Question

If the pressure of hydrogen gas is increased from $1$ atm. to $100$ atm, keeping the hydrogen ion concentration constant at $1$M, the reduction potential of the hydrogen half cell is at ${25}^o$C will be:

• A. $0.059$V
• B. $-0.059$V
• C. $0.295$V
• D. $0.118$V

Answer 1

Answer: (B)

$-0.059$V

$H^{+}+e^{-}⟶1/2H_2$

$E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{[P_{H_2}{]}^{1/2}}{\left[H_2^{+}\right]}$

$E_{{cell}_1}=1-0.0591\log \frac{[1{]}^{1/2}}{\left[1\right]}=-0.0591\times 0=0$

$E_{{cell}_2}=-0.0591\log \frac{[100{]}^{1/2}}{\left[1\right]}=-0.0591V$

Thus, the reduction potential at $100$ atm is

$E_{{cell}_2}=-0.0591V$