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)

For the cells in opposition, Zn(s) | ZnCl_(2)(sol).|AgCl(s)|Ag|AgCl(s)

1.	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» Correct Answer - 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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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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