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Chemistry · 2014
JEE Main 2014 · 2014-04-06 Offline · Q48
Given below are the half - cell reactions: Mn^2++2 e^- → Mn; E^°=-1.18 V; 2( Mn^3++ e^- → Mn^2+); E^°=+1.51 V The E^° for 3 Mn^2+ → Mn +2 Mn^3+ will…
Given below are the half - cell reactions :
$$\begin{aligned}
& \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} ; \mathrm{E}^{\circ}=-1.18 \mathrm{~V} \\
& 2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}\right) ; \mathrm{E}^{\circ}=+1.51 \mathrm{~V}
\end{aligned}
$$The $\displaystyle \mathrm{E}^{\circ}$ for $\displaystyle 3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}$ will be :
Official answer
(1)
$\displaystyle 2.69$ V ; the reaction will not occur
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JEE Main 2014 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.