Calculate concentration in the solution:

Problem:

A solution is 0.l50 M in Co²⁺ and 0.05 in Cd²⁺, Calculate

a) The Co²⁺ concentration in the solution when Cd²⁺ begins to deposit

b) The cathode potential required lowering the Co²⁺ concentration to l × l0^-5 M

Answer:

a) Co²⁺ +2e  ?  Co(s)  Eo   = - 0.277V

                       Cd²⁺ +2e ?   Cd(s) Eo   = - 0.403V

Among the two cations, cobalt will deposit first since its Eo = - 0.277V

E_cathode (Cd)  = E^ocd²⁺     - 0.0592/2 log 1/0.05

=  - 0.403 - 0.0296 log 1/0.05 = - 0.403 - 0.0385

E_cathode = - 0.44l5V

- 0.44l5 =  E^oCo²⁺ - 0.0592/2 log 1/X

- 0.44l5 = - 0.277 - 0.0592/2 log 1/X

- 0.l645/0.0296 =   log 1/X

X  =   2.77 × l0^-6 M

The concentration of Co²⁺ when cadmium begins to deposit

=  2.77 × l0^-6 M.

b) Cathode potential needed to lower Co²⁺ concentration to 1 × l0^-5M

E_cathode =  E^o   - 0.0592/2 log l/1 × l0^-5

=   - 0.277 - 0.0296 log 1/10^-5 = - 0.277 - 0.148

E_cathode  = - 0.425V

Cathode potential needed is - 0. 425 V