The emf of a cell corresponding to the reaction
`Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1` atm is `0.28` volt at `25^(@)C`. Calculate the `pH` of the solution at the hydrogen electrode.
`E_(Zn^(2+)//Zn)^(@) =- 0.76` volt and `E_(H^(+)//H_(2))^(@) = 0`

Options

A. `10^(1.23)M`
B. `10^(0.56)M`
C. `10^(0.66)M`
D. `10^(0.12)M`

Answer

C

Explanation

`(e^(-) +2H^(+) +NO_(3)^(-) rarr NO_(2) +H_(2)O) xx 2`
`Cu rarr Cu^(+2) +2e^(-)`
`4H^(+) +2NO_(3)^(-) +Cu rarr Cu^(+2) +2NO_(2) +2H_(2)O`
`E = 0.79-0.34 -(0.0591)/(2) log.([NO_(2)]^(2)[Cu^(+2)])/([NO_(3)^(-)]^(2)[H^(+)]^(4))`..(ii)
Let `[HNO_(3)] = xM` so `[H^(+)] = [NO_(3)^(-)] = x`
equation (i) & (ii)
`0.62 -(0.0591)/(2) log.(10^(-9))/(x^(10)) = 0.45 -(0.0591)/(2)log.(10^(-7))/(x^(6))`
`0.17 =(0.591)/(6) [-9-10 logx] -(0.0591)/(2)[-7-6logx]`
`0.0518 = 0.0788 logx`
`x = 10^(0.657)M`