How to solve CoCl2+HO2- ------> Co(OH)3+ Cl- by half reaction method
Co^2+ ==> Co^3+ + e
HO2^- + 2e + H2O ==> 3OH^-
Multiply eqn 1 by 2 and add. Cancel the 2e on each side.
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2Co^2+ +HO2^- +H2O=> 3OH^- +2Co^3+
You can add in the Cl^- here as
2CoCl2 + HO2^- + H2O ->2Co^3+ + 3OH^- + 4Cl^-
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You can make that 2Co^3+ on the right as 2Co(OH)3 BUT you have only 3 OH^- there to do that so you must add 3OH^- on the left side, too, as
2CoCl2 + HO2^- + H2O + 3OH^- ==> 2Co(OH)3 + 4Cl^-
1. Atoms balance on each side.
2. change e balance
3. charge balances on each side.
To solve the given chemical equation using the half-reaction method, follow these steps:
Step 1: Separate the overall reaction into two half-reactions: oxidation and reduction.
The given reaction:
CoCl2 + HO2- → Co(OH)3 + Cl-
Oxidation half-reaction:
CoCl2 → Co(OH)3
Reduction half-reaction:
HO2- → Cl-
Step 2: Balance the atoms other than oxygen and hydrogen in each half-reaction.
Balancing the oxidation half-reaction:
CoCl2 → Co(OH)3
There is one cobalt (Co) atom on both sides, so it is already balanced. However, there are two chlorine (Cl) atoms on the left side and none on the right side. To balance the chlorine, add a coefficient of 2 in front of HCl on the right side:
CoCl2 → Co(OH)3 + 2Cl-
Step 3: Balance the oxygen atoms by adding water (H2O) to each half-reaction.
Balancing the reduction half-reaction:
HO2- → Cl-
There are two oxygen (O) atoms on the left side and none on the right side. To balance the oxygen, add a coefficient of 2 in front of the water (H2O) on the right side:
HO2- → Cl- + 2H2O
Step 4: Balance the hydrogen (H) atoms by adding hydrogen ions (H+) to each half-reaction.
Balancing the reduction half-reaction:
HO2- → Cl- + 2H2O
There are two hydrogen (H) atoms on both sides, so it is already balanced.
Step 5: Balance the charges by adding electrons (e-) to each half-reaction.
Balancing the oxidation half-reaction:
CoCl2 + 2e- → Co(OH)3 + 2Cl-
Balancing the reduction half-reaction:
HO2- + 2H+ + 2e- → Cl- + 2H2O
Note that in this step, the number of electrons added to one half-reaction must be the same as the number of electrons gained in the other half-reaction.
Step 6: Balance the electrons by multiplying one or both half-reactions as necessary.
In this case, the electrons are already balanced (both reactions have 2e- on each side).
Step 7: Combine the half-reactions by multiplying them so that the number of electrons gained in one half-reaction equals the number of electrons lost in the other half-reaction.
Multiply the oxidation half-reaction by 2:
2CoCl2 + 4e- → 2Co(OH)3 + 4Cl-
Step 8: Cancel out the common species on both sides of the equation.
The electrons cancel out, leaving us with the balanced overall reaction:
2CoCl2 + 4HO2- → 2Co(OH)3 + 4Cl-