1.

Forthe followingcell reaction Ag|Ag^(+)|AgCI|CI^(-)|CI_(2) Pt triangle G_(f)^(@)(Ag^(+))=78 Kj // mol E^(@) of the cellis E^(@) ofthe cell is

Answer»

`-0.50 V`
`0.60 V`
6
None of these

Solution :Forthe given CELL `Ag|Ag^(+)|AgCI|CI^(-)|CI_(2)|CI_(2)`PT
The cell reactionare 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`


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