Balance the following redox equations. All occur in Acidic solutions.
1. Cr2O7^2-^ + C2O4^2-^ -> Cr^3+^ + CO2
2. Cu + NO3^-^ -> CU^2+^ + NO
3. MnO2+ HNO2 -> Mn^2+^ + NO3^-^
4. PbO2 + Mn^2+^ + SO4^2-^ -> PbSO4 + MnO4^-^
5. HNO2 + Cr2O7^2-^ -> Cr^3+^ + NO3^-^
Step1: assign oxidation numbers.
Cr2O72- + C2O42- → Cr3+ + CO2
+6/-2 +3/-2 +3 +4/-2
Step2: Separate the overall reaction into two separate half reactions. One will be the oxidation reaction(where the oxidation number increased) and the other will be the reduction reaction( where the oxidation numbers decreased).
Oxidation:
C2O42- →CO2
Reduction:
Cr2O72- → Cr3+
Step 3: Balance each half-reaction in the following order:
First, balance all elements other than Hydrogen and Oxygen.
C2O42- →2CO2
Cr2O72- → 2Cr3+
Second, balance Oxygen by adding H2O.
C2O42- →2CO2
Cr2O72- → 2Cr3+ + 7H2O
Third, balance Hydrogen by adding H+.
C2O42- →2CO2
14H+ + Cr2O72- → 2Cr3+ + 7H2O
Step 4: balance each half reaction with respect to charge by adding electrons. The sum of the charges on both sides of the equation should be equal.
C2O42- →2CO2 + 2e–
6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O
Step 5: Make sure the number of electrons in both half reactions equal by multiplying one or both half reactions by a small whole number.
3 x [C2O42- →2CO2 + 2e–] = 3C2O42- →6CO2 + 6e–
6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O
Step 6: Add the two half reactions together canceling electrons and other species as necessary.
3C2O42- →6CO2 + 6e–
+
6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O
3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2
Final Answer: 3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2
Hope this helps.
-CoolChemTutor
Balancing oxidation-reduction (redox) reaction is a complex process. You must know how to assign some oxidation numbers and how to calculate others no matter which method you use. The best method in my opinion is the “Ion-Electron” method. This is not a lesson on how to do it. Your textbook and your class notes should take care of that. It is only a demonstration using your first reaction:
A. Determine which element is reduced and which is oxidized.
1. Find out which element is reduced (Its oxidation number decreases). The oxidation number of Cr in Cr2O7^-2 is found by assigning -2 as the oxidation number of O, and x to Cr:
2x + (-2)(7) = -2 (the -2 on the right side is the ionic charge)
Solving for x we get x = +6. So Cr on the left side = +6 on the right side it is +3.
2. Find out which element is oxidized. The only possibility is C. Why? Looking at C2O4^-2, we can set up:
2y + (-2)(4) = -2, and y = 3.
B. Write incomplete half reaction for reduction and for oxidation, then complete them and balance them for number of atoms and electrical charge:
1. Reduction
Cr2O7^-2 —> Cr^+3
(balancing atoms and charge by adding H+, H2O, and e^- as needed)
Cr2O7^-2(aq) + 14H^+(aq) + 6e^- —> 2Cr^+3 + 7H2O
(check number of atoms of each kind and total charge on both sides)
2. Oxidation
C2O4^2- —> CO2
(balancing atoms and charge)
C2O4^2- —> 2CO2 + 2e^-
C. Rewrite and reconcile the two half reactions, then add them and simplify:
Cr2O7^-2(aq) + 14H^+(aq) + 6e^- —> 2Cr^+3 + 7H2O
3C2O4^2- —> 6CO2 + 6e^-
(I multiplied the oxidation by 3 so that the total number of electrons transferred will be 6 as it is in the reduction)
Cr2O7^-2(aq) + 14H^+(aq) + 6e^- + 3C2O4^2- —> 2Cr^+3 + 7H2O + 6CO2 + 6e^-
(added left and right sides ... then we simplify through cancellations)
Cr2O7^-2(aq) + 14H^+(aq) + 3C2O4^2- —> 2Cr^+3 + 7H2O + 6CO2
Step1: assign oxidation numbers.
