Chemistry - Electrochemistry Question with Solution | TestHub

The cell reaction is ${\mathrm{Cd}}_{(\mathrm{s})}+{\mathrm{Cu}}_{(\mathrm{aq})}^{+2}\to {\mathrm{Cd}}_{(\mathrm{aq})}^{+2}+{\mathrm{Cu}}_{(\mathrm{s})}$.

The Nernst equation for the cell is given by:

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

Given that the standard emf $E_{\mathrm{cell}}^{\circ }=0.74\mathrm{V}$ and the number of electrons transferred $n=2$.

When the concentration of both ${\mathrm{Cd}}^{+2}$ and ${\mathrm{Cu}}^{+2}$ decreases by 10 times, their new concentrations become $0.1\mathrm{M}$ each.

Substituting these values into the Nernst equation:

$E_{\mathrm{cell}}=0.74-\frac{0.0592}{2}{\log }_{10}\left(\frac{0.1}{0.1}\right)$

$E_{\mathrm{cell}}=0.74-\frac{0.0592}{2}{\log }_{10}(1)$

Since ${\log }_{10}(1)=0$, we get:

$E_{\mathrm{cell}}=0.74-0=+0.740\mathrm{V}$

Answer: $+0.740\mathrm{V}$