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(B) The cell potential $E^{\circ}_{cell}$ is calculated as $E^{\circ}_{cathode} - E^{\circ}_{anode}$.To maximize $E^{\circ}_{cell}$,we need the highes

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$(i)$ and $(iv)$

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(B) The cell potential $E^{\circ}_{cell}$ is calculated as $E^{\circ}_{cathode} - E^{\circ}_{anode}$.To maximize $E^{\circ}_{cell}$,we need the highest possible $E^{\circ}_{cathode}$ (reduction potential) and the lowest possible $E^{\circ}_{anode}$ (oxidation potential).First,convert all given values to standard reduction potentials $(E^{\circ}_{red})$:$(i) \, A^{3-}$ $\rightarrow A^{2-} + e^{-}; E^{\circ}_{ox} = 1.5 \, V$ $\Rightarrow E^{\circ}_{red} = -1.5 \, V$$(ii) \, B^{+} + e^{-} \rightarrow B; E^{\circ}_{red} = 0.5 \, V$$(iii) \, C^{2+} + e^{-} \rightarrow C^{+}; E^{\circ}_{red} = 0.5 \, V$$(iv) \, D$ $\rightarrow D^{2+} + 2e^{-}; E^{\circ}_{ox} = -1.15 \, V$ $\Rightarrow E^{\circ}_{red} = 1.15 \, V$Comparing the reduction potentials,the highest is $E^{\circ}_{red} = 0.5 \, V$ (from $ii$ or $iii$) and the lowest is $E^{\circ}_{red} = -1.5 \, V$ (from $i$).However,the question asks for the combination of two half-cells. Using $(i)$ as the anode $(E^{\circ}_{ox} = 1.5 \, V)$ and $(ii)$ or $(iii)$ as the cathode $(E^{\circ}_{red} = 0.5 \, V)$,$E^{\circ}_{cell} = 0.5 - (-1.5) = 2.0 \, V$.Using $(iv)$ as the anode $(E^{\circ}_{ox} = -1.15 \, V)$ and $(ii)$ or $(iii)$ as the cathode $(E^{\circ}_{red} = 0.5 \, V)$,$E^{\circ}_{cell} = 0.5 - (-1.15) = 1.65 \, V$.Using $(i)$ as the anode $(E^{\circ}_{ox} = 1.5 \, V)$ and $(iv)$ as the cathode $(E^{\circ}_{red} = 1.15 \, V)$,$E^{\circ}_{cell} = 1.15 - (-1.5) = 2.65 \, V$.Thus,the combination of $(i)$ and $(iv)$ gives the maximum potential.

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