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For the cellls in opposition, Zn(s)|Zn...

For the cellls in opposition,
`Zn(s)|ZnCl_(2)(sol.)|AgCl(s)|Ag|AgCl(s)|ZnCl_(2)(sol.)|Zn(s)`
`C_(1)=0.02 M , C_(2)=0.5 M`
Find out of emf (in millivolt) of the resultant cell ? (Take log 2= 0.3, `(RT)/F` at 298 K=0.060)

Text Solution

Verified by Experts

The correct Answer is:
42
As cell reaction is
1st cell: `Zn+2AgCltoZnCl_(2)+2Ag`
2nd cell: `underline(ZnCl_(2)+2Agto2AgCl+Zn)`
overall `ZnCl_(2)(C_(2))toZnCl_(2)(C_(1))`
`E=(RT)/(2F)ln ((0.5)/(0.02))V=[(0.059)/(2)log((0.5)/(0.02))]V=42`
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