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

The EMF of a cell corresponding to the reaction : `Zn_((s)) + 2H_((aq))^(+) toZn^(2+) (0.1 M) + H_(2(g)) ` ( 1 atm ) is 0.28 volt at `15^@C`.
The pH of the solution at the hydrogen electrode is (Given : `E_(Zn^(2+)//Zn)^@`=-0.76 volt , `E_(H^+ // H_2)^@`= 0 volt )

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`E_(cell)^(@) = 0.76` volt
Applying Nernst equation `E_(cell)^(@) =E_(cell)^(@) - (0.0591)/(2)log.([Zn^(2+)][H_(2)])/([H^(+)]^(2))`
`0.28 = 0.76 -(-0.0591)/(2) log. ((0.1)xx1)/([H^(+)]^(2))`
`log.(0.1)/([H^(+)]^(2)) = (2xx 0.48)/(0.0591)`
`log 0.1 - log [H^(+)]^(2) = 16.24` [since, `-log[H^(+)] = pH]`
`2pH = 16.24 - log 0.1`
`pH = (17.24)/(2) = 8.62`
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