Question

Calculate the pressure of hydrogen passed through cathode half cell at $298K$.
$Pt\left(s\right)|H_2(g,800torr)|HCl\left(aq\right)|H_2(g,p_2)|Pt\left(s\right)$
Given:
$E_{cell}=0.018V{10}^{2.3}\approx 200$

Answer 1

From the cell representation,
Anode reaction:
$H_2\left(800torr\right)\to 2H^{+}+2e^{-}$
Cathode reaction:
$2H^{+}+2e^{-}\to H_2\left(p_2\right)$
Thus, the cell reaction is
$H_2\left(800torr\right)\to H_2\left(p_2\right)$

By Nernst equation,
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{p_2}{800}$
$E_{cell}^0=0$
$E_{cell}=-\frac{0.0591}{2}log\frac{p_2}{800}$

$0.018=\frac{0.0591}{2}log\frac{800}{p_2}$

$0.61=log\frac{800}{p_2}$

$0.61=log800-log{p}_2$

$0.61=2.90-log{p}_2$

$2.3\approx log{p}_2$

$p_2\approx 200torr$