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  • Given below are half cell reactions: <img alt="\begin{aligned} &\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \\ \end{al

Given below are half cell reactions:
\begin{aligned} &\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \\ \end{aligned}

\begin{aligned} &\mathrm{E}_{\mathrm{Mn}^{2+}} /\mathrm{MnO}_{4}=-1.510 \mathrm{~V} \\ \end{aligned}

\begin{aligned} &\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} ,\\ \end{aligned}

\mathrm{E}_{\mathrm{_{O2}}}/\mathrm{H_{2}O}=1.223 \mathrm{~V}
Will the permanganate ion,\mathrm{MnO}_{4}^{-} liberate \mathrm{O}_{2} from Water in the presence of an acid?
 

Option: 1

No, because \mathrm{E^{0}}_{\text {cell }}=-0.287 \mathrm{~V}
 


Option: 2

Yes, because \mathrm{E^{o}}_{\text {cell }}=+2.733 \mathrm{~V}
 


Option: 3

No, because \mathrm{E^{o}}_{\text {cell }} =-2.733 \mathrm{~V}


Option: 4

Yes, because \mathrm{E^{0}}_{\text {cell }}=+0.287 \mathrm{~V}


Answers (1)

best_answer

The \mathrm{E^{0}} values corresponding to the reactions are given below:

\mathrm{MnO}^{-}+8 \mathrm{H}^{+}+\mathrm{Se}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, E^{0}=1.510 \\

\mathrm{H_{2}O} \longrightarrow \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 e^{-}, E^{0}=-1.223 \\

\therefore \mathrm{E}^{\circ} \mathrm{Cell} =(1.570-1.223) \mathrm{V} \\

                  =0.287 \mathrm{~V}

Thus, \mathrm{MnO_{4}^{-}} will be able to liberate \mathrm{O_{2}} from water in the presence of \mathrm{H^{+}}

Hence correct option is 4

Posted by

Devendra Khairwa

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