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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 :
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JEE Main 2014 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.