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

Answer

42

Explanation

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`