A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure, The oxidation potential of electrode would be

1- 0.59V 2- 0.118V

3- 1.18V 4- 0.059V

Ans is - 0.59 V

As pH = 10

and pH = - log[H⁺]

      10 = - log[H⁺]

 [H⁺] = ${10}^{-10}$

Now according to Nernst equation,

E_cell​ = E^ocell - $\frac{2.303 RT}{nF}\log \frac{\left[H_2\right]}{\left[H^{+}\right]}$

Half cell reaction is 

H⁺(aq)  +   e^- $\to$ $\frac{1}{2}{H}_2$ (g)

Thus n = 1 and $\frac{RT}{F}= 0.0591$

Now for this reaction E⁰_cell = 0 V

So   E_cell = 0 - 0.0591 log(1/10^-10)

              = - 0.0591 log 10¹⁰

 ​= - 0.0591 x 10
 ​E_cell = - 0.591 V
The correct option is (1).