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(D) The standard Gibbs free energy change $\Delta G^0$ is related to the standard electrode potential $E^0$ by the equation $\Delta G^0 = -nFE^0$.For

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$2$

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(D) The standard Gibbs free energy change $\Delta G^0$ is related to the standard electrode potential $E^0$ by the equation $\Delta G^0 = -nFE^0$.For the reaction $Pb^{2+} + 2e^{-} \rightarrow Pb$,$\Delta G^0_1 = -2Fm$.For the reaction $Pb^{4+} + 4e^{-} \rightarrow Pb$,$\Delta G^0_2 = -4Fn$.We want the potential for $Pb^{2+} \rightarrow Pb^{4+} + 2e^{-}$,which is the reverse of the second reaction added to the first.$\Delta G^0_3 = \Delta G^0_1 - \Delta G^0_2 = -2Fm - (-4Fn) = -2Fm + 4Fn$.Since $\Delta G^0_3 = -2FE^0_{(Pb^{2+}/Pb^{4+})}$,we have $-2FE^0_{(Pb^{2+}/Pb^{4+})} = -2Fm + 4Fn$.Dividing by $-2F$,we get $E^0_{(Pb^{2+}/Pb^{4+})} = m - 2n$.Comparing this with $m - xn$,we find $x = 2$.

In an electrochemical reaction of lead,at standard temperature,if $E^0_{(Pb^{2+}/Pb)} = m \ V$ and $E^0_{(Pb^{4+}/Pb)} = n \ V$,then the value of $E^0_{(Pb^{2+}/Pb^{4+})}$ is given by $m - x n$. The value of $x$ is $...........$. (Nearest integer)

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