Calculate $\mathrm{E}_{\text {cell }}^{\circ}$ if the equilibrium constant for following reaction is $1.2…

Calculate $\mathrm{E}_{\text {cell }}^{\circ}$ if the equilibrium constant for following reaction is $1.2 \times 10^6$. $2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{++}+\mathrm{Cu}_{(\mathrm{s})}$
  1. $0.36 \mathrm{~V}$
  2. $-0.36 \mathrm{~V}$
  3. $-0.18 \mathrm{~V}$
  4. $0.18 \mathrm{~V}$

Solution

$\begin{aligned} & \mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K} \text { at } 298 \mathrm{~K} \\ & \mathrm{Cu}_{(\mathrm{mq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \\ & \frac{\mathrm{Cu}_{\text {(aq) }}^{+} \longrightarrow \mathrm{Cu}_{\text {(aq) }}^{2+}+\mathrm{e}^{-}}{2 \mathrm{Cu}_{\text {(aq) }}^{+} \longrightarrow \mathrm{Cu}_{(\text {aq) }}^{2+}+\mathrm{Cu}_{(\mathrm{s})}} \\ & \therefore \quad \mathrm{n}=1 \\ & \therefore \quad \mathrm{E}_{\text {eell }}^0=\frac{0.0592}{1} \log _{10}\left(1.2 \times 10^6\right) \\ & =0.0592\left(\log 1.2+\log 10^6\right) \\ & =0.0592(0.079+6) \\ & =0.0592 \times 6.079 \\ & =0.36 \mathrm{~V} \\ & \end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

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