Question

An oxidation-reduction reaction in which $3$ electrons are transferred has a $∆{\mathrm{G}}^0$ of $17.37{\mathrm{KJmol}}^{-1}$ at ${25}^0\mathrm{C}$. The value of ${{\mathrm{E}}^0}_{\mathrm{cell}}$ (in $\mathrm{V}$) is ______ $\times {10}^{-2}$. $(1\mathrm{F}=96500{\mathrm{Cmol}}^{-1})$

Answer 1

Step 1: Given Data

$∆{\mathrm{G}}^0$ is $17.37{\mathrm{KJmol}}^{-1}$

$1\mathrm{F}=96500{\mathrm{Cmol}}^{-1}$

Step 2: $∆{\mathrm{G}}^0=-{{\mathrm{nFE}}^0}_{\mathrm{cell}}$

$\mathrm{n}$ is the number of electrons $\left(\mathrm{here}\mathrm{n}=3\right)$

$17.37\times 1000=-3\times 96500\times {{\mathrm{E}}^0}_{\mathrm{cell}}$

${{\mathrm{E}}^0}_{\mathrm{cell}}=\frac{17.37\times 1000}{-3\times 96500}\mathrm{V}$

${{\mathrm{E}}^0}_{\mathrm{cell}}=-6\times {10}^{-2}\mathrm{V}$

Hence the answer is $\mathbf{-}\mathbf{6}\times {10}^{-2}\mathrm{V}$.