Question

Balance the following equation in basic conditions. Phases are optional.

CoCl2+Na2O2 = Co(OH)3+Cl- +Na+

Answer 1

Ans : 2CoCl_{2 (aq)} + 1Na₂O_{2 (aq)} + 2 OH^{- (aq)} + 2 H₂O (l) ----> 2Co(OH)_{3 (s)} +4Cl^-_(aq) + 2Na⁺_(aq)

CoCl2 --> Co(OH)3 + 1 e-
cobalt loses 1 e- going from Co+2 to Co+3

Na2O2 + 2e- --> Co(OH)3
while O₂^-2 take a total of 2 e- going from O^-1 each to O^-2 each

so we have to double the Cobalt to balance the electrons taken and lost:
2CoCl2 --> 2 Co(OH)3 & 2 e-
Na2O2 & 2e- --> Co(OH)3

combining these:
2CoCl2 (s) + 1Na2O2 (ag) ----> 2Co(OH)3 (ag)+Cl- (ag) + Na+ (ag)

let's balance the Na's:
2CoCl2 (s) + 1Na2O2 (ag) ----> 2Co(OH)3 (ag)+Cl- (ag) + 2Na+ (ag)

let's balance the Cl's:
2CoCl2 (s) + 1Na2O2 (ag) ----> 2Co(OH)3 (ag)+ 4Cl- (ag) + 2Na+ (ag)

let's balance the net -2 on the right by adding 2 OH- to the left:
2CoCl2 + 1Na2O2 + 2 OH- ----> 2Co(OH)3 + 4Cl- + 2Na+

we need 4 more H's on the left to balance the 6 H's on the right, so add neutral water:
2CoCl2 + 1Na2O2 + 2 OH- + 2 H2O ----> 2Co(OH)3 +4Cl- + 2Na+