Balance the following equation in an acidic and basic medium
* Cr^(+3)+BrO^(-)--->CrO4^(2-)+Br^(-)
Cr^3+ + 4H2O==> CrO4^2- + 3e + 8H^+
BrO^- + 2e + 2H^+ ==> Br^- + H2O
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Multiply eqn 1 by 2 and eqn 2 by 3, then add to obtain the complete balanced redox equation.
Is it in basic medium
To balance the given equation in both acidic and basic mediums, we will follow the steps below:
Balancing in acidic medium:
1. Separate the redox half-reactions:
Cr^(+3) → CrO4^(2-)
BrO^(-) → Br^(-)
2. Balance the atoms other than oxygen and hydrogen.
For the reduction half-reaction:
Cr^(+3) → CrO4^(2-)
The Cr atom is already balanced.
For the oxidation half-reaction:
BrO^(-) → Br^(-)
The Br atom is already balanced.
3. Balance the oxygen atoms using water (H2O):
For the reduction half-reaction:
Cr^(+3) → CrO4^(2-)
There are four oxygen atoms on the right side, but none on the left side. Add four H2O molecules to the left side to balance the oxygen atoms:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
For the oxidation half-reaction:
BrO^(-) → Br^(-)
There are four oxygen atoms on the left side and none on the right side. Add four H2O molecules to the right side to balance the oxygen atoms:
BrO^(-) + 4H2O → Br^(-) + 4H^+
4. Balance the hydrogen atoms using H^+ (acidic medium) or OH^-(basic medium):
For the reduction half-reaction:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
There are eight hydrogen atoms on the left side and four on the right side. Add four H^+ ions to the right side to balance the hydrogen atoms:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
For the oxidation half-reaction:
BrO^(-) + 4H2O → Br^(-) + 4H^+
There are four hydrogen atoms on the left side and four on the right side. No additional balancing is required for the hydrogen atoms.
5. Balance the charges by adding electrons (e^-):
For the reduction half-reaction:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
The charge on the left side is 3+ and on the right side is 2-. To balance the charges, add 6 electrons (e^-) to the left side:
Cr^(+3) + 6e^- + 4H2O → CrO4^(2-) + 4H^+
For the oxidation half-reaction:
BrO^(-) + 4H2O → Br^(-) + 4H^+
There is no charge on either side, so no electrons need to be added.
6. Multiply each half-reaction by an integer to equalize the number of electrons transferred. In this case, we need to multiply the oxidation half-reaction by 6:
6BrO^(-) + 24H2O → 6Br^(-) + 24H^+
The balanced equation in acidic medium is:
Cr^(+3) + 6H2O + 6BrO^(-) → CrO4^(2-) + 6H^+ + 6Br^(-)
Balancing in basic medium:
1. Follow steps 1 and 2 from the acidic medium.
2. Balance the oxygen atoms using water (H2O):
For the reduction half-reaction:
Cr^(+3) → CrO4^(2-)
Add four H2O molecules to the left side to balance the oxygen atoms:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
For the oxidation half-reaction:
BrO^(-) → Br^(-)
Add four OH^- ions to the right side to balance the oxygen atoms:
BrO^(-) + 4OH^- → Br^(-) + 2H2O
3. Balance the hydrogen atoms using H^+ (acidic medium) or OH^-(basic medium):
For the reduction half-reaction:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
There are eight hydrogen atoms on the left side and four on the right side. Add four H^+ ions to the right side to balance the hydrogen atoms:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
For the oxidation half-reaction:
BrO^(-) + 4OH^- → Br^(-) + 2H2O
There are four hydrogen atoms on the left side and four on the right side. No additional balancing is required for the hydrogen atoms.
4. Balance the charges by adding electrons (e^-):
For the reduction half-reaction:
Cr^(+3) + 4H2O → CrO4^(2-) + 4H^+
The charge on the left side is 3+ and on the right side is 2-. To balance the charges, add 1 electron (e^-) to the left side:
Cr^(+3) + e^- + 4H2O → CrO4^(2-) + 4H^+
For the oxidation half-reaction:
BrO^(-) + 4OH^- → Br^(-) + 2H2O
There is no charge on either side, so no electrons need to be added.
5. Multiply each half-reaction by an integer to equalize the number of electrons transferred. In this case, we need to multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2:
2BrO^(-) + 8OH^- → 2Br^(-) + 4H2O
The balanced equation in basic medium is:
3Cr^(+3) + 3e^- + 12H2O + 2BrO^(-) + 8OH^- → 3CrO4^(2-) + 12H^+ + 2Br^(-) + 4H2O