Chemistry - Electrochemistry Question with Solution | TestHub

$$\begin{array}{rl}& \text{ As }{\mathrm{E}}_{\mathrm{Cr}/{\mathrm{cr}}^{3+}}^0=-0.72\mathrm{V}\text{ and }{\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^0=-0.42\mathrm{V}\\ & 2\mathrm{Cr}+3{\mathrm{Fe}}^{2+}⟶3\mathrm{Fe}+2{\mathrm{Cr}}^{3+}\\ & {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^0-\frac{0.0591}{6}\log \frac{{\left({\mathrm{Cr}}^{3+}\right)}^2}{{\left({\mathrm{Fe}}^{2+}\right)}^3}\\ & =(-0.42+0.72)-\frac{0.0591}{6}\log \frac{(0.1{)}^2}{(0.01{)}^3}=0.30-\frac{0.0591}{6}\log \frac{(0.1{)}^2}{(0.01{)}^3}\\ & =0.30-\frac{0.0591}{6}\log \frac{10^{-2}}{10^{-6}}=0.30-\frac{0.0591}{6}\log 10^4\\ & {\mathrm{E}}_{\text{cell }}=0.2606\mathrm{V}\end{array}$$