The `e.m.f.` of a cell corresponding to the reaction :
`Zn_((s))+2H_((aq.))^(+) rarr underset((0.1 M))(Zn^(2+))+underset((1 atm))(H_(2(g))`
is `0.28 V` at `25^(@)C` and `E_(Zn//Zn^(2+))^(@) = 0.76V`.
(i) Write half cell reactions.
(ii) Calculate `pH` of the solution at `H` electrode.

Explanation

(i) Anode: `Zn rarr Zn^(2+)+2e`
Cathode: `2H^(+)+2e rarr H_(2)`
(ii) `E_(cell) = E_(OP_(Zn))^(@) - (0.059)/(2)"log"[Zn^(2+)]+E_(RP)^(@)`
`-(0.059)/(2)"log"([H^(+)]^(2))/(P_(H_(2)))`
or `E_(cell) = E_(OP_(Zn))^(@) + 0 +(0.059)/(2)"log"([H^(+)]^(2))/(P_(H_(2)) xx [Zn^(2+)])`
`= 0.28 = 0.76 + (0.059)/(2)"log"([H^(+)]^(2))/(0.1 xx 1)`
or `([H^(+)]^(2))/(1) = - (0.48 xx 2)/(0.059) = -16.271`
`([H^(+)]^(2))/(1) = 5.3556 xx 10^(-17)`
or `[H^(+)] = 2.3142 xx 10^(-9)`
`:. pH = -log[H^(+)] = 8.635`