`H_2` (gas) is bubbled through an aqueous solution of `HCl(pH = 2.5)` at `25^@C`. If at `P_(H_2(g)) = 10^(-x))` bar the electrode potential will be zero. Then find value of x.

Answer

5

Explanation

Reaction : `H_(2) (g) = 2H^(+) + 2e^(-)`
`E_("Rxn") = E_("Rxn")^(@) - (0.0591)/2 log ([H^+]^2)/(P_(H_2)) (at 298 k)`
To get `E_("Rxn") = 0 (or) ([H^+]^(2))/(P_(H_2)), PH_(2) = 10^(-5) , x = 5`.