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"`.

Calculate the value of equilibrium constant for the reaction taking pl

1.	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"`.
Answer» 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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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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