Calculate potential of the cathode:

Problem:

A solution is 0.05 M in BiO+ and 0.0l M in Co²⁺ and has a pH of 2.50

a) What is the potential of the cathode when the concentration of the more easily reduced species is l.00 × l0^-6 M

b) What is the concentration of the more readily reduced cation when the less reduced cation begins to deposit.

Answer:

BiO⁺   + 2H⁺   + 3e  ? Bi(s ) + H₂O   E^o   =  + 0.320 V

Co²⁺ + 2e ?  Co(s)   E^o   = - 0.277 V

Considering the two cations,  BiO⁺ will deposit first.

E_cathode   =  E^o - 0.0592/3 log 1/[BiO]⁺

=   + 0.320 - 0.0592/3 log l/l × l0^-6

=  + 0.320 -  0.118   =   0.202 V

Potential of cathode to reduce the concentration of [BiO]⁺ to l × l0^-6 M

= + 0.202 V

E_cathode   to deposit Co²⁺

E_c  =  E_o  -  0.0592/2 log 1/Co²⁺

= - 0.277 - 0.0592/2 log 1/0.0l

=   - 0.277 - 0.0592   =   - 0.336 V

- 0.336 =   + 0.320 - 0.0592/3 log 1/X

- 0.656   =   - 0.0592/3 log 1/X

- 0.656/0.0l97 =   log 1/X

X = 5 × l0^-34 M

Concentration of bismuth when Co²⁺   begins to deposit is 5 × l0^-34 M.