Interpret chlorine with hot concentrated sodium hydroxide
Write the hot alkali reaction and compare its chlorate product and oxidation-number change with cold conditions.
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
What oxyanion forms with hot concentrated NaOH?
What you will learn
- Balance the hot concentrated chlorine–alkali equation
- Identify chlorate as the hot-condition product
- Compare oxidation numbers with the cold reaction
Hot concentrated alkali forms chlorate
With hot, concentrated aqueous sodium hydroxide, chlorine forms sodium chloride and sodium chlorate: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. The ionic equation is 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O.
Chlorine changes from 0 to −1 in chloride and to +5 in chlorate. It is simultaneously reduced and oxidised; the reaction remains disproportionation.
Do not switch cold and hot product formulas
Cold dilute alkali gives ClO⁻; hot concentrated alkali gives ClO₃⁻. The different oxyanion is a condition-dependent product, not a different halogen.
Balance chlorine atoms carefully: six chlorine atoms on the left become five in chloride and one in chlorate.
Pause and explain
What is chlorine's oxidation number in ClO₃⁻?
Worked example
Write the hot concentrated NaOH equation and show the chlorine oxidation-number change.
Show the worked solution
- The products are chloride and chlorate.
- Three Cl₂ molecules supply six chlorine atoms; five form NaCl and one forms NaClO₃.
- Chlorine changes from 0 to −1 and +5.
Answer 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O; chlorine is reduced 0 → −1 and oxidised 0 → +5.
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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.
Balance the equation for chlorine reacting with hot concentrated sodium hydroxide.
Give me a hint
Use five NaCl and one NaClO₃ for six chlorine atoms.
Compare my reasoning
- Three Cl₂ provide six chlorine atoms.
- Six NaOH supply six sodium ions and six OH groups.
- Three H₂O balance hydrogen and oxygen.
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.
Look for these in your work
- Condition attached
- Chlorine balanced
- Na/H/O balanced
Compare the oxidation state of chlorine in the oxyanion from cold conditions and hot conditions.
Give me a hint
Cold gives hypochlorite; hot concentrated gives chlorate.
Compare my reasoning
- ClO⁻ has chlorine at +1.
- ClO₃⁻ has chlorine at +5.
- The more oxidised chlorine product is chlorate.
Cold: ClO⁻ contains Cl at +1; hot: ClO₃⁻ contains Cl at +5.
Look for these in your work
- Both species correctly named
- Oxidation numbers charge-balanced
- Conditions connected
A student uses Cl₂ + 2NaOH → NaCl + NaClO + H₂O for hot concentrated alkali. Correct the answer.
Give me a hint
The given equation is for cold dilute conditions; substitute the hot-condition product and rebalance.
Compare my reasoning
- Hot concentrated conditions form chlorate, NaClO₃.
- Three Cl₂ provide six chlorine atoms.
- Balance as 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.
For hot concentrated NaOH use 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O; the earlier equation applies to cold dilute conditions.
Look for these in your work
- Product corrected
- Equation balanced
- Condition contrast explicit
Common mistakes
- Using NaClO instead of NaClO₃ for hot concentrated conditions
- Using coefficients from the cold equation
- Calling chlorate chlorine +3
- Leaving the equation unbalanced
What can you explain now?
What oxyanion forms with hot concentrated NaOH?
Compare with the explanation
Chlorate, ClO₃⁻
Hot concentrated conditions form chlorate.
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Make it stick
How do hot and cold alkali conditions change the chlorine oxyanion and its oxidation number?
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