Question

Sol. [Key Concept: The Nernst plot has slope $- \frac{2.303}{n}$; the incorrect statement is that reversible electrical work equals charge divided by potential (it is charge $\times$ potential).]

Anode : $\left\lbrack {Red}_{1} \rightarrow {Ox}_{1}^{n_{1}^{+}} + n_{1}e^{-} \right\rbrack \times n_{2}$

Cathode : $\left\lbrack {Ox}_{2} + n_{2}e^{-} \rightarrow {Red}_{2}^{n_{2} -} \right\rbrack \times n_{1}$

Overall reaction :

$$n_{2}{Red}_{1} + n_{1}{Ox}_{2} \rightarrow n_{2}{Ox}_{1}^{n_{1}^{+}} + n_{1}{Red}_{2}^{n_{2}^{-}} $$The electrons does not appear in overall reaction

$${E = E^{o} - 2.303\frac{RT}{nF}logQ }{\frac{\left( E - E^{\circ} \right)}{RT/F} = - \frac{2.303}{n}logQ }$$Slope $= \frac{- 2.303}{n}$

Straight line passing through origin.

Electrical work $=$ Charge × Potential difference