Balance the following redox reaction and identify the oxidizing and reducing agents
A- Fe2+(aq) + Cr2072-(aq) -->Fe3+(aq) + Cr3+(aq) (acidic)
B- N2H4(g) + ClO3-(aq) -->NO2(g) + Cl- (basic)
To balance a redox equation, you need to follow these steps:
Step 1: Assign oxidation numbers to each atom in the equation.
Step 2: Identify the atoms that undergo oxidation and reduction.
Step 3: Divide the equation into two half-reactions: one for reduction and one for oxidation.
Step 4: Balance the atoms in each half-reaction by adding the appropriate number of electrons.
Step 5: Balance the charge in each half-reaction by adding H+ ions (for acidic medium) or OH- ions (for basic medium).
Step 6: Combine the two half-reactions, canceling out electrons on both sides.
Step 7: Verify that both mass and charge are balanced.
Let's go through each reaction and balance them step by step:
A) Fe2+(aq) + Cr2O7^2-(aq) → Fe3+(aq) + Cr3+(aq) (acidic)
Step 1:
Fe2+ → +2
Cr2O7^2- → +6 (Cr) and -2 (Oxygen)
Step 2:
Fe2+ is oxidized to Fe3+. It loses one electron.
Cr2O7^2- is reduced to Cr3+. It gains three electrons.
Step 3:
Oxidation Half-Reaction: Fe2+ → Fe3+
Reduction Half-Reaction: Cr2O7^2- → Cr3+
Step 4:
Oxidation Half-Reaction: Fe2+ → Fe3+ + 1e-
Reduction Half-Reaction: Cr2O7^2- + 6e- → 2Cr3+
Step 5:
Oxidation Half-Reaction: Fe2+ → Fe3+ + 1e-
Reduction Half-Reaction: Cr2O7^2- + 6e- + 14H+ → 2Cr3+ + 7H2O
Step 6:
Multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1 to balance the electrons:
6Fe2+ → 6Fe3+ + 6e-
Cr2O7^2- + 6e- + 14H+ → 2Cr3+ + 7H2O
Step 7:
Combine the two half-reactions and cancel out the electrons:
6Fe2+ + Cr2O7^2- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
Now the equation is balanced.
The oxidizing agent is Cr2O7^2-, which gains electrons and gets reduced.
The reducing agent is Fe2+, which loses electrons and gets oxidized.
B) N2H4(g) + ClO3-(aq) → NO2(g) + Cl- (basic)
The balancing steps are the same as above, but in this case, we have a basic medium. So we will follow the steps with OH- ions instead of H+ ions.
Step 1:
N2H4 → 0
ClO3- → +5 (Cl) and -2 (Oxygen)
NO2 → +4 (N) and -2 (Oxygen)
Cl- → -1
Step 2:
N2H4 is oxidized to NO2. It loses four electrons.
ClO3- is reduced to Cl-. It gains five electrons.
Step 3:
Oxidation Half-Reaction: N2H4 → NO2
Reduction Half-Reaction: ClO3- → Cl-
Step 4:
Oxidation Half-Reaction: N2H4 → NO2 + 4e-
Reduction Half-Reaction: ClO3- + 5e- → Cl-
Step 5:
Oxidation Half-Reaction: N2H4 → NO2 + 4e-
Reduction Half-Reaction: ClO3- + 5e- + 6OH- → Cl- + 3H2O
Step 6:
Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 4 to balance the electrons:
5N2H4 → 5NO2 + 20e-
4ClO3- + 20e- + 24OH- → 4Cl- + 12H2O
Step 7:
Combine the two half-reactions and cancel out the electrons:
5N2H4 + 4ClO3- + 20OH- → 5NO2 + 4Cl- + 12H2O
Now the equation is balanced.
The oxidizing agent is ClO3-, which gains electrons and gets reduced.
The reducing agent is N2H4, which loses electrons and gets oxidized.