For the disproportion of copper:
`2 Cu^(+) to Cu^(+2) + Cum E^(0)` is :- Given `E^(0)` for `Cu^(+2)//Cu` is 0.34 V & `E^(0)` for `Cu^(+2)//Cu^(+)` is 0.15 V:

For the disproportion of copper: 2 Cu^(+) to Cu^(+2) + Cu E^(0) is :- Given E^(0) for Cu^(+2)//Cu is 0.34 V & E^(0) for Cu^(+2)//Cu^(+) is 0.15 V:

E^(@) for Fe//Fe^(2+) is +0.44 V and E^(@) for Cu//Cu^(2+) is -0.32 V . Then, in the cell,

(CU^(+) ((aq)) is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction 2Cu_((aq))^(+) Cu_((aq))^(2+) +Cu_((s)) choose E^(@) for the above reaction if E_(Cu^(2+)//Cu^(@))= 0.34V and E_(Cu^(2+)//Cu^(+) )= 0.15V

Calculate equilibrium constant for tha disproportionation reaction : 2Cu^(2+)(aq)rarrCu(s)+Cu^(2+)(aq)" at " 25^(@)C ("Given " E_((Cu^(+)//Cu))^(@)=0.52 V, E_((Cu^(2+)//Cu^(+)))^(@)=0.16 V) .

E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

Cu^(2+)(aq.) is unstable in solution and under goes simultaneous oxidation and reduction according to the reaction 2Cu^(+)(aq.)hArr Cu^(2+)(aq.)+Cu(s) Choose the correct E^(@) for the above reaction if E_(Cu^(2+))^(@)//Cu = 0.34 V and E_(Cu^(2+))^(@)//Cu^(+) = 0.15 V

Civen E^@ for Cu^(2+) rarr Cu^+ is + 0.15 V and Cu^+ rarr Cu is +0.05 V Calculate E^@ for Cu^(2+) rarr Cu .