Question

From the following $E^{^\circ }$ values of half-cells;
i. $A+e \rightarrow A^{-};E^\circ =-0.24\,V$
ii. $B^{-}+e \rightarrow B^{2 -}:E^\circ =+1.25\,V$
iii. $C^{-}+2e \rightarrow C^{3};E^{0}=-1.25\,V$
iv. $D+2e \rightarrow D^{2 -};E^{0}=+0.68\,V$
If you were to construct a cell using combination of two half-cells from above that gives the largest cell potential, then value of the largest cell potential would be $\frac{x}{2}$ $V.$ Value of x will be.

Answer 1

Larger the difference between the electrode potentials, greater is the cell potential. Hence, combination of half-cells $\left(i i\right)$ and $\left(i i i\right)$ will give the largest cell potential.
$E_{cll}^\circ =E_{calsade}^\circ -E_{rmak}^\circ =1.25-\left(\right.-1.25\left.\right)$
$=+2.50\,V=\frac{5}{2}V$