : $$\begin{aligned} & \mathrm{Cu}(\mathrm{~s})+\mathrm{Sn}^{2+}(\mathrm{aq} .) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq} .)+\mathrm{Sn}(\mathrm{~s}) ; \\ & \left(\mathrm{E}_{\mathrm{Sn}^{2+} \mid \mathrm{Sn}}^0=-0.16 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} \mid \mathrm{Cu}}^0=0.34 \mathrm{~V},\right. \end{aligned} $$ Take $\displaystyle \mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}$ )
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JEE Main 2020 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.