Answer:
ClO₃⁻ + 6I⁻ + 6H⁺ ⟶ Cl⁻ + 3I₂ + 3H₂O
Explanation:
Step 1. Write the skeleton equation
ClO₃⁻ + I⁻ ⟶ I₂ + Cl⁻
Step 2. Separate into two half-reactions.
ClO₃⁻ ⟶ Cl⁻
I⁻ ⟶ I₂
Step 3. Balance all atoms other than H and O
ClO₃⁻ ⟶ Cl⁻
2I⁻ ⟶ I₂
Step 4. Balance O by adding H₂O molecules to the deficient side.
ClO₃⁻ ⟶ Cl⁻ + 3H₂O
2I⁻ ⟶ I₂
Step 5. Balance H by adding H⁺ ions to the deficient side.
ClO₃⁻ + 6H⁺ ⟶ Cl⁻ + 3H₂O
2I⁻ ⟶ I₂
Step 6. Balance charge by adding electrons to the deficient side.
ClO₃⁻ + 6H⁺ + 6e⁻ ⟶ Cl⁻ + 3H₂O
2I⁻ ⟶ I₂ + 2e⁻
Step 7. Multiply each half-reaction by a number to equalize the electrons transferred.
1 × [ClO₃⁻ + 6H⁺ + 6e⁻ ⟶ Cl⁻ + 3H₂O]
3 × [ 2I⁻ ⟶ I₂ + 2e⁻]
Step 8. Add the two half-reactions.
ClO₃⁻ + 6H⁺ + 6e⁻ ⟶ Cl⁻ + 3H₂O
6I⁻ ⟶ 3I₂ + 6e⁻
ClO₃⁻ + 6I⁻ + 6H⁺ + 6e⁻ ⟶ Cl⁻ + 3I₂ + 3H₂O + 6e⁻
Step 9. Cancel species that occur on each side of the equation
ClO₃⁻ + 6I⁻ + 6H⁺ + 6e⁻ ⟶ Cl⁻ + 3I₂ + 3H₂O + 6e⁻
becomes
ClO₃⁻ + 6I⁻ + 6H⁺ ⟶ Cl⁻ + 3I₂ + 3H₂O
Step 10. Check that all atoms are balanced.
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Cl} & 1 & 1\\\text{O} & 3 & 3\\\text{I} & 6 & 6\\\text{H} & 6 & 6\\\end{array}[/tex]
Step 11. Check that charge is balanced
[tex]\begin{array}{rl}\textbf{On the left} & \textbf{On the right}\\6+ + \; 7- = & 1-\\1- =& 1-\\\end{array}[/tex]
Everything checks. The balanced equation is
ClO₃⁻ + 6I⁻ + 6H⁺ ⟶ Cl⁻ + 3I₂ + 3H₂O
