Question

Given the half-cells:

Ni/Ni2+ and Fe/Fe2+

a) Chose the anode and cathode. b)Calculate Eo and DG for this reaction. c) Calculate what the new cell potential (E) would be if the non equilibrium concentrations were [Ni2+] =.40 M and [Fe2+] = 0.30 M. Calculate the how long you would have to run the battery at 6 amps to make this battery die at zero volts.

Answer 1

The standard reduction potential values are

Ni²⁺ (aq) + 2e^-   --------------------> Ni(s), E0(Red) = - 0.25V

Fe²⁺ (aq) + 2e^-   --------------------> Fe(s), E0(Red) = - 0.44V

Since the standard reduction potential value of Ni²⁺ (aq) is higher than Fe²⁺ (aq), hence Ni²⁺ (aq) will be reduced to to Ni(s) and Fe(s) will be oxidised to Fe²⁺ (aq).

(a) Since oxidation takes place at anode, hence Fe(s)/ Fe²⁺ (aq) will act as anode.

Oxidation Half-cell: Fe(s) -------------> Fe²⁺ (aq) + 2e- , E0(oxi) = - (- 0.44V) = +0.44V

Since reduction takes place at cathode, hence Ni²⁺ (aq) / Ni(s), will act as cathode.

Reduction half - cell: Ni²⁺ (aq) + 2e- ------------> Ni(s), E0(red) = - 0.25V

(b) E0(cell) =  E0(oxi) + E0(red) = +0.44V + (- 0.25V) = 0.44V - 0.25V = 0.19V (answer)

DeltaG0 = - nxFxE0(cell)

Since 2 electrons are transferred during the reaction, n = 2

=> DeltaG0 = - nxFxE0(cell) = - 2x96500Cx0.19V = - 36670 J = 36.67 KJ (answer)

(c) Given  [Ni2+] =0.40 M , [Fe2+] = 0.30M

The complete cell reaction is

Fe(s) + Ni²⁺ (aq)  -------------> Fe²⁺ (aq) + Ni(s)

Applying Nernst equation

E = E0(cell) - (0.059/n)xlog[Fe²⁺ (aq)] / [Ni²⁺ (aq)]

=> E = 0.19V - (0.059/2)xlog(0.30/0.40) = 0.194V (answer)