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For the cells in opposition, Zn(s) | Z...

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)

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