Question

Calculate the cell potential of following cell

$Pt\left(s\right)\left|H_{2} \left(g\right) \left(0.1 \text{ bar}\right)\right|\text{BOH}\left(\text{0.1M}\right)\left|\right.\left|\text{HA} \left(\text{0.1M}\right)\right|H_{2}\left(g\right)\left(1 \text{ bar}\right)\left|\right.\text{Pt}$

Given $K_{a}\left(H A\right)=10^{- 7},K_{b}\left(B O H\right)=10^{- 5}$

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

Answer 1

Answer: (A)

$E_{c e l l}=\frac{0.059}{2}log \frac{\left[\right. H^{+} \left]\right._{c a t h o d e}^{2} \times \left[\right. P_{H_{2}} \left]\right._{a n o d e}}{\left[\right. H^{+} \left]\right._{a n o d e}^{2} \times \left[\right. P_{H_{2}} \left]\right._{c a t h o d e}}$

$\left[O H^{-}\right]=\sqrt{C K_{b}}=\sqrt{0.1 \times 1 0^{- 5}}=10^{- 3}M$

$\left[\right.H^{+}\left]\right._{a n o d e}=\frac{1 0^{- 14}}{1 0^{- 3}}=10^{- 11}M$

$\left[\right.H^{+}\left]\right._{c a t h o d e}=\sqrt{C K_{a}}=\sqrt{0.1 \times 1 0^{- 7}}=10^{- 4}M$

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

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