Question

For the cell Pt | ${\text{H}}_{\text{2}}$ (0.4 atm) | ${\text{H}}^{+}$(pH = 1) || (pH = 2) | ${\text{H}}_{\text{2}}$ 90.1 atm) | Pt. The measured potential at ${\text{25}}^{\circ }\text{C}$ is:

• A. $-0.1V$
• B. $-0.5V$
• C. $-0.041V$
• D. $-0.030V$

Answer 1

The correct option is C $-0.041V$

$answerisC$

$atanode\left(0.4atm\right)H_2\to 2H^{+}+2e^{-}$

$atcathode2H^{+}+2e^{-}\to H_2\left(0.1atm\right)$

$pH=1=-\log H^{+}$

$atanode\left[H^{+}\right]={10}^{-1}=0.1M$

$atcathode\left[H^{+}\right]={10}^{-2}=0.01M$

$E_{cell}=E_{cell}^0-\frac{0.0591}{2}\log \frac{[H^{+}{]}^2\times P_{H_2}}{[H^{+}{]}^2\times P_{H_2}}$

$E_{cell}=0-\frac{0.0591}{2}\log \frac{{10}^{-2}\times 0.1}{{10}^{-4}\times 0.4}$

$E_{cell}=-0.041V$