Chemistry - Electrochemistry Question with Solution | TestHub

The given cell reaction is:

${\mathrm{Cd}}_{(\mathrm{s})}+{\mathrm{Cu}}_{(\mathrm{aq})}^{+2}⟶{\mathrm{Cd}}_{(\mathrm{aq})}^{+2}+{\mathrm{Cu}}_{(\mathrm{s})}$

The number of electrons transferred in the reaction is $n=2$.

According to the Nernst equation at $298\text{ K}$:

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0592}{n}{\log }_{10}\frac{[{\mathrm{Cd}}^{+2}]}{[{\mathrm{Cu}}^{+2}]}$

Given that the concentration of ${\mathrm{Cd}}^{+2}$ is 10 times greater than the concentration of ${\mathrm{Cu}}^{+2}$, we have $\frac{[{\mathrm{Cd}}^{+2}]}{[{\mathrm{Cu}}^{+2}]}=10$.

Substituting the values into the equation:

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0592}{2}{\log }_{10}(10)$

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-0.0296\times 1$

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-0.0296\text{ V}$

Thus, the emf of the cell is less than $E_{\text{cell}}^{\circ }$ by $0.0296\text{ V}$.

Answer: less than ${\mathrm{E}}^{\circ }{}_{\text{cell }}$ by 0.0296 V