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Calculate the value of equilibrium constant for the reaction taking place between Cu(II) and Sn (II) ions in aqueous solution at 298 K.
Given : `E_(cu^(2+)//cu)^(@)=0.34" V" , E_(Sn^(2+)//Sn^(4+))^(@)=-0.154" V"`.

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Calculation of `E^(@)` for the redox reaction
The redox reaction that takes place in aqueous solution is :
`(Sn^(2+)(aq) to Sn^(4+)(aq)+2e^(-) , E^(@)=-0.154" V".)/(Cu^(2+)(aq)+2e^(-) to Cu(s) ," " E^(@)=+0.34" V"`.
Add : `Sn^(2+)(aq)+Cu^(2+)(aq) to Sn^(4+)(aq)+Cu(s) , E_(cell)^(@)=+0.186" V"`.
Step II. Calculation of equilibrium constant
`E_(cell)^(@)=(0.0591)/(n)logK`
`E_(cell)^(@)=0.186" V"`, n=2,
`:. " " logK=(2xx(0.186))/(0.0591)=6.294" or "K="Antilog " 6.294=1.97xx10^(6)`
The `E_(cell)` comes out to negative. Therefore, the reaction given above will process spontaneously.
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