Chemistry - Electrochemistry Question with Solution | TestHub

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

The number of electrons transferred is $n=2$.

Using the Nernst equation at 298 K:

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

Substituting the given concentrations $[{\mathrm{Ni}}^{2+}]=0.01\text{ M}$ and $[{\mathrm{Ag}}^{+}]=0.01\text{ M}$:

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0592}{2}{\log }_{10}\frac{0.01}{(0.01{)}^2}$

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-0.0296{\log }_{10}(100)$

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-0.0296\times 2$

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-0.0592\text{ V}$

Thus, the cell emf is less than $E_{\text{cell}}^{\circ }$ by $0.0592\text{ V}$.

Answer: less than $E^{\circ }{}_{\text{cell }}$ by 0.0592 V