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The emf of a cell corresponding to the r...

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`

Text Solution

Verified by Experts

The half cell reaction are: `ZntoZn^(2+)+e^(-)` (Oxidation half reaction)
`2H^(+)+2e^(-)toH_(2)` (Reduction half reaction)
The cell may be represented as: `Zn|Zn^(2+)||H^(+)|H_(2)`
For the given cell reaction, `E_(cell)^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-E_(Zn^(2+)//Zn)^(@)=0-(-0.76V)=0.76V`
Applying Nernst eqn. to the given cell reaction, `E_(cell)=E_(cell)^(@)=(0.0591)/(2)"log"([Zn^(2+)])/([H^(+)]^(2))`
`therefore0.28=0.76-(0.0591)/(2)"log"((0.1))/([H^(+)]^(2))=0.76-(0.0591)/(2)[log0.1-2log(H^(+))]`
`=0.76-0.02955(-1+2pH)` `[becausepH=-log[H^(+)]]`
`=0.76+0.2095-0.0591pH`
or `pH=(0.5095)?(0.591)=8.62`
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