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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

`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"((0.1xx1))/([H^(+)]^(2))`
`"log"0.1/([H^(+)]^(2))=(2xx0.48)/0.0591`
`log 0.1-log [H^(+)]^(2)=16.2436" "["Since,"-log [H^(+)]=pH]`
`2 pH =16.2436-log 0.1`
`pH=17.2436/2=8.6218`
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