## Balance the following equation in basic conditions. Phases are optional CoCl_2 + Na_2O_2

Question

### Question:

Balance the following equation in basic conditions. Phases are optional CoCl_2 + Na_2O_2

CoCl_2 + Na_2O_2 –> Co(OH)_3 + Cl^- + Na^+

Adjusting a condition(equation) in Basic medium:

1) First of all, partition a condition into two half-responses, for example, oxidation and decrease response.
2) Balance two half-responses independently.
3) First balance particles other than O and H
4) Then balance O by including H2O atom that side of the condition where O particles are less.
5) Then balance H by including H+ particles and a similar amount of OH− particles on the two sides of the response.
6) Then we can compose H+ particles and OH− particles consolidated as H2O.
7) At last parity charge by including electrons.
8) Then adjusted two half-responses are added to discover the last balance condition to such an extent that the quantity of electrons is counterbalanced by increasing the condition with a proportionate number.

Explanation:

Lopsided response:
CoCl2+Na2O2→Co(OH)3+Cl−+Na+
First half-response:
CoCl2→Co(OH)3+Cl−
CoCl2→Co(OH)3+2Cl− ( Cl balance)
CoCl2+3H2O→Co(OH)3+2Cl− ( O balance)
CoCl2+3H2O+3OH−→Co(OH)3+2Cl−+3H++3OH− ( H balance)
CoCl2+3H2O+3OH−→Co(OH)3+2Cl−+3H2O
CoCl2+3OH−→Co (OH)3+2Cl−+e− (charge balance) – (1)

Second half-response:

Na2O2→Na+
Na2O2→2Na+ (Na balance)
Na2O2→2Na++2H2O (O balance)
Na2O2+4H++4OH−→2Na++2H2O+4OH− (H balance)
Na2O2+4H2O→2Na++2H2O+4OH−
Na2O2+2H2O+2e−→2Na++4OH− (charge balance) – (2)
apply 2*equation (1) and add it to condition (2)-
2CoCl2+6OH−+Na2O2+2H2O+2e−→2Co(OH)3+4Cl−+2e−+2Na++4OH−