Question

Calculate the cell potential of following cell

$Pt\left(s\right)\left|H_2\left(g\right)\left(0.1\text{ bar}\right)\right|\text{BOH}\left(\text{0.1M}\right)|\left|\text{HA}\left(\text{0.1M}\right)\right|H_2\left(g\right)\left(1\text{ bar}\right)|\text{Pt}$

Given $K_a\left(HA\right)=10^{-7},K_b\left(BOH\right)=10^{-5}$

• A. 0.39 V
• B. 0.36 V
• C. 0.93 V
• D. 0.63 V

Answer 1

Answer: (A)

$E_{cell}=\frac{0.059}{2}log\frac{[H^{+}{]}_{cathode}^2\times [P_{H_2}{]}_{anode}}{[H^{+}{]}_{anode}^2\times [P_{H_2}{]}_{cathode}}$

$\left[O{H}^{-}\right]=\sqrt{C{K}_b}=\sqrt{0.1\times 10^{-5}}=10^{-3}M$

$[H^{+}{]}_{anode}=\frac{10^{-14}}{10^{-3}}=10^{-11}M$

$[H^{+}{]}_{cathode}=\sqrt{C{K}_a}=\sqrt{0.1\times 10^{-7}}=10^{-4}M$

So $E_{\text{cell }}=\frac{0.059}{2}\log \frac{{\left(10^{-4}\right)}^2\times 0.1}{{\left(10^{-11}\right)}^2\times 1}$

$=0.\text{39}\text{V}$ .