Chemistry · 2013
NEET 2013 · 18 May · Code Y · Q169
Consider the half-cell reduction reaction: Mn^2++2 e^- → Mn, E^0=-1.18 V; Mn^2+ → Mn^3++ e^-, E^0=-1.51 V The E^0 for the reaction 3 Mn^2+ → Mn^0+2…
Consider the half-cell reduction reaction :
$$\begin{aligned}
& \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}, \mathrm{E}^0=-1.18 \mathrm{~V} \\
& \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+}+\mathrm{e}^{-}, \mathrm{E}^0=-1.51 \mathrm{~V}
\end{aligned}
$$
The $\displaystyle \mathrm{E}^0$ for the reaction $\displaystyle 3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^0+2 \mathrm{Mn}^{3+}$, and possibility of the forward reaction are, respectively :
Official answer
(2)
-$\displaystyle 2.69$ V and no
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NEET (UG) 2013 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.