`E^(c-)._(cell)` of the reaction above in Question is

Options

A. `0.59 V`
B. `-0.59 V`
C. `0.295 V`
D. `-0.295 V`

Answer

a

Explanation

At anode `: Ag(s)+Cl^(c-) rarr AgCl(s)+e^(c-)`
At cathode `: Ag^(o+)(aq)+e^(c-)rarr Ag(s)`
`ul bar(Cell reaction: Ag^(o+)(aq)+Cl^(c-)(aq) rarr AgCl(s))`
`Deltaa_(r)G^(c-)=-nFE^(c-)._(cell)=-1xx96500xxE^(c-)._(cell)`
`:. E^(c-)._(cell)=(-57xx10^(3))/(-96500)=0.59V`