Question

Which of the following concentration cells with produce maximum $E_{cell}$ at $298$K?

[Take $P_{H_2}=1.0$ atm in each case.]

• A. $Pt|H_2\left(g\right)|H^{+}\left(0.01M\right)||H^{+}\left(0.1M\right)|H_2\left(g\right)|Pt$
• B. $Pt|H_2\left(g\right)|N{H}_{4}Cl\left(0.01M\right)||HCl\left(0.1M\right)|H_2\left(g\right)|Pt$
• C. $Pt|H_2\left(g\right)|H^{+}\left(0.1M\right)||H^{+}\left(0.2M\right)|H_2\left(g\right)|Pt$
• D. $Pt|H_2\left(g\right)|H^{+}(pH=0.0)||H^{+}(pH=0.0)|H_2\left(g\right)|Pt$

Answer 1

The correct option is A $Pt|H_2\left(g\right)|H^{+}\left(0.01M\right)||H^{+}\left(0.1M\right)|H_2\left(g\right)|Pt$

Cathode : $H^{+}+e^{-}\to \frac{1}{2}{H}_2$

$\underset{–}{Anode:\frac{1}{2}{H}_2\to H^{+}+e^{-}}$

$\underset{–}{cellequation:H^{+}\frac{1}{2}{H}_{2}Cathode\to H^{+}\frac{1}{2}{H}_2}anode$

$E_{cell}=E_{cell}^0-\frac{0.0591}{1}log\frac{\left[H^{+}\right]anode}{\left[H^{+}\right]cathode}$

$=\frac{-0.0591}{1}log\frac{\left[H^{+}\right]A}{\left[H^{+}\right]C}$ $E_{cell}^0=0$ due to identical half cells.

Given : ${\left[H^{+}\right]}_A=0.01M$

${\left[H^{+}\right]}_C=0.1M$

$E_{cell}=-0.0591log\frac{0.01}{0.1}=0.0591V$