Question

If the pressure of $H_2$ gas is increased from 1 atm to 100 atm. Keeping $H^{+}$ concentration constant at 1 M, the voltage of hydrogen half cell at $25°C$ will be

• A. $0.0591$ V
• B. $0.59$ V
• C. $0.0295$ V
• D. $0.118$ V

Answer 1

Answer: (A)

$H_2\to 2H^{+}+2e^{-}$ $\therefore E_{H_2{H}^{+}}=E_{H_2{H}^{+}}^0-\frac{0.0591}{2}log\frac{{\left[H^{+}\right]}^2}{P{H}_2}$
$=0.00-\frac{0.0591}{2}log\frac{{\left(1\right)}^2}{1}=0.00V$
Now, pressure is increased from 1 atm to $100$ atm
$\therefore E_{H_2/H^{+}}=E_{H_2/H^{+}}^o-\frac{0.0591}{2}log\frac{{\left[H^{+}\right]}^2}{P_{H_2}}$
$0-\frac{0.0591}{2}log\frac{{\left(1\right)}^2}{100}\left(\left[H^{+}\right]isconstant\right)$
$=-\frac{0.0591}{2}log10^{-2}=0.0591V$