Chemistry - Electrochemistry Question with Solution | TestHub

The reduction half-reaction is ${\text{Sn}}^{2+}+2e^{-}\to \text{Sn(s)}$.

Using the Nernst equation:

$E=E^{\circ }-\frac{0.0591}{n}\log \frac{1}{[{\text{Sn}}^{2+}]}$

Substituting the given values ($n=2$, $[{\text{Sn}}^{2+}]=0.01\text{ M}$, $E^{\circ }=-0.136\text{ V}$):

$E=-0.136-\frac{0.0591}{2}\log \frac{1}{0.01}$

$E=-0.136-0.02955\log (100)$

$E=-0.136-0.02955\times 2$

$E=-0.136-0.0591$

$E=-0.1951\text{ V}\approx -0.195\text{ V}$

Answer: $-0.195$ V