Question

Using these standard reduction potentials:

Reduction Reaction

(1) H2O2 + 2e- --> 2OH- E (under std conditions) (V) = 1.77

(2) [Co(H₂O)₆]³⁺ + e^- --> [Co(H₂O)_6]²⁺    E (under std conditions) (V) = 1.84

(3) [Co(NH₃)₆]³⁺ + e^- --> [Co(NH₃)₆]²⁺ E (under std conditions) (V) = 0.10

Show that one can prepare an ammine complex from CoCl₂ and hydrogen peroxide in the presence of ammonia but not in its abscene. You will need to write two redox reactions, calculate standard potentials for the reactions, and make conclusions. That is, set up an equation to calculate E (under std conditions) (V) using one cobalt complex half-cell with the peroxide half-cell, then calculate E (under std conditions) (V) again using the other cobalt complex and peroxide. Compare the two E (under std conditions) values.

Answer 1

For a spontaneous reaction E should be positive. The reaction with positive E will produce the complex

H2O2 + 2e- --> 2OH- (cathode)   E(V) = 1.77

[Co(H₂O)_6]2+ ----->[Co(H₂O)₆]³⁺ + e^- (anode) E(V) =1.84

E=E_cathode-E_anode

E=1.77-1.84=-0.07 negative potential reaction will not occur

H2O2 + 2e- --> 2OH- (cathode)   E(V) = 1.77

[Co(NH₃)₆]2+ -----> [Co(NH₃)₆]³⁺ + e^- (anode) E (V) = 0.10

E=E_cathode-E_anode

E=1.77-0.10=1.67

The reaction with positive E will produce the complex