Question

The oxidation potential of $H_2$ electrode $H_2/H_3{O}^{+}\left(aq\right)$ will be more than zero, if:

(keeping all other terms unity)

• A. $\left[H_3{O}^{+}\right]=1M$
• B. $\left[H_3{O}^{+}\right]=2M$
• C. pressure $H_2=2$ bar
• D. $E_{ox}$ can never be +ve

Answer 1

The correct option is A $\left[H_3{O}^{+}\right]=1M$

$H_2+{2H}_{2}O⟶2H_3{O}^{+}+{2e}^{-}$

$E_{cell}=E^0-\frac{0.059}{n}log\frac{{\left[H_3{O}^{+}\right]}^2}{\left[{PH}_2\right]}$ (Nernst equation)

$E_{cell}=E^0-\frac{0.059}{2}log\frac{{\left[H_3{O}^{+}\right]}^2}{\left[{pH}_2\right]}$

$E_{cell}=E^0-0.059log\frac{\left[H_3{O}^{+}\right]}{{\left[P_{H_2}\right]}^{1/2}}$

if $\left(H_3{O}^{+}\right)=1M$, no change in $E^0$

$\left[H_3{O}^{+}\right]=2m,E^0$ will decrease

if pressure $H_2=2bar$

$E_{cell}$ will increase i.e. more than zero.