Cr2O72- + C2O42- → Cr3+ + CO2
+6/-2 +3/-2 +3 +4/-2
Step2: Separate the overall reaction into two separate half reactions. One will be the oxidation reaction(where the oxidation number increased) and the other will be the reduction reaction( where the oxidation numbers decreased).
Oxidation:
C2O42- →CO2
Reduction:
Cr2O72- → Cr3+
Step 3: Balance each half-reaction in the following order:
First, balance all elements other than Hydrogen and Oxygen.
C2O42- →2CO2
Cr2O72- → 2Cr3+
Second, balance Oxygen by adding H2O.
C2O42- →2CO2
Cr2O72- → 2Cr3+ + 7H2O
Third, balance Hydrogen by adding H+.
C2O42- →2CO2
14H+ + Cr2O72- → 2Cr3+ + 7H2O
Step 4: balance each half reaction with respect to charge by adding electrons. The sum of the charges on both sides of the equation should be equal.
C2O42- →2CO2 + 2e–
6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O
Step 5: Make sure the number of electrons in both half reactions equal by multiplying one or both half reactions by a small whole number.
3 x [C2O42- →2CO2 + 2e–] = 3C2O42- →6CO2 + 6e–
6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O
Step 6: Add the two half reactions together canceling electrons and other species as necessary.
3C2O42- →6CO2 + 6e–
+
6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O
3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2
Final Answer: 3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2
Cr2O7^-2 + 12H+ +C2O4^-2 -----> 2Cr^2+ + 2CO2 + 6H2O
To balance redox equations in acidic solutions, you need to follow the steps given below:
Step 1: Write the unbalanced equation.
Write the unbalanced equation, showing the reactants and products with their respective charges.
Step 2: Separate the equation into two half-reactions.
Split the equation into two half-reactions: one for the reduction half-reaction (gaining electrons) and one for the oxidation half-reaction (losing electrons). Identify the oxidized and reduced species.
Step 3: Balance atoms other than hydrogen and oxygen.
Balance the atoms other than hydrogen and oxygen in each half-reaction, making sure that the number of atoms is the same on both sides.
Step 4: Balance oxygen atoms.
Add water (H2O) molecules to balance the oxygen atoms. Add the same number of water molecules to both sides of the equation, as necessary.
Step 5: Balance hydrogen atoms.
Add hydrogen ions (H+) to balance the hydrogen atoms. Add the same number of hydrogen ions to both sides of the equation, as necessary.
Step 6: Balance charges.
Balance the charges by adding electrons (e-) to the side that needs charge balance. Add the same number of electrons to both sides of the equation, as necessary.
Step 7: Multiply the half-reactions.
Multiply each half-reaction by a whole number so that the number of electrons gained in one half-reaction is equal to the number of electrons lost in the other half-reaction. This is done to make sure that the number of electrons cancels out when the two half-reactions are combined.
Step 8: Combine the half-reactions.
Add the two half-reactions together, cancelling out the electrons. This will give you the balanced redox equation.
Here are the balanced redox equations for the provided reactions:
1. Cr2O7^2- + 3C2O4^2- + 14H+ -> 2Cr^3+ + 6CO2 + 7H2O
(Oxidation half-reaction: Cr2O7^2- -> 2Cr^3+)
(Reduction half-reaction: C2O4^2- -> 2CO2)
2. 3Cu + 8NO3^- + 4H+ -> 3Cu^2+ + 4NO + 2H2O
(Oxidation half-reaction: Cu -> Cu^2+)
(Reduction half-reaction: NO3^- -> NO)
3. MnO2 + 2HNO2 -> Mn^2+ + 2NO3^- + H2O
(Oxidation half-reaction: MnO2 -> Mn^2+)
(Reduction half-reaction: HNO2 -> NO3^-)
4. PbO2 + 4H+ + Mn^2+ + 2SO4^2- -> PbSO4 + MnO4^- + 2H2O
(Oxidation half-reaction: PbO2 + 4H+ + 2SO4^2- -> PbSO4 + 2H2O)
(Reduction half-reaction: Mn^2+ -> MnO4^-)
5. HNO2 + Cr2O7^2- + 8H+ -> 2Cr^3+ + 3NO3^- + 4H2O
(Oxidation half-reaction: HNO2 -> NO3^-)
(Reduction half-reaction: Cr2O7^2- -> 2Cr^3+)