Question

When two half-cells of electrode potential of $E_{1}^{o}$ and $E_{2}^{o}$ are combined to form a cell of electrode potential $E_{3}^{o}\left(\right.n_{1},n_{2},n_{3}$ are the number of electrodes exchanged in first, second and third half-cell reactions), then

• A. $E_{3}^{o}=E_{1}^{o}+E_{2}^{o}$
• B. $E_{3}^{o}=E_{2}^{o}-E_{1}^{o}$
• C. $E_{3}^{o}=\frac{n_{1} E_{1}^{o} + n_{2} E_{2}^{o}}{n_{3}}$
• D. $E_{3}^{o}=\frac{n_{1} E_{1}^{o} - n_{2} E_{2}^{o}}{n_{3}^{2}}$

Answer 1

Answer: (C)

As $E^\circ $ is non additive but $\Delta G^\circ $ is additive

$\Delta G_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}$

$-n_{3}FE_{3}^{o}=-nFE_{1}^{o}-n_{2}FE_{2}^{o}$

$E_{3}^{o}=\frac{n_{1} E_{1}^{o} + n_{2} E_{2}^{o}}{n_{3}}$