Determine the value of `E^0` cell for the following reaction
`Cu^(+2)+Sn^(+2)toCu+Sn^(+4)`
Equilibrium constant is `10^6`
`Cu^(++)+Sn^(++)toCu+Sn^(+4)`

At 25^@C, E^@ for reaction, Cu^(2+) + Sn(s) to Cu(s) + Sn^(2+) is 0.118 V, the equilibrium constant for the reaction is:

Find the equilibrium constant of the following reaction Cu^(2+) (aq) +Sn^(2+) (aq) hArr Cu(s) +Sn^(4+) (aq) at 25^(@)C, E_(Cu^(2+)//Cu)^(Theta)=0.34V, E_(Sn^(4+)//Sn^(2+))^(Theta)=0.155V

The Gibbs energy change (in J) for the given reaction at [Cu^(2+) ] = [Sn^(2+) ] = 1 M and 298 K is : Cu (s) + Sn^(2+) (aq) to Cu^(2+) (aq.) + Sn(s), (E_(Sn^(2+) // Sn)^(0) = -0.16 V E_(Cu^(2+) |Cu)^(0) = 0.34 V , Take F = 96500 C mol^(-1) )

The Gibbs energy change (in J) for the given reaction at [Cu^(2+) ] = [Sn^(2+) ] = 1 M and 298 K is : Cu (s) + Sn^(2+) (aq) to Cu^(2+) (aq.) + Sn(s), (E_(Sn^(2+) // Sn)^(0) = -0.16 V E_(Cu^(2+) |Cu)^(0) = 0.34 V , Take F = 96500 C mol^(-1) )

Sn1 and Sn2 reaction

SN1 and SN2 reaction

The standard emf for the cell reaction, 2Cu^(+)(aq)toCu(s)+Cu^(2+)(aq) is 0.36V at 298K . The equilibrium constant of the reaction is

When Sn^(2+) changes to Sn^(4+) in a reaction