From the following {E^o} values of half cells | vedclass

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(B) The cell potential ${E^o}_{cell}$ is calculated as ${E^o}_{cathode} - {E^o}_{anode}$.To obtain the largest potential,we must choose the half-cell

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$II, III$

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(B) The cell potential ${E^o}_{cell}$ is calculated as ${E^o}_{cathode} - {E^o}_{anode}$.To obtain the largest potential,we must choose the half-cell with the highest ${E^o}$ value as the cathode and the half-cell with the lowest ${E^o}$ value as the anode.Comparing the given values:$I: +0.24 \ V$$II: +1.25 \ V$$III: +0.15 \ V$$IV: +0.68 \ V$The highest value is $II$ $(+1.25 \ V)$ and the lowest value is $III$ $(+0.15 \ V)$.Therefore,the combination of $II$ and $III$ will result in the largest potential: ${E^o}_{cell} = 1.25 - 0.15 = 1.10 \ V$.

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