The cell potential for the following cell notation is approximately $\begin{aligned} &…

The cell potential for the following cell notation is approximately $\begin{aligned} & \mathrm{M}(\mathrm{s})\left|\mathrm{M}^{3+}(\mathrm{aq}, 0.01 \mathrm{M}) \| \mathrm{N}^{2+}(\mathrm{aq}, 0.1 \mathrm{M})\right| \mathrm{N}(\mathrm{s}) \\ & \mathrm{E}_{\mathrm{M}^{3+} / \mathrm{M}}^0=0.6 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{N}^{2+} / \mathrm{N}}^0=0.1 \mathrm{~V}\end{aligned}$
  1. $0.51 \mathrm{~V}$
  2. $1.5 \mathrm{~V}$
  3. $2.0 \mathrm{~V}$
  4. $2.5 \mathrm{~V}$

Solution

$\left(\mathrm{M} \rightarrow \mathrm{M}^{3+}+3 \mathrm{e}\right) \times 2 ;\left(\mathrm{N}^{2+}+2 \mathrm{e} \rightarrow \mathrm{N}\right) \times 3$ Overall reaction $: 2 \mathrm{M}(\mathrm{s})+3 \mathrm{~N}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{M}^{3+}(\mathrm{aq})+3 \mathrm{~N}(\mathrm{~s})$ $\therefore \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.059}{6} \log \frac{\left[\mathrm{M}^{3+}\right]^2}{\left[\mathrm{~N}^{2+}\right]^3}$ $\begin{aligned} & =-0.6+0.1-\frac{0.059}{6} \log \frac{\left(10^{-2}\right)^2}{\left(10^{-1}\right)^3} \\ & =-0.5-\frac{0.059}{6} \log 10^{-1}=0.51 \mathrm{~V}\end{aligned}$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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