Balance the following redox reactions in acidic solution by the half-reaction method Br–(aq) + I–(aq) + HClO(aq) = Br2(l) + IO3 + Cl–(aq)
To balance redox reactions in acidic solution using the half-reaction method, you need to follow these steps:
Step 1: Split the overall reaction into two half-reactions, one representing oxidation and the other representing reduction. In this case, we have Br–(aq) being oxidized to Br2(l) and I–(aq) being reduced to IO3–(aq).
Half-reaction for oxidation: Br–(aq) ⟶ Br2(l)
Half-reaction for reduction: I–(aq) ⟶ IO3–(aq)
Step 2: Balance the elements that are not balanced in each half-reaction, except for hydrogen and oxygen.
For the oxidation half-reaction: Br–(aq) ⟶ Br2(l)
- Since the bromines (Br) are not balanced, we need to add a coefficient of 2 in front of Br–(aq) to balance them.
2Br–(aq) ⟶ Br2(l)
For the reduction half-reaction: I–(aq) ⟶ IO3–(aq)
- Since the iodines (I) and oxygens (O) are not balanced, we need to add coefficients of 3 in front of both I–(aq) and IO3–(aq) to balance them.
3I–(aq) ⟶ IO3–(aq)
Step 3: Balance the charges in each half-reaction by adding electrons (e–).
For the oxidation half-reaction: 2Br–(aq) ⟶ Br2(l)
- The bromide ions (Br–) have a charge of -1, and to balance the charge, we need to add 2e– on the right side.
2Br–(aq) + 2e– ⟶ Br2(l)
For the reduction half-reaction: 3I–(aq) ⟶ IO3–(aq)
- The iodide ions (I–) have a charge of -1, and to balance the charge, we need to add 6e– on the left side.
6I–(aq) ⟶ IO3–(aq) + 6e–
Step 4: Multiply each half-reaction by integers to ensure that the number of electrons (e–) transferred is the same in both half-reactions. In this case, multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2 achieves this.
Multiplying the oxidation half-reaction by 3:
6Br–(aq) + 6e– ⟶ 3Br2(l)
Multiplying the reduction half-reaction by 2:
12I–(aq) ⟶ 2IO3–(aq) + 12e–
Step 5: Combine the half-reactions, canceling out the electrons, and adding the resulting equations together.
Adding the combined half-reactions:
6Br–(aq) + 6e– + 12I–(aq) ⟶ 3Br2(l) + 2IO3–(aq) + 12e–
Simplifying the equation:
6Br–(aq) + 12I–(aq) ⟶ 3Br2(l) + 2IO3–(aq)
Finally, balance any remaining atoms and charges on both sides of the equation as needed. In this case, we need to balance the charges by adding protons (H+) to the appropriate side of the equation.
Final balanced equation in acidic solution:
6Br–(aq) + 12I–(aq) + 6H+(aq) ⟶ 3Br2(l) + 2IO3–(aq) + 6H2O(l)
To balance the redox reaction Br–(aq) + I–(aq) + HClO(aq) = Br2(l) + IO3– + Cl–(aq) in acidic solution using the half-reaction method, follow the steps below:
Step 1: Divide the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Oxidation half-reaction: Br–(aq) → Br2(l)
Reduction half-reaction: I–(aq) → IO3–(aq)
Step 2: Balance atoms other than hydrogen and oxygen in each half-reaction.
Oxidation half-reaction: Br–(aq) → Br2(l) (Already balanced)
Reduction half-reaction: 3I–(aq) → IO3–(aq)
Step 3: Balance the oxygen atoms by adding water molecules (H2O) to the appropriate side of each half-reaction.
Oxidation half-reaction: Br–(aq) → Br2(l) (Already balanced)
Reduction half-reaction: 3I–(aq) → IO3–(aq) + 6H2O(l)
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side of each half-reaction.
Oxidation half-reaction: Br–(aq) → Br2(l) + 2H+(aq)
Reduction half-reaction: 3I–(aq) + 6H+(aq) → IO3–(aq) + 6H2O(l)
Step 5: Balance the charge by adding electrons (e–) to the appropriate side of each half-reaction.
Oxidation half-reaction: Br–(aq) → Br2(l) + 2H+(aq) + 2e–
Reduction half-reaction: 3I–(aq) + 6H+(aq) + 6e– → IO3–(aq) + 6H2O(l)
Step 6: Make the total number of electrons equal in both half-reactions by multiplying the half-reactions appropriately.
Multiplying the oxidation half-reaction by 3:
3Br–(aq) → 3Br2(l) + 6H+(aq) + 6e–
Step 7: Add the half-reactions together and cancel out the electrons on both sides to obtain the balanced overall redox reaction.
3Br–(aq) + 3I–(aq) + 6H+(aq) → 3Br2(l) + IO3–(aq) + 6H2O(l) + 3Cl–(aq)
Finally, the balanced redox reaction in acidic solution is:
3Br–(aq) + 3I–(aq) + 6H+(aq) = 3Br2(l) + IO3–(aq) + 6H2O(l)+3Cl–(aq)