Q. 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
Electrochemistry
Solution:
$H_{2}\rightarrow2H^{+} +2e^{-}$
$\therefore E_{H_2H^{+}}=E^{0}_{H_2H^{+}}- \frac{0.0591}{2}log \frac{\left[H^{+}\right]^{2}}{PH_{2}}$
$=0.00-\frac{0.0591}{2}log \frac{\left(1\right)^{2}}{1}=0.00 V$
Now, pressure is increased from 1 atm to $100$ atm
$\therefore E_{H_2/H^{+}}=E^{o}_{H_2/H^{+}}-\frac{0.0591}{2}log \frac{\left[H^{+}\right]^{2}}{P_{H_2}} $
$0-\frac{0.0591}{2}log\frac{\left(1\right)^{2}}{100} \left(\left[H^{+}\right]is\, constant\right)$
$=-\frac{0.0591}{2}log 10^{-2}=0.0591 V$
$=0.00-\frac{0.0591}{2}log \frac{\left(1\right)^{2}}{1}=0.00 V$
Now, pressure is increased from 1 atm to $100$ atm
$\therefore E_{H_2/H^{+}}=E^{o}_{H_2/H^{+}}-\frac{0.0591}{2}log \frac{\left[H^{+}\right]^{2}}{P_{H_2}} $
$0-\frac{0.0591}{2}log\frac{\left(1\right)^{2}}{100} \left(\left[H^{+}\right]is\, constant\right)$
$=-\frac{0.0591}{2}log 10^{-2}=0.0591 V$