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. 7.62
B. 8.62
C. 9
D. 9.62

Answer

B

Explanation

`E_(Cell)^(@)=0.76"volt"`
Applying Nernst equation
`E_(Cell)=E_(Cell)^(@)-(0.0591)/(2)log.([Zn^(2+)][H_(2)])/([H^(+)]^(2))`
`log.(0.1)/([H^(+)]^(2))=(2xx0.48)/(0.591)`
`log0.1-log[H^(+)]^(2)=16.2436`
`2pH=16.2436-log0.1`
`pH=(17.2436)/(2)=8.62`