Question

Calculate the emf of the concentration cell at $298K$.
$Pt\left(s\right)|H_2\left(g\right)\left(p_1\right)|HCl\left(aq\right)|H_2\left(g\right)\left(p_2\right)|Pt\left(s\right)$

$p_1=600torr{p}_2=400torr$

Take:
$log\frac{3}{2}=0.176$

• A. $0.005V$
• B. $0.025V$
• C. $0.15V$
• D. $1.02V$

Answer 1

The correct option is A $0.005V$

From the cell representation,
Anode reaction:
$H_2\left(p_1\right)\to 2H^{+}+2e^{-}$
Cathode reaction:
$2H^{+}+2e^{-}\to H_2\left(p_2\right)$
Thus, the cell reaction is
$H_2\left(p_1\right)\to H_2\left(p_2\right)$

By Nernst equation,
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{p_2}{p_1}$
$E_{cell}^0=0$
$E_{cell}=-\frac{0.0591}{2}log\frac{p_2}{p_1}$

$E_{cell}=\frac{0.0591}{2}log\frac{p_1}{p_2}$

$E_{cell}=\frac{0.0591}{2}log\frac{600}{400}$

$E_{cell}=\frac{0.0591}{2}log\frac{3}{2}$

$E_{cell}=\frac{0.0591}{2}\times 0.176$

$E_{cell}=5.2\times {10}^{-3}V$