If the pressure of ${\mathrm{H}}_2$ gas is increased from 1 atm to 100 atm. Keeping ${\mathrm{H}}^{+}$ concentration constant at 1 M, the voltage of hydrogen half cell at ${25}^{\mathrm{o}}\mathrm{C}$ will be 

$\underset{\left(1\text{ atm}\right)}{{\mathrm{H}}_2}\to \underset{\left(1\mathrm{M}\right)}{2{\mathrm{H}}^{+}}+2{\mathrm{e}}^{-}$

$\begin{array}{l}\therefore {\mathrm{E}}_{{\mathrm{H}}_2/{\mathrm{H}}^{+}}={\mathrm{E}}_{{\mathrm{H}}_2/{\mathrm{H}}^{+}}^{\mathrm{o}}-\frac{0.0591}{2}\log \frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}}\\ =0.00-\frac{0.0591}{2}\log \frac{{\left(1\right)}^2}{1}=0.00\mathrm{V}\end{array}$

Now, pressure is increased from 1 atm to 100 atm

$\begin{array}{l}\therefore {\mathrm{E}}_{{\mathrm{H}}_2/{\mathrm{H}}^{+}}={\mathrm{E}}_{{\mathrm{H}}_2/{\mathrm{H}}^{+}}^{\mathrm{o}}-\frac{0.0591}{2}\log \frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}}\\ =0-\frac{0.0591}{2}\log \frac{{\left(1\right)}^2}{100} ( \left[{\mathrm{H}}^{+}\right]\text{is constant })\\ =-\frac{0.0591}{2}\log {10}^{-2}=0.0591\mathrm{V}\end{array}$