Question

A concentration cell Pt $\left|H_2\left(g\right)\right|\left|HCl\left(aq\right)\right|\left|H_{2}S{O}_4\left(aq\right)\right|\left|H_2\left(g\right)\right|$ is constructed using equal concentration of acids and equal number of moles of $H_2$ gas in both the compartments at the same temperature. If volume of $H_2$ gas at the anodic compartment is $\frac{1}{9}$ times the volume of $H_2$ gas at cathodic compartment.

Which of the following is/are correct for the given cell (log 2 = 0.3, log 3 = 0.48) $\frac{2.303RT}{F}=0.06V$

• A. Pressure of $H_2$ gas in both the cell compartments are equal
• B. Concentration of $H^{+}$ ion in both the cell compartments are unequal
• C. $E_{cell}^o$ = 0 for the above cell
• D. $E_{cell}\ne$ 0 for the above cell

Answer 1

Answer: (B), (C), (D)

The correct options are
B $E_{cell}^o$ = 0 for the above cell
C Concentration of $H^{+}$ ion in both the cell compartments are unequal
D $E_{cell}\ne$ 0 for the above cell

$\underset{P_1}{H_2\left(g\right)}\left|\underset{C}{H^{+}}\right|\left|\underset{2C}{H^{+}}\right|\underset{P_2}{H_2\left(g\right)}$

Anodic reaction,

$\frac{1}{2}{H}_2⟶H^{+}+e$

Cathodic reaction -

$H^{+}+e⟶\frac{1}{2}{H}_2$

$\underset{\left(anode\right)}{\frac{1}{2}{H}_2+}\frac{H^{+}}{\left(cathode\right)}⟶\underset{\left(cathode\right)}{\frac{1}{2}{H}_2+}\frac{H^{+}}{\left(anode\right)}$

$E_{cell}^o=0$

$E_{cell}=E_{cell}^o-\frac{0.06}{1}log\frac{\sqrt{{\left(P_{H_2}\right)}_{cathode}}\times {\left[H^{+}\right]}_{anode}}{\sqrt{{\left(P_{H_2}\right)}_{anode}}\times {\left[H^{+}\right]}_{cathode}}$

$E_{cell}=-0.06log(\sqrt{\frac{1}{9}}\times \frac{C}{2C})=-0.06log\left(\frac{1}{2\times 3}\right)=0.06(log2+log3)=0.06(0.3+0.48)=0.0468volt$