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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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Verified by Experts

The correct Answer is:
42
as cell reaction is
`({:("1st cell:" Zn+2AgCl to ZnCl_2+2Ag),("2nd cell:" ZnCl_2+2Ag to 2AgCl+Zn):})/{:"Overall:" ZnCl_2(Cl_2) to ZnCl_2(C_1))`
`E=(RT)/(2F)` In `(2.5/0.02)V=[0.06/2 "log" (0.5/0.02)]V=42 mV`
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