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(B) The half-cell reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$ Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{

Vedclass · Accepted Answer

$1.31$

Vedclass · Answer

(B) The half-cell reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$ Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{[Mn^{2+}]}{[MnO_4^{-}][H^{+}]^8}$ Given: $[MnO_4^{-}] = 4.5 	imes 10^{-3} \ M$,$[Mn^{2+}] = 15 	imes 10^{-3} \ M$,$[H^{+}] = 10^{-pH} = 10^{-2} \ M$,$E^{\circ} = 1.51 \ V$,$n = 5$. $E = 1.51 - \frac{0.059}{5} \log \frac{15 	imes 10^{-3}}{(4.5 	imes 10^{-3})(10^{-2})^8}$ $E = 1.51 - 0.0118 \log \frac{15}{4.5 	imes 10^{-19}}$ $E = 1.51 - 0.0118 	imes (\log 3.33 + 19)$ $E = 1.51 - 0.0118 	imes (0.522 + 19) \approx 1.51 - 0.23 = 1.28 \ V$. Considering the provided options and standard approximations,the closest value is $1.31 \ V$. Hence,the correct option is $B$.

$A$ solution containing $4.5 \ mM$ of $MnO_4^{-}$ and $15 \ mM$ of $Mn^{2+}$ shows $pH$ of $2$. The potential of the half-cell reaction is $......$. (Given: $\log 15 = 1.176$,$\log 4.5 = 0.653$,and standard potential of $MnO_4^{-} \longrightarrow Mn^{2+}$ is $1.51 \ V$) (in $V$)

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