Question

Use standard reduction potentials to calculate the equilibrium constant for the reaction:

Cu²⁺ (aq) + Ni(s) → Cu(s) + Ni²⁺ (aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.

Answer 1

Value of equilibrium constant is
`1.92* 10^(19)`

Step-by-step explanation:

Oxidation:
`Ni-2e^(-)\rightarrow Ni^(2+)`;
`E_(Ni^(2+)\mid Ni)^(0)=-0.23 V`

Reduction:
`Cu^(2+)+2e^(-)\rightarrow Cu`;
`E_(Cu^(2+)\mid Cu)^(0)=0.34V`

-----------------------------------------------------------------------------------------------------------

Overall:
`Ni+Cu^(2+)\rightarrow Ni^(2+)+Cu`

`E_(cell)^(0)=E_(Cu^(2+)\mid Cu)^(0)-E_(Ni^(2+)\mid Ni)^(0)=(0.34+0.23)V=0.57V`

We know,
`k=e^{(nFE_(cell)^(0))/(RT)}`

where, k is equilibrium constant, n is no. of electron exchanged, 1F = 96500 C/mol, R is gas constant and T = 298 K

So,
`k=e^{(2* 96500(C)/(mol)* 0.57V)/(8.314(J)/(mol.K)* 298K)}=1.92* 10^(19)`