Using the half-reaction method, balance the redox reaction below. Show your work; partial credit will be given.
I- + Br2 → IO3- + Br-
Write the reduction and oxidation half-reactions. Balance them for atoms. Balance each of the half-reactions for charge. Multiply the half-reactions by factors that make the total charge transfer equal. Add the half-reactions, canceling electrons and species.
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To balance the given redox reaction using the half-reaction method, follow these steps:
Step 1: Identify the reduction and oxidation half-reactions.
The reduction half-reaction occurs when an element gains electrons, while the oxidation half-reaction occurs when an element loses electrons. In this case:
Reduction: Br2 → 2Br-
Oxidation: I- → IO3-
Step 2: Balance the half-reactions for atoms.
In this step, balance the elements other than hydrogen and oxygen in each half-reaction, making sure that the number and type of atoms are the same on both sides.
Reduction: Br2 → 2Br-
Oxidation: 6I- → IO3-
Step 3: Balance the half-reactions for charge.
Balancing the half-reactions for charge involves adjusting the number of electrons gained or lost in each half-reaction to make the total charge the same on both sides.
Reduction: Br2 + 2e- → 2Br-
Oxidation: 6I- → IO3- + 6e-
In the reduction half-reaction, 2 electrons are gained, while in the oxidation half-reaction, 6 electrons are lost.
Step 4: Multiply the half-reactions by factors to equalize the total charge transfer.
To balance the total charge transfer, multiply the reduction half-reaction by 6 and the oxidation half-reaction by 1.
6(Br2 + 2e- → 2Br-)
6(6I- → IO3- + 6e-)
This yields:
6Br2 + 12e- → 12Br-
6(6I- → IO3- + 6e-)
Simplified:
6Br2 + 12e- → 12Br-
6I- → IO3- + 6e-
Step 5: Add the half-reactions, canceling electrons and species.
Combine the two half-reactions by adding them together, ensuring that the number of electrons transferred is equal on both sides.
6Br2 + 12e- + 6I- → 12Br- + 6I- + IO3- + 6e-
Simplify and cancel out the common species:
6Br2 + 6I- → 12Br- + 6I- + IO3-
Final balanced redox equation:
6Br2 + 6I- → 12Br- + 6I- + IO3-