Given below are half-cell reactions:MnO_{4}^{ | vedclass

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(D) For the reaction to be spontaneous,the cell potential $E_{cell}^{o}$ must be positive.Reduction (Cathode): $2MnO_{4}^{-} + 16H^{+} + 10e^{-} \righ

Vedclass · Accepted Answer

Yes,because $E_{cell}^{o} = +0.287 \, V$

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(D) For the reaction to be spontaneous,the cell potential $E_{cell}^{o}$ must be positive.Reduction (Cathode): $2MnO_{4}^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_{2}O$ $(E^{o} = +1.510 \, V)$Oxidation (Anode): $5H_{2}O \rightarrow \frac{5}{2} O_{2} + 10H^{+} + 10e^{-}$ $(E^{o} = +1.223 \, V)$Overall reaction: $2MnO_{4}^{-} + 6H^{+} \rightarrow 2Mn^{2+} + \frac{5}{2} O_{2} + 3H_{2}O$$E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$$E_{cell}^{o} = 1.510 \, V - 1.223 \, V = +0.287 \, V$Since $E_{cell}^{o} > 0$,the reaction is spontaneous,and $MnO_{4}^{-}$ will liberate $O_{2}$ from water in the presence of an acid.

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+} \| Cu$	$(II)$ $0.059$
$(C)$ $Zn \| Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303 RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

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