Given $\mathrm{E}_{\mathrm{Cr}^3+/ \mathrm{Cr}}^{\circ}=-0.72 \mathrm{~V}, \mathrm{E}_{\mathrm{Fe}^{2+/ /…

Given $\mathrm{E}_{\mathrm{Cr}^3+/ \mathrm{Cr}}^{\circ}=-0.72 \mathrm{~V}, \mathrm{E}_{\mathrm{Fe}^{2+/ / \mathrm{Fe}}}^{\circ}=-0.42 \mathrm{~V}$. The potential for the cell $\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M})\right|\left|\mathrm{Fe}^{2+}(0.01 \mathrm{M})\right| \mathrm{Fe}$ is
  1. $0.26 \mathrm{~V}$
  2. $0.399 \mathrm{~V}$
  3. $-0.339 \mathrm{~V}$
  4. $-0.26 \mathrm{~V}$

Solution

$ \begin{aligned} & \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+} \longrightarrow 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{aligned} $

Asked in: JEE Main 2008

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