Yes, in order to balance this equation in a basic solution, it is helpful to use half-reactions. Let's proceed step-by-step:
Step 1: Separate the equation into two half-reactions.
The first half-reaction involves the oxidation of bromide ions (Br-) to bromine (Br2):
Br- → Br2
The second half-reaction involves the reduction of chromate ions (CrO4{2-}) to chromium hydroxide (Cr(OH)3):
CrO4{2-} → Cr(OH)3
Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction.
In the first half-reaction, there is already one bromine atom on both sides.
In the second half-reaction, there is one chromium (Cr) atom on both sides.
Step 3: Balance the oxygen atoms by adding water (H2O) molecules.
In the first half-reaction, there are zero oxygen atoms on both sides, so no water molecules are needed.
In the second half-reaction, there are four oxygen atoms on the left side (CrO4{2-}) and zero on the right side (Cr(OH)3). Therefore, four water molecules need to be added to the right side:
CrO4{2-} + 4H2O → Cr(OH)3
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+).
In the first half-reaction, there are two hydrogen atoms on the left side, so two hydrogen ions (H+) need to be added on the right side:
2Br- → Br2 + 2H+
In the second half-reaction, there are already twelve hydrogen atoms on both sides.
Step 5: Balance the charges by adding electrons (e-).
In the first half-reaction, the charge on both sides is balanced.
In the second half-reaction, there is a total charge of 8- on the left side (CrO4{2-}) and 3+ on the right side (Cr(OH)3). Therefore, 11 electrons need to be added to the left side:
CrO4{2-} + 4H2O + 11e- → Cr(OH)3
Step 6: Equalize the number of electrons in both half-reactions.
Multiply the first half-reaction by 11:
22Br- → 11Br2 + 22H+
Now the number of electrons in both half-reactions is the same.
Step 7: Combine both half-reactions.
Add the two half-reactions together:
22Br- + CrO4{2-} + 4H2O + 11e- → 11Br2 + 22H+ + Cr(OH)3
Step 8: Simplify the equation.
To simplify and convert this equation into the basic solution, you need to neutralize the hydrogen ions (H+) by adding hydroxide ions (OH-). The number of hydroxide ions added should be equal to the number of hydrogen ions.
You can combine the hydrogen ions and hydroxide ions to form water molecules (H2O). In this case, 22 hydroxide ions need to be added:
22Br- + CrO4{2-} + 4H2O + 11e- + 22OH- → 11Br2 + 22H2O + Cr(OH)3
Now the equation is balanced in a basic solution.