In which of the following Galvanic cells emf is maximum? (Given: $\mathrm{E}_{\mathrm{Mg}^{2+} \mid…

In which of the following Galvanic cells emf is maximum? (Given: $\mathrm{E}_{\mathrm{Mg}^{2+} \mid \mathrm{Mg}}^0=-2.36 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Cl}_2 \mid 2 \mathrm{Cl}^{-}}^0=+1.36 \mathrm{~V}$)
  1. $\mathrm{Mg}\left|\mathrm{Mg}^{2+}(1 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(1 \mathrm{M})\right| \mathrm{Cl}_2(1 \mathrm{~atm}), \mathrm{Pt}$
  2. $\mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(1 \mathrm{M})\right| \mathrm{Cl}_2(1 \mathrm{~atm}), \mathrm{Pt}$
  3. $\mathrm{Mg}\left|\mathrm{Mg}^{2+}(1 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(0.01 \mathrm{M})\right| \mathrm{Cl}_2(1 \mathrm{~atm}), \mathrm{Pt}$
  4. $\mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(0.01 \mathrm{M})\right| \mathrm{Cl}_2(1 \mathrm{~atm}), \mathrm{Pt}$

Solution

$\begin{gathered} \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \\ \mathrm{Cl}_2 \rightarrow+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \\ \hline \mathrm{Mg}+\mathrm{Cl}_2 \rightarrow \mathrm{Mg}^{2+} 2 \mathrm{Cl}^{-} \\ \hline \end{gathered}$ (a) $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{MgH}_2\right]}{\left[\mathrm{Cl}^{-}\right]^2} \\ & \mathrm{E}_{\text {cell }}=3.72-\frac{\mathrm{RT}}{\mathrm{nF}} \ln (1)^2 \times 1 \\ & \mathrm{E}_{\text {cell }}=3.72-\frac{\mathrm{RT}}{\mathrm{nF}} \ln 1 \end{aligned}$ (b) $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln (1)^2 \times 10^{-2} \\ & =\mathrm{E}^{\circ}+2 \frac{\mathrm{RT}}{\mathrm{nF}} \ln 10 \end{aligned}$ (c) $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \left(10^{-2}\right)^2 \times 1 \\ & =\mathrm{E}^{\circ}+4 \frac{\mathrm{RT}}{\mathrm{nF}} \ln 10 \end{aligned}$ (d) $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nf}} \ln \left(10^{-2}\right)^2 \times 10^{-2} \\ & =\mathrm{E}^{\circ}=6 \frac{\mathrm{RT}}{\mathrm{nF}} \ln 10 \end{aligned}$ $\therefore \quad$ So 'd' option have highest emf.

Asked in: AP EAMCET 2024 (22 May Shift 1)

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