J.R. S. answered • 04/27/24
Ph.D. University Professor with 10+ years Tutoring Experience
CoCl2 + Na2O2 ==> Co(OH)3 + Cl- + Na+
Co goes from ox.no. +2 to ox.no. +3, so it has been oxidized:
Co2+ ==> Co(OH)3 .. unbalanced oxidation half reaction
Co2+ + 3H2O ==> Co(OH)3 .. balanced for Co and O
Co2+ + 3H2O + 3OH- ==> Co(OH)3 + 3H2O .. balanced for Co, O and H using base (OH-)
Co2+ + 3H2O + 3OH- ==> Co(OH)3 + 3H2O + e- .. balanced rxn
Co2+ + 3OH- ==> Co(OH)3 + e- .. balanced oxidation reaction
O goes from ox.no. -1 to ox.no. -2, so it has been reduced:
O2-2 ==> 2OH- .. unbalanced reduction half reaction
O2-2 ==> 2OH- .. balanced for O
O2-2 + 2H2O ==> 2OH- + 2OH- .. balanced for O and H using base (OH-)
O2-2 + 2H2O + 2e- ==> 2OH- + 2OH- .. balanced rxn
O2-2 + 2H2O + 2e- ==> 4OH- .. balanced reduction reaction
Multiply oxidation by 2 to equalize electrons and add the 2 rxns together. Then combine/cancel like terms;
2Co2+ + 6OH- ==> 2Co(OH)3 + 2e-
O2-2 + 2H2O + 2e- ==> 4OH-
------------------------------------------------------------
2Co2+ + 2OH- + O2-2 + 2H2O ==> 2Co(OH)3 .. BALANCED REDOX EQUATION (absent spectator ions)
2CoCl2 + 2OH- + Na2O2 + 2H2O ==> 2Co(OH)3 + 2Na+ + 4Cl- (including spectator ions)
