Question

Use the half-reaction method to balance the following equation which is in an acidic

solution:
ClO3-(aq) + I- (aq)
+I2 (s) + Cl- (aq)

Answer 1

Answer:

ClO₃⁻ + 6I⁻ + 6H⁺ ⟶ Cl⁻ + 3I₂ + 3H₂O

Step-by-step 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.

`\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}`

Step 11. Check that charge is balanced

`\begin{array}{rl}\textbf{On the left} & \textbf{On the right}\\6+ + \; 7- = & 1-\\1- =& 1-\\\end{array}`

Everything checks. The balanced equation is

ClO₃⁻ + 6I⁻ + 6H⁺ ⟶ Cl⁻ + 3I₂ + 3H₂O