Question 19.1: Permanganate ion and iodide ion react in basic solution to p...
Permanganate ion and iodide ion react in basic solution to produce manganese(IV) oxide and molecular iodine. Use the half-reaction method to balance the equation:
\text{MnO}_4^– + \text{I}^– → \text{MnO}_2 + \text{I}_2
Strategy The reaction takes place in basic solution, so apply steps 1 through 9 to balance for mass and for charge.
Setup Identify the oxidation and reduction half-reactions by assigning oxidation numbers.
\underset{\fbox{+7}\ \fbox{–2}}{\text{MnO}_4^–}+ \underset{\fbox{–1}}{ \text{ I}^– } → \underset{\fbox{+4}\ \fbox{–2}}{\text{MnO}_2} + \underset{\fbox{0}}{\text{I}_2}
Step 1. Separate the unbalanced reaction into half-reactions
Oxidation : \text{I}^– → \text{I}_2
Reduction : \text{MnO}_4^– → \text{MnO}_2
Step 2. Balance each half-reaction for mass, excluding \text{O} and \text{H}
2\text{I}^– → \text{I}_2
\text{MnO}_4^– → \text{MnO}_2
Step 3. Balance both half-reactions for \text{O} by adding \text{H}_2\text{O}.
2\text{I}^– → \text{I}_2
\text{MnO}_4^– → \text{MnO}_2 + 2\text{H}_2\text{O}
Step 4. Balance both half-reactions for \text{H} by adding \text{H}^+.
2\text{I}^– → \text{I}_2
4\text{H}^+ + \text{MnO}_4^–→ \text{MnO}_2 + 2\text{H}_2\text{O}
Step 5. Balance the total charge of both half-reactions by adding electrons
2\text{I}^– → \text{I}_2 + 2e^–
3e^– + 4\text{H}^+ + \text{MnO}_4^– → \text{MnO}_2 + 2\text{H}_2\text{O}
Step 6. Multiply the half-reactions to make the numbers of electrons the same in both.
3(2\text{I}^– → \text{I}_2 + 2e^–)
2(3e^– + 4\text{H}^+ + \text{MnO}_4^– → \text{MnO}_2 + 2\text{H}_2\text{O})
Step 7. Add the half-reactions back together, canceling electrons.
6\text{I}^– → 3\text{I}_2 + \cancel{6e^–}
\underline{\cancel{6e^–} + 8\text{H}^+ + 2\text{MnO}_4^– → 2\text{MnO}_2 + 4\text{H}_2\text{O}}
8\text{H}^+ + 2\text{MnO}_4^– + 6\text{I}^– → 2\text{MnO}_2 + 3\text{I}_2 + 4\text{H}_2\text{O}
Step 8. For each \text{H}^+ ion in the final equation, add one \text{OH}^– ion to each side of the equation, combining the \text{H}^+ and \text{OH}^– ions to produce \text{H}_2\text{O}.
8\text{H}^+ + 2\text{MnO}_4^– + 6\text{I}^– → 2\text{MnO}_2 + 3\text{I}_2 + 4\text{H}_2\text{O}
\underline{+ 8\text{OH}^– + 8\text{OH}^– }
\boxed{8\text{H}_2\text{O}} + 2\text{MnO}_4^– + 6\text{I}^– → 2\text{MnO}_2 + 3\text{I}_2 + \boxed{4\text{H}2\text{O}} + 8\text{OH}^–
Step 9. Carry out any cancellations made necessary by the additional \text{H}_2\text{O} molecules.
4\text{H}_2\text{O} + 2\text{MnO}_4^– + 6\text{I}^– → 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^–