Chemistry - Electrochemistry Question with Solution | TestHub

The reaction is $$\begin{array}{rl}& 3{\mathrm{Zn}}_{(\mathrm{s})}+2{\mathrm{Cr}}_{(0.1\mathrm{M})}^{3+}⟶3{\mathrm{Zn}}_{(0.1)}^{2+}+2{\mathrm{Cr}}_{(\mathrm{s})}\\ & {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^{\circ }-\frac{0.0592\mathrm{V}}{\mathrm{n}}{\log }_{10}\frac{{\left[{\mathrm{Zn}}^{2+}\right]}^3}{{\left[{\mathrm{Cr}}^{3+}\right]}^2}\end{array}$$ $\begin{array}{rl}\therefore {\mathrm{E}}_{\text{cell }}& =0.02\mathrm{V}-\frac{0.0592\mathrm{V}}{6}{\log }_{10}\frac{(0.1{)}^3}{(0.1{)}^2}\\ & =0.02\mathrm{V}-\frac{0.0592\mathrm{V}}{6}{\log }_{10}0.1\\ & =0.02\mathrm{V}+\frac{0.0592\mathrm{V}}{6}\times 1\\ & =0.02\mathrm{V}+0.0099\mathrm{V}\\ & =0.03\mathrm{V}\end{array}$