Balance the following equation in basic solution. Phases are optional.
I{-}+IO4{-} --> I3{-}+IO3{-}
An equation like this is confusing because I^-(on the left) is going to I2 (on the right) but it then dissolves in KI to make KI3 so let's simplify it by taking the KI out of it, balance it, then put the KI back in.
2I^- --> I2
I is -2 on left and 0 on right. Balance with e change.
2I^- ==> I2 + 2e
Charge on left is 2- and 2- on right. No OH^- needed.
IO4^- --> IO3^-
I on left is +7 and +5 on right. Add e
IO4^- + 2e ==> IO3^-
Charge on left is -3 and on right -1. Add OH to balance
IO4^- + 2e ==> IO3^- + 2OH^-
Now add H2O to balance.
IO4^- + 2e + H2O ==> IO3^- + 2OH^-
Both half rxns have 23 change; therefore, we can add the two to obtain
2I^- + IO4^- + H2O ==> I2 + IO3^- + 2OH^-
Now we add the I^- back in to make that I2 on the right I3^- and the final becomes
3I^- + IO4^- + H2O ==> I3^- + IO3^- + 2OH^-
To balance the given equation in basic solution, we need to follow these steps:
Step 1: Separate the equation into two half-reactions. One for the oxidation half-reaction and the other for the reduction half-reaction.
Oxidation half-reaction: I⁻ → I₃⁻
Reduction half-reaction: IO₄⁻ → IO₃⁻
Step 2: Balance the atoms that are not oxygen or hydrogen. In this case, we only have iodine (I) atoms.
Oxidation half-reaction: I⁻ → I₃⁻ (already balanced in terms of iodine atoms)
Reduction half-reaction: IO₄⁻ → IO₃⁻ (already balanced in terms of iodine atoms)
Step 3: Balance the oxygen atoms by adding water molecules to the side that is deficient in oxygen. Each water molecule (H₂O) contributes one oxygen atom.
Oxidation half-reaction: I⁻ → I₃⁻ + 3H₂O
Reduction half-reaction: IO₄⁻ + H₂O → IO₃⁻
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H⁺) to the side that is deficient in hydrogen.
Oxidation half-reaction: I⁻ → I₃⁻ + 3H₂O + 6H⁺
Reduction half-reaction: IO₄⁻ + H₂O → IO₃⁻ + 2H⁺
Step 5: Balance the charges by adding electrons (e⁻) to the side that is more positive in charge.
Oxidation half-reaction: I⁻ → I₃⁻ + 3H₂O + 6H⁺ + 6e⁻
Reduction half-reaction: IO₄⁻ + H₂O + 6e⁻ → IO₃⁻ + 2H⁺
Step 6: Multiply the half-reactions by appropriate coefficients, so the number of electrons in both reactions cancels out.
6I⁻ → I₃⁻ + 3H₂O + 6H⁺ + 6e⁻
6IO₄⁻ + 6H₂O + 6e⁻ → 6IO₃⁻ + 12H⁺
Step 7: Finally, combine both half-reactions and cancel out common species on both sides.
6I⁻ + 6IO₄⁻ + 6H₂O → 6I₃⁻ + 6IO₃⁻ + 12H⁺
Therefore, the balanced equation in basic solution is:
6I⁻ + 6IO₄⁻ + 6H₂O → 6I₃⁻ + 6IO₃⁻ + 12H⁺
To balance the given equation in a basic solution, follow these steps:
Step 1: Write the unbalanced equation:
I{-} + IO4{-} → I3{-} + IO3{-}
Step 2: Determine the non-oxygen and non-hydrogen atoms and balance them first. In this case, we have only iodine (I) and iodate (IO3) atoms.
I{-} + IO4{-} → I3{-} + IO3{-}
Step 3: Balance the atoms by adding coefficients:
2I{-} + IO4{-} → I3{-} + IO3{-}
Step 4: Balance the oxygen atoms by adding water (H2O) molecules:
2I{-} + 8H2O + IO4{-} → I3{-} + 4H2O + IO3{-}
Step 5: Balance the hydrogen atoms by adding hydroxide (OH-) ions:
2I{-} + 8H2O + IO4{-} → I3{-} + 4H2O + IO3{-} + 10OH-
Step 6: Balance the charge by adding electrons (e-) where necessary:
2I{-} + 8H2O + IO4{-} → I3{-} + 4H2O + IO3{-} + 10OH- + 6e-
Step 7: Combine the half-reactions and cancel out common species:
I- + 3OH- → I3- + 2H2O + 6e-
IO4- + 12e- + 4H2O → IO3- + 8OH-
Multiply the first half-reaction by 2 and the second half-reaction by 3:
2I- + 6OH- → 2I3- + 4H2O + 12e-
3IO4- + 36e- + 12H2O → 3IO3- + 24OH-
Finally, add the two half-reactions together:
2I- + 6OH- + 3IO4- + 12e- + 12H2O → 2I3- + 4H2O + 12e- + 3IO3- + 24OH-
Cancel out the electrons and water molecules:
2I- + 6OH- + 3IO4- + 12H2O → 2I3- + 3IO3- + 24OH-
Simplify the equation:
2I- + 3IO4- + 6OH- → 2I3- + 3IO3- + 12H2O