Sagot :
Solution:
Step 1: Separate the equation into two half-reactions.
Reduction: [tex]\text{Br}_{2} → \text{Br}^{-}[/tex]
Oxidation: [tex]\text{Br}_{2} → \text{Br}\text{O}_{3}^{-}[/tex]
Step 2: Balance each half-reaction for number and type of atoms and charges. For reactions in an acidic medium, add [tex]H_{2}O[/tex] to balance the O atoms and add [tex]H^{+}[/tex] to balance the H atoms.
For reduction half-reaction:
[tex]\text{Br}_{2} → \text{Br}^{-}[/tex]
[tex]\text{Br}_{2} → 2\text{Br}^{-}[/tex]
[tex]\text{Br}_{2} + 2e^{-} → 2\text{Br}^{-}[/tex]
For oxidation half-reaction:
[tex]\text{Br}_{2} → \text{Br}\text{O}_{3}^{-}[/tex]
[tex]\text{Br}_{2} + 6\text{H}_{2}\text{O} → 2\text{Br}\text{O}_{3}^{-}[/tex]
[tex]\text{Br}_{2} + 6\text{H}_{2}\text{O} → 2\text{Br}\text{O}_{3}^{-} + 12\text{H}^{+} + 10e^{-}[/tex]
Step 3: Multiply both half-reactions by a certain number to equalize the number of electrons.
For reduction half-reaction:
[tex]\text{Br}_{2} + 2e^{-} → 2\text{Br}^{-}[/tex]
[tex](\text{Br}_{2} + 2e^{-} → 2\text{Br}^{-}) × 5[/tex]
[tex]5\text{Br}_{2} + 10e^{-} → 10\text{Br}^{-}[/tex]
For oxidation half-reaction:
[tex]\text{Br}_{2} + 6\text{H}_{2}\text{O} → 2\text{Br}\text{O}_{3}^{-} + 12\text{H}^{+} + 10e^{-}[/tex]
[tex](\text{Br}_{2} + 6\text{H}_{2}\text{O} → 2\text{Br}\text{O}_{3}^{-} + 12\text{H}^{+} + 10e^{-}) × 1[/tex]
[tex]\text{Br}_{2} + 6\text{H}_{2}\text{O} → 2\text{Br}\text{O}_{3}^{-} + 12\text{H}^{+} + 10e^{-}[/tex]
Step 4: Add the two half-reactions.
[tex]5\text{Br}_{2} + 10e^{-} → 10\text{Br}^{-}[/tex]
[tex]\underline{\text{Br}_{2} + 6\text{H}_{2}\text{O} → 2\text{Br}\text{O}_{3}^{-} + 12\text{H}^{+} + 10e^{-}}[/tex]
[tex]6\text{H}_{2}\text{O} + 6\text{Br}_{2} → 2\text{Br}\text{O}_{3}^{-} + 10\text{Br}^{-} + 12\text{H}^{+}[/tex]
Reducing the coefficients to lowest terms
[tex]3\text{H}_{2}\text{O} + 3\text{Br}_{2} → \text{Br}\text{O}_{3}^{-} + 5\text{Br}^{-} + 6\text{H}^{+}[/tex]
Step 5: For reactions in basic medium, we need to add equal [tex]OH^{-}[/tex] ions for every [tex]H^{+}[/tex] ion to both sides of the equation.
[tex]3\text{H}_{2}\text{O} + 3\text{Br}_{2} → \text{Br}\text{O}_{3}^{-} + 5\text{Br}^{-} + 6\text{H}^{+}[/tex]
[tex]3\text{H}_{2}\text{O} + 3\text{Br}_{2} + 6\text{OH}^{-} → \text{Br}\text{O}_{3}^{-} + 5\text{Br}^{-} + 6\text{H}^{+} + 6\text{OH}^{-}[/tex]
Step 6: Combine the [tex]H^{+}[/tex] and [tex]OH^{-}[/tex] ions to form water.
[tex]3\text{H}_{2}\text{O} + 3\text{Br}_{2} + 6\text{OH}^{-} → \text{Br}\text{O}_{3}^{-} + 5\text{Br}^{-} + 6\text{H}^{+} + 6\text{OH}^{-}[/tex]
[tex]3\text{H}_{2}\text{O} + 3\text{Br}_{2} + 6\text{OH}^{-} → \text{Br}\text{O}_{3}^{-} + 5\text{Br}^{-} + 6\text{H}_{2}\text{O}[/tex]
[tex]\boxed{6\text{OH}^{-} + 3\text{Br}_{2} → \text{Br}\text{O}_{3}^{-} + 5\text{Br}^{-} + 3\text{H}_{2}\text{O}}[/tex]
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