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(D) The cell reaction is: $Mg(s) + Cl_2(g) \rightarrow Mg^{2+}(aq) + 2 Cl^{-}(aq)$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.36

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$Mg \mid Mg^{2+}(0.01 \ M) \parallel 2 Cl^{-}(0.01 \ M) \mid Cl_2(1 \ atm), Pt$

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(D) The cell reaction is: $Mg(s) + Cl_2(g) \rightarrow Mg^{2+}(aq) + 2 Cl^{-}(aq)$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.36 - (-2.36) = 3.72 \ V$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$,where $Q = [Mg^{2+}][Cl^{-}]^2$.For option $A$: $Q = (1)(1)^2 = 1$,$E_{cell} = 3.72 - 0 = 3.72 \ V$.For option $B$: $Q = (0.01)(1)^2 = 10^{-2}$,$E_{cell} = 3.72 - \frac{0.0591}{2} \log(10^{-2}) = 3.72 + 0.0591 = 3.7791 \ V$.For option $C$: $Q = (1)(0.01)^2 = 10^{-4}$,$E_{cell} = 3.72 - \frac{0.0591}{2} \log(10^{-4}) = 3.72 + 0.1182 = 3.8382 \ V$.For option $D$: $Q = (0.01)(0.01)^2 = 10^{-6}$,$E_{cell} = 3.72 - \frac{0.0591}{2} \log(10^{-6}) = 3.72 + 0.1773 = 3.8973 \ V$.Since $Q$ is smallest for option $D$,the $emf$ is maximum.

In which of the following Galvanic cells is the $emf$ maximum? (Given: $E_{Mg^{2+} \mid Mg}^0 = -2.36 \ V$ and $E_{Cl_2 \mid 2 Cl^{-}}^0 = +1.36 \ V$)

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