Question

Be sure to answer all parts.

Answer 1

Answer:

a. 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

b. Br₂

c. Br₂

Explanation:

In order to balance a redox reaction, we use the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: Br₂(l) → Br⁻(aq)

Oxidation: Br₂(l) → BrO₃⁻(aq)

Step 2: Perform the mass balance. This reaction takes place in basic medium, so we have to add OH⁻ and H₂O where appropriate.

0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O

Step 3: Perform the electrical balance adding electrons where appropriate.

1 e⁻ + 0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same.

5 × (1 e⁻ + 0.5 Br₂(l) → Br⁻(aq))

1 × (6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻)

Step 5: Add both half-reactions and simplify when appropriate.

5 e⁻ + 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O + 5 e⁻

3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

The species that is reduced is the oxidizing agent. Ther species that is oxidized is the reducing agent. In this case, Br₂ is both.

Answer 2

Answer:

the oxidizing agent is $Br_{2}$ and the reducing agent is  $Br_{2}$

Explanation:

Generally, the reduction-oxidation reactions involve the transfer of electrons from one compound to another.  In the reduction reaction, there is a reduction in the oxidation number while in the oxidation reaction, there is an increase in the oxidation number.  Reducing agent loses electrons and its oxidation number increases while the oxidizing agent gains electrons and its oxidation number decreases.

The unbalanced chemical reaction is:

$Br_{2(l)}$ → $BrO^{-} _{3(aq)}$ + $Br_{(aq)} ^{-}$

For the oxidation reaction:

$Br_{2} + 6H_{2}O$ → $2BrO^{-} _{3(aq)}$ + 12$H^{+}$ + $10e^{-}$

For the reduction reaction:

$Br_{2(l)}$  + $2e^{-}$ → $2Br^{-}$

The next step is to make the number of electrons gained equal to the number of electrons lost:

$Br_{2} + 6H_{2}O$ → $2BrO^{-} _{3(aq)}$ + 12$H^{+}$ + $10e^{-}$

$5Br_{2(l)}$  + $10e^{-}$ → $10Br^{-}$

Then we add up the two equations:

$3Br_{2(aq)}+3H_{2}O_{(l)}$⇌  $2BrO^{-} _{3(aq)}$ +$5Br_{2(l)}$ +$6H_{(aq)} ^{+}$

Therefore, the oxidizing agent is $Br_{2}$ and the reducing agent is  $Br_{2}$