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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 )

Text Solution

Verified by Experts

The cell reaction is
`Zn_((s))+2H_((aq))^(+) Leftrightarrow (0.10M)+H_(2(g))+"1 atm"`
From questions `E_(Zn^(+2)//Zn)^(0)=0.75" volt"`
`E^(0)=E_(Zn//Zn^(++))+E_(H^(+)//H_(2))^(0)`
=+0.76V+0.0V
From Nernst equation
`E_("cell")=E_("cell")^(0)-(0.052)/(n) log ([Zn^(++)].P_(H_(2)))/([H^(+)]^(2))`
`0.2 =E_("cell")^(0) -(0.0592)/(2) log (0.1)/([H^(2)]^(2)) xx 1`
`0.28=0.76-(0.0592)/(2) log"" (0.1)/([H^(+)]^(2))`
`or, log""(0.1)/([H^(+)]^(2))=2 xx ((0.76-0.28))/(0.0592)`
`log_(0.1)-2log[H^(+)]=16.216`
`or, 2pH =17.216`
`pH=8.608`
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