Question

potassium chlorate reacts with sulfur dioxide in acidic solution according to the skeleton equation below. balance the reaction using the half - reaction method.
\ce{clo^{-}_{3} + so_{2} -> so^{2-}_{4} + cl^{-}}
what is the coefficient for the number of water molecules in the reduction half - reaction?
\ce{clo^{-}_{3} -> cl^{-}}

Explanation:

Step1: Balance Cl atoms (already balanced, 1 each)

$\ce{ClO^{-}_{3} -> Cl^{-}}$

Step2: Balance O atoms by adding $\ce{H_{2}O}$

$\ce{ClO^{-}_{3} -> Cl^{-} + 3H_{2}O}$ (3 O on left, so 3 $\ce{H_{2}O}$ on right)

Step3: Balance H atoms by adding $\ce{H^{+}}$

$\ce{ClO^{-}_{3} + 6H^{+} -> Cl^{-} + 3H_{2}O}$ (6 H on left from $\ce{6H^{+}}$, 6 H on right from $\ce{3H_{2}O}$)

Step4: Balance charge by adding $\ce{e^{-}}$ (not needed for H₂O coefficient)

Answer:

3