Given below are the half-cell reactions:Mn^{2 | vedclass

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(A) The given half-cell reactions are:$(i) \ Mn^{2+} + 2e^{-} \rightarrow Mn; E^{\circ}_{red} = -1.18 \ V$$(ii) \ Mn^{3+} + e^{-} \rightarrow Mn^{2+};

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$-2.69 \ V$; the reaction will not occur

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(A) The given half-cell reactions are:$(i) \ Mn^{2+} + 2e^{-} \rightarrow Mn; E^{\circ}_{red} = -1.18 \ V$$(ii) \ Mn^{3+} + e^{-} \rightarrow Mn^{2+}; E^{\circ}_{red} = +1.51 \ V$The target reaction is $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$.This reaction is the sum of the reduction of $Mn^{2+}$ to $Mn$ and the oxidation of $Mn^{2+}$ to $Mn^{3+}$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$Here,$Mn^{2+} \rightarrow Mn$ acts as the cathode $(E^{\circ} = -1.18 \ V)$ and $Mn^{2+} \rightarrow Mn^{3+} + e^{-}$ acts as the anode ($E^{\circ}_{ox} = -1.51 \ V$,so $E^{\circ}_{red} = +1.51 \ V$).$E^{\circ}_{cell} = (-1.18 \ V) - (+1.51 \ V) = -2.69 \ V$.Since $E^{\circ}_{cell}$ is negative,$\Delta G^{\circ} = -nFE^{\circ}_{cell}$ will be positive,meaning the reaction is non-spontaneous and will not occur.

Given below are the half-cell reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$
$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$
The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be:

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Given below are the half-cell reactions:$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be: