A silver electrode dipping in `AgNO_(3)` solution `(0.1M)` is combined salt bridge with a hydrogen electrode dipping in a solution of `pH=3(at 25^(@)C)`. If the standard reduction potential of the silver electrode is `0.799V`, what is the `EMF` of the cell ?

Explanation

`E^(c-)._(Ag^(o+)|Ag)=0.99V,E_(cell)=?`
`E_(c-)._(cell)=E^(c-)._(H_(2))-E^(c-)._(Ag)`
`=0-0.799V=-0.799V`
Since the reduction potential of `Ag^(o+)gt` Reduction potential of `H_(2)`
If polartiy reversed, `E^(c-)=0.799V`
So cell would be
`H_(2)|(H^(o+))(pH=3)||AgNO_(3)(0.1M)|Ag`
Cell reaction `:`
`(1)/(2) H_(2) rarr H^(o+)+e^(-)`
`Ag^(o+)+e^(-)rarr Ag`
`ulbar((1)/(2)H_(2)+Ag^(o+)rarrH^(o+)+Ag)`
`E_(cell)=E^(c-)._(cell)-(0.059)/(1)log .([H^(o+)])/([Ag^(o+)])`
`=0.799-(0.059)/(1)log .(10^(-3))/(0.1)`
`=0.799-0.059log 10^(-2)`
`=0.799+0.059xx2 = 0.917V`