Question

. Consider the following reduction-oxidation “skeleton” equation,

Br2(l) = BrO3 ‒ (aq) + Br‒ (aq)

a. Write a balanced equation for the reduction-oxidation reaction that occurs in basic solution.

b. Using your balanced equation, identify the chemical species that is: (1) oxidized, (2) reduced, (3) the reducing agent, and (4) the oxidizing agent.

Answer 1

Br₂ (l) ----------- > BrO₃^‒ (aq.) + Br^‒ (aq.)

Oxidation half reaction: Br₂ (l) ------------ > BrO₃^‒ (aq.)

Reduction half reaction: Br₂ (l) ------------ > Br^‒ (aq.)

1) Balancing of atoms other than O and H,

OHR: Br₂ (l) --------- > 2BrO₃^‒ (aq.)

RHR: Br₂ (l) ------------ > 2Br^‒ (aq.)

2)Balancing of O and H,

OHR: Br₂ (l) + 6H₂O --------- > 2BrO₃^‒ (aq.) + 12H⁺ (aq.)

RHR: Br₂ (l) ------------ > 2Br^‒ (aq.) no need.

3) Balancing of charge:

OHR: Br₂ (l) + 6H₂O --------- > 2BrO₃^‒ (aq.) + 12H⁺ (aq.) + 10 e^-

RHR: Br₂ (l) + 2e^- ------------ > 2Br^‒ (aq.)

4) Balancing of electron number

RHR : 5Br₂ (l) + 10e^- ------------ > 10 Br^‒ (aq.)

5) Addition of OHR and RHR

Br₂ (l) + 6H₂O + 5Br₂ (l) + 10e^- --------- > 2BrO₃^‒ (aq.) + 12H⁺ (aq.) + 10 Br^‒ (aq.) + 10 e^-

10 e^- cancelled out as then,

Br₂ (l) + 6H₂O + 5Br₂ (l) --------- > 2BrO₃^‒ (aq.) + 12H⁺ (aq.) + 10 Br^‒ (aq.)

For the radox reaction in basic medium let us add 12 OH^- ions to both side to neutralize 12H⁺ ions

Br₂ (l) + 6H₂O + 5Br₂ (l) + 12 OH^‑ --------- > 2BrO₃^‒ (aq.) + 12H⁺ (aq.) + 10 Br^‒ (aq.) + 12 OH^-

On right 12 H2O molecules will be formed and 6 of which are cancelled by 6H2O on left.

Br₂ (l) + 5Br₂ (l) + 12 OH^‑ --------- > 2BrO₃^‒ (aq.) + 6H₂O + 10 Br^‒ (aq.)

i.e. 6Br₂ (l) + 12 OH^‑ --------- > 2BrO₃^‒ (aq.) + 6H₂O + 10 Br^‒ (aq.)

is the balanced equation for the Redox that occurs in basic solution.

(1) oxidized: Br₂(O.S.= 0) is oxidized to BrO₃^‒ (O.S. = +5)

(2) Reduced: Br₂(O.S.= 0) is reduced to Br^- .

(3) the reducing agent: Br₂

(4) the oxidizing agent: Br₂.

No other atom in the reaction changed oxidation state.

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