Balance the following oxidation-reduction reaction using the half-reaction method:Al(s) + IO3-(aq) → Al3+(aq) + I-(aq)

Sure, here are the steps to balance the given redox reaction:

Step 1: Identify the oxidation and reduction half-reactions

• Al(s) → Al3+(aq) is the oxidation half-reaction (Al is losing electrons)
• IO3-(aq) → I-(aq) is the reduction half-reaction (I is gaining electrons)

Step 2: Balance the atoms other than O and H in each half-reaction

• The Al and I atoms are already balanced in both half-reactions

Step 3: Balance the O atoms by adding H2O

• For the reduction half-reaction: IO3-(aq) → I-(aq) + 3H2O(l)

Step 4: Balance the H atoms by adding H+

• For the reduction half-reaction: IO3-(aq) + 6H+(aq) → I-(aq) + 3H2O(l)

Step 5: Balance the charge by adding electrons

• For the oxidation half-reaction: Al(s) → Al3+(aq) + 3e-
• For the reduction half-reaction: IO3-(aq) + 6H+(aq) + 5e- → I-(aq) + 3H2O(l)

Step 6: Make the electron loss equal to the electron gain in the two half-reactions

• Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 3
• 5Al(s) → 5Al3+(aq) + 15e-
• 3IO3-(aq) + 18H+(aq) + 15e- → 3I-(aq) + 9H2O(l)

Step 7: Add the half-reactions together and simplify

• 5Al(s) + 3IO3-(aq) + 18H+(aq) → 5Al3+(aq) + 3I-(aq) + 9H2O(l)

So, the balanced redox reaction is: 5Al(s) + 3IO3-(aq) + 18H+(aq) → 5Al3+(aq) + 3I-(aq) + 9H2O(l)