Question

A solution is 0.0450 M in BiO⁺ and 0.040 M in Co²⁺ and has a pH of 2.50. What is the concentration of the more easily reduced cation at the onset of deposition of the less reducible one?

Can you show work and explain please?

Answer 1

the reduction equations for given cations will be

BiO⁺ + 2H⁺ + 3e- --> Bi(s) + H2O(l) ,E0 = +0.32 V

Co⁺² + 2e --> Co(s)                      E0 = -0.277 V

So here Co+2 will act as reducing agent and will get oxidized as the reduction potential is low as compared to BiO+

The overall equation will be

Co --> Co⁺² + 2e                    ]X 3

BiO⁺ + 2H⁺ + 3e- --> Bi(s) + H2O(l) ]X 2

.................................................................

3Co + 2BiO⁺ + 4H⁺ + 6e- --> 2Bi(s) + 2H2O(l) + 3Co⁺² + 6e

final equation

3Co(s) + 2BiO⁺ + 4H⁺ --> 2Bi(s) + 2H2O(l) + 3Co⁺²

For a redox reaction we use Nernst equation for calculating

E = E0 - RT / nF log Q

F = 96485

R = 8.314

T = 298 K

For Cobalt

E = -0.277- 0.0592 / 2 log [1/[Co+2]]

E = -0.277 - 0.0296 log [1/ 0.040]

E = -0.318 V

The Nernst equation for the BiO+ will be

E = E0 - 0.0592 / n log [1/[BiO⁺][H⁺]²]

When cobalt starts deposition

E_Co = E_BiO+ = -0.318 V

pH = 2.5

[H+] = 0.00316

Putting values

-0.318 = 0.32 - 0.0592 / 3 log [1 / [BiO+ ] [ 0.00316]^2

-0.638 = -0.0592 / 3 log [1 / [BiO+ ] [ 0.00316]^2

32.33 = log [1 / [BiO+ ] [ 0.00316]^2 ]

Taking antilog

2.137 X 10^32 = [1 / [BiO+ ] [ 0.00316]^2

4.68 X 10^-33 = [BiO+ ] [ 0.00316]^2

4.68 X 10^-28 = [BiO+]