Question

Calculate emf of the following cell at ${25}^{\circ }C$:
$Sn|S{n}^{2+}\left(0.001M\right)||H^{+}\left(0.01M\right)|H_2$ (g)(1 bar)|Pt(s)
$E_{S{n}^{2+}/Sn}^0=-0.14V{E}_{H+/H_2}^0=0.00V$

Answer 1

Cell reaction,
$Sn+2H^{+}\to S{n}^{2+}+H_2;n=2$

$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{\left[S{n}^{2+}\right]}{[H^{+}{]}^2}$

$=0-\left(0.14\right)-\frac{0.0591}{2}log\frac{0.001}{(0.01{)}^2}$

$=0.14-\frac{0.0591}{2}log10=0.14-\frac{0.0591}{2}$

$=0.14-\frac{0.0591}{2}=0.14-0.02955=0.11056V$