# Calculate Kc for the system, Ni2+ + Co <====> Ni + Co2+. at 25 C? Ni2+ (aq) + 2e === Ni (s) E = - 0.25 V Co2+ (aq) + 2e === Co (s) E = - 0.28 First we have to determine which is the reduction half reaction and the oxidation half reaction.

In the chemical equation

Ni2+ + Co <====> Ni + Co2+

Ni2+ gained 2e to form Ni therefore it is the reduction half reaction

On the other hand,

Co loses 2 electrons thus...

First we have to determine which is the reduction half reaction and the oxidation half reaction.

In the chemical equation

Ni2+ + Co <====> Ni + Co2+

Ni2+ gained 2e to form Ni therefore it is the reduction half reaction

On the other hand,

Co loses 2 electrons thus forming Co2+; this is the oxidation half reaction.

Ecell = Ered - Eox

Ered = E of the reduction half reaction = -0.25 V

Eox = E of the oxidation half reaction = -0.28 V

Ecell = -0.25 - (-0.28)

Ecell = 0.03V

Remember that:

`Delta G = -nF E^(o) = -RTlnK`

therefore,

`lnKc = ((nFE^(o))/(RT))`

`Kc = e^(((nFE^(o))/(RT)))`

` `

n = 2 ( number of electrons shared/gained)

F = 96, 485

E = 0.03

R = 8.314 (gas constant)

T = 298 (25 + 273)K

`Kc = e^(((2*96485*0.03E)/(8.314*298)))`

K = 10.346

Approved by eNotes Editorial Team