Question

If the pressure of $H_2$ gas is increased from $1$atm to $100$atm keeping $H^{+}$ concentration constant at $1M$, the change in reduction potential of hydrogen half cell at ${25}^{o}C$ will be:

• A. $0.059V$
• B. $0.59V$
• C. $0.0295V$
• D. $0.118V$

Answer 1

The correct option is A $0.059V$

Half Cell reaction will be

${2H}^{+}+{2e}^{-}\to H_2$

Here, pressure changed from 1 atm to 100 atm.

So, $Q$ = 100

So, reduction in potential of hydrogen half cell will be

$=\frac{0.059}{2}\log Q=\frac{0.059}{2}\log 100=0.059V$

So, correct answer is option A.