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Calculate DeltaG(r)^(@) of the following...

Calculate `DeltaG_(r)^(@)` of the following reaction
`Ag^(+)(aq)+cI^(-)(aq)rarrAgCI(s)`
Given
`DeltaG_(r)^(@)(AgCI)rarr-109 kJ Mol^(-1)`
`DeltaG_(r)^(@)(CI^(-))rarr-129 kJ Mol^(-1)`
`DeltaG_(r)^(@)(ag^(-))rarr77 kJ Mol^(-1)`
(i)Represent the above reaction in form of a cel
(ii) Calcualte `E^(@)` of the cel
(iii) Find `log_(10)K_(sp)` of AgCI

Text Solution

Verified by Experts

The correct Answer is:
A, B
(a). `triangleG_(r)^(@)=-109+129-77=-57kJ//mol`
cell representation `Ag|AgCl||Cl^(-)|Ag^(+)|Ag`.
`-1xx96500xxE^(@)=057xx10^(3)`
`E^(@)=0.59"volt"`
`0=0.59-(0.059)/(1)log((1)/(K_(SP)))`
`logK_(SP)=-10`.
(b). `ZntoZn^(2+)+2e^(-)," "0.76"volt"`
`underline(2Ag^(+)+2e^(-)to2Ag," "0.80"volt")`
`underline(Zn+2Ag^(+)toZn^(2+)+2Ag,) " "E_(cell)^(@)=1.56"volt"`
`n_(2n)=(6.539)/(65.39)=10^(-3)mol" "[Ag^(+)]=sqrt(K_(sp))=10^(-5)M`
`0=1.56-(0.059)/(2)logK" "n_(Ag^(+))=10^(-5)xx0.1=10^(-6)M`.
`n_(Ag)=10^(-6)molimplieslogK=52.8`
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