Question

Consider the cadmium electrode, whose half-reaction is shown below. Calculate the potential of the electrode at 25 °C under the given conditions. A) Immersed in a 0.0120 M solution of CdBr2. B) Immersed in a 0.0490 M NaOH solution that is saturated with Cd(OH)2. Ksp for Cd(OH)2 is 7.2 × 10–15. C) Immersed in a 0.0140 M Cd(NH3)42 and 0.375 M NH3 solution. β4 for Cd(NH3)42 is 3.63 × 106.

Answer 1

A ) Immersed in a 0.0120 M solution of CdBr2.

CdBr2----------------------> Cd+2 + 2 Br-

Cd+2   + 2e- ------------------------> Cd (s) , E^o = -0.4022 V

[Cd+2 ] = 0.0120 M

E^ocell = E^o   - (0.0591 / n ) x log [1/Cd+2]

           = -0.4022 - (0.0591 /2) x log[1/0.0120]

           = -0.459 V

potential = -0.459 V

(B)   Immersed in a 0.0490 M NaOH solution that is saturated with Cd(OH)2. Ksp for Cd(OH)2 is 7.2 × 10–15. C

Cd(OH)2 <-----------------------> Cd+2 + 2OH-

                                                 S          0.0490M

Ksp = [Cd+2] [OH-]^2

Ksp = [Cd+2] (0.0490)^2

7.2 x 10^-15 = [Cd+2] (0.0490)^2

[Cd+2] = 3.12 x 10^-12 M

E^ocell = E^o   - (0.0591 / n ) x log [1/Cd+2]

           = -0.4022 - (0.0591 /2) x log[1/3.12 x 10^-12 ]

            = 0.742 V

potential = - 0.742 V