Chemistry - Electrochemistry Question with Solution | TestHub

The cell reaction is:

${\mathrm{Zn}}_{(s)}+2{\mathrm{Ag}}_{(aq)}^{+}\to {\mathrm{Zn}}_{(aq)}^{2+}+2{\mathrm{Ag}}_{(s)}$

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

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

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

Initially, $[{\mathrm{Zn}}^{2+}]=1\mathrm{M}$ and $[{\mathrm{Ag}}^{+}]=1\mathrm{M}$.

$E_1=E_{\mathrm{cell}}^{\circ }-\frac{0.0592}{2}{\log }_{10}(1)=E_{\mathrm{cell}}^{\circ }$

When $[{\mathrm{Zn}}^{2+}]$ decreases to $0.1\mathrm{M}$:

$E_2=E_{\mathrm{cell}}^{\circ }-\frac{0.0592}{2}{\log }_{10}\left(\frac{0.1}{1^2}\right)$

$E_2=E_{\mathrm{cell}}^{\circ }-\frac{0.0592}{2}{\log }_{10}(10^{-1})$

$E_2=E_{\mathrm{cell}}^{\circ }-\frac{0.0592}{2}(-1)$

$E_2=E_{\mathrm{cell}}^{\circ }+0.0296\mathrm{V}$

Change in emf $=E_2-E_1=0.0296\mathrm{V}$

Therefore, the emf of the cell increases by $0.0296\mathrm{V}$.

Answer: increases by 0.0296 V