For the following cell reaction
`Ag|Ag^(+)|AgCI|CI^(-)|CI_(2)` Pt
`triangle G_(f)^(@)(Ag^(+))=78 Kj // mol`
`E^(@)` of the cell is
`E^(@)` of the cell is

Options

A. `-0.50 V`
B. `0.60 V`
C. 6
D. None of these

Answer

A

Explanation

For the given cell `Ag|Ag^(+)|AgCI|CI^(-)|CI_(2)|CI_(2)`pt
The cell reaction are as follows
At anode `Ag rarr Ag^(+) +e^(-)`
At cathode `AgCI+e^(-) rarr Ag(s) +CI^(-)`
net cell reaction `AgCI rarr Ag^(+) +CI^(-)`
`therefore triangle G_("reaction")^(@) = Sigma triangle G_(R )^(@)`
`=(78-129 )-(-109)=+58 kJ // "mol"`
`58xx10^(3) j=-1xx96500xxE_("cell")^(@)`
`E_("cell")^(@)=(-58xx1000)/(96500)=-0.6 V`