Explain bromide and iodide reactions with concentrated sulfuric acid
Balance the redox products formed when bromide and iodide ions reduce concentrated sulfuric acid.
Section 11 · Group 17
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. State conditions for halogen reactions, distinguish halogen oxidising strength from halide-ion reducing strength, and balance redox equations. Chlorine, bromine, iodine and concentrated acids are hazardous; examples teach chemistry, not practical instructions. Written practice is self-checked
Before you begin
- Use molecular halogen formulas X₂, oxidation numbers, ionic equations and electron-transfer ideas.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is the stronger reducing agent, Br⁻ or I⁻?
What you will learn
- Explain why Br⁻ and I⁻ reduce concentrated sulfuric acid
- Write balanced equations with multiple reduction products
- Compare halide reducing strength
Bromide is oxidised to bromine and sulfuric acid is reduced
When a bromide salt reacts with concentrated sulfuric acid, an initial acid–base step can release HBr, then bromide reduces sulfuric acid. A representative overall equation is 2NaBr + 3H₂SO₄ → 2NaHSO₄ + Br₂ + SO₂ + 2H₂O.
Bromine changes from −1 to 0; sulfur in sulfuric acid changes from +6 to +4 in SO₂. Bromide is the reducing agent and sulfuric acid the oxidising agent.
Iodide is a stronger reducing agent and can reduce sulfuric acid further
Iodide is oxidised to iodine. Depending on conditions, sulfuric acid may be reduced to SO₂, sulfur or H₂S. Balanced sodium-iodide examples are: 2NaI + 3H₂SO₄ → 2NaHSO₄ + I₂ + SO₂ + 2H₂O; 6NaI + 7H₂SO₄ → 6NaHSO₄ + 3I₂ + S + 4H₂O; 8NaI + 9H₂SO₄ → 8NaHSO₄ + 4I₂ + H₂S + 4H₂O.
Do not claim a single sulfur product in every reaction mixture. Iodide is a stronger reducing agent than bromide, so it can transfer more electrons and reduce sulfur to lower oxidation states.
Pause and explain
What oxidation-number change shows sulfuric acid is reduced to SO₂?
Worked example
For the reaction 2NaBr + 3H₂SO₄ → 2NaHSO₄ + Br₂ + SO₂ + 2H₂O, identify the oxidised and reduced species.
Show the worked solution
- Br⁻ in NaBr changes from −1 to 0 in Br₂.
- Sulfur changes from +6 in H₂SO₄ to +4 in SO₂.
- Br⁻ is oxidised and sulfuric acid is reduced.
Answer Br⁻ is oxidised to Br₂; sulfuric acid is reduced to SO₂. Br⁻ is the reducing agent.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Complete the balanced equation for bromide reducing concentrated sulfuric acid: NaBr + H₂SO₄ → NaHSO₄ + Br₂ + SO₂ + H₂O.
Give me a hint
Balance bromine as Br₂, then balance sodium hydrogensulfate and the sulfur dioxide reduction.
Compare my reasoning
- Use 2NaBr to make Br₂.
- Use 2NaHSO₄ for sodium and sulfate balance.
- Three H₂SO₄, one SO₂ and two H₂O balance H, S and O.
2NaBr + 3H₂SO₄ → 2NaHSO₄ + Br₂ + SO₂ + 2H₂O.
Look for these in your work
- Br₂ balanced
- Sulfur products balanced
- H and O checked
Write the balanced equation for sodium iodide reducing sulfuric acid to sulfur.
Give me a hint
Use the smallest integer equation for NaHSO₄, I₂, S and water.
Compare my reasoning
- Six NaI give six NaHSO₄ and three I₂.
- The sulfur product is one S atom.
- Seven H₂SO₄ and four H₂O balance hydrogen, sulfur and oxygen.
6NaI + 7H₂SO₄ → 6NaHSO₄ + 3I₂ + S + 4H₂O.
Look for these in your work
- Iodine balanced
- Sulfur product stated
- All atoms balanced
Explain why iodide can produce H₂S in a concentrated sulfuric acid reaction while chloride does not show the same redox behavior.
Give me a hint
Compare the reducing strengths and sulfur oxidation-state changes.
Compare my reasoning
- I⁻ is a much stronger reducing agent than Cl⁻.
- I⁻ can reduce sulfur from +6 to −2 in H₂S while being oxidised to I₂.
- Cl⁻ is too weak a reducing agent in the standard test and forms HCl by acid–base reaction.
I⁻ is strong enough to reduce H₂SO₄ to H₂S as it is oxidised to I₂; Cl⁻ is weaker and gives the non-redox HCl reaction.
Look for these in your work
- Reducing strengths compared
- Oxidation states linked
- Chloride contrast correct
Common mistakes
- Saying bromide reduces sulfuric acid to H₂S as the only product
- Missing that iodide can form S or H₂S as well as SO₂
- Forgetting the initial acid–base HBr step in a mechanism description
- Leaving equations unbalanced or omitting water
What can you explain now?
Which is the stronger reducing agent, Br⁻ or I⁻?
Compare with the explanation
I⁻
Reducing strength increases down the group.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How do the products and electron transfers differ for chloride, bromide and iodide with concentrated sulfuric acid?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.