1.

Calculate the equilibrium constant of the reaction: Cu_((s))+2Ag_((ag))^(+)toCu_((aq))^(2+)+2Ag_((s))""[E_(cell)^(Theta)=0.46V]

Answer»

SOLUTION :
ELECTRON CHANGE in this reaction =2=n
`E_((cell))^(Theta)=0.46V`
`K_(C)=`EQUILIBRIUM constant
`E_((cell))^(Theta)=(0.059V)/(n)"log "K_(C)=0.46V`
`therefore logK_(C)=(0.46Vxx2)/(0.059)=15.5932`
`therefore K_(C)=`Antilog 15.5932
`=3.9194xx10^(15)=3.92xx10^(15)`.


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