Balance the redox reaction below using the half-reaction method.
Bi3+(aq) + Co(s)Bi(s) + Co2+(aq)
I can do this and you want learn anything. What is it you don't understand about how to do it.
Bi goes from 3+ on the left to zero on the right. You should find the arrow key and use it.
Co goes from zero on the left to 2| on the right.
To balance the redox reaction using the half-reaction method, we need to separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Step 1: Identify the oxidation and reduction half-reactions.
In the given reaction, Bi3+ is being reduced to Bi and Co is being oxidized to Co2+.
The oxidation half-reaction is: Co → Co2+
The reduction half-reaction is: Bi3+ → Bi
Step 2: Balance the atoms other than hydrogen and oxygen in the half-reactions.
In the oxidation half-reaction, there is only one Co atom on both sides, so it is already balanced.
In the reduction half-reaction, there is one Bi atom on both sides, so it is already balanced.
Step 3: Balance the charge in the half-reactions.
In the oxidation half-reaction, Co has an oxidation state of 0 on the left side and an oxidation state of +2 on the right side. To balance the charge, we need to add two electrons (e-) to the left side:
Co → Co2+ + 2e-
In the reduction half-reaction, Bi3+ has an oxidation state of +3 on the left side and an oxidation state of 0 on the right side. To balance the charge, we need to add three electrons (3e-) to the right side:
3Bi3+ + 3e- → 3Bi
Step 4: Balance the number of electrons in both half-reactions.
To make the number of electrons equal in both half-reactions, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
3Co → 3Co2+ + 6e-
6Bi3+ + 6e- → 6Bi
Step 5: Combine the balanced half-reactions.
Add the two balanced half-reactions together:
3Co + 6Bi3+ + 6e- → 3Co2+ + 6Bi
Finally, simplify the equation by canceling out the electrons and any common species on both sides. The final balanced redox reaction is:
3Co + 6Bi3+ → 3Co2+ + 6Bi