Topic 8 · page 26 of 29

Chlorine with alkali: bleach and chlorate(V)

Chlorine disproportionates in sodium hydroxide. Cold, dilute alkali gives chloride and chlorate(I), which is bleach; hot alkali gives chloride and chlorate(V). Bromine and iodine behave in the same way.

13 min read8.268.26(iii)8.26(iv)8.26(v)

Read first: Halogens with metals, and chlorine with water

Cold, dilute alkali: bleach 8.268.26(iii)

Chlorine reacts with cold, dilute aqueous sodium hydroxide to form sodium chloride, sodium chlorate(I) and water.

ClX2(g)+2 NaOH(aq)→NaCl(aq)+NaClO(aq)+HX2O(l) \ce{Cl2(g) + 2NaOH(aq) -> NaCl(aq) + NaClO(aq) + H2O(l)}

The ionic equation, with the spectator Na⁺ left out:

ClX2(g)+2 OHX−(aq)→ClX−(aq)+ClOX−(aq)+HX2O(l) \ce{Cl2(g) + 2OH-(aq) -> Cl-(aq) + ClO-(aq) + H2O(l)}
A vertical scale of chlorine oxidation numbers with four marked levels: chloride at minus 1, chlorine at 0, chlorate(I) at plus 1 and chlorate(V) at plus 5. Two sets of arrows start from chlorine at 0. In water or cold dilute alkali, one arrow goes down to chloride at minus 1 and one goes up to chlorate(I) at plus 1, one chlorine atom each. In hot alkali, five arrows go down to chloride and one arrow goes up to chlorate(V) at plus 5.
Chlorine goes down to chloride and up to chlorate(I) or chlorate(V), depending on the conditions.

Chlorine goes from 0 in Cl₂ to −1 in NaCl (reduced) and to +1 in NaClO (oxidised): a disproportionation. The mixture of sodium chloride and sodium chlorate(I) in water is bleach. The chlorate(I) ion, ClO⁻, is the active part: it kills bacteria and oxidises coloured compounds.

Uses of bleach: household cleaning and disinfection, treating swimming pools and bleaching paper and textiles.

The half-equations show the electron transfer:

ClX2+2 eX−→2 ClX−ClX2+4 OHX−→2 ClOX−+2 HX2O+2 eX− \ce{Cl2 + 2e- -> 2Cl-} \qquad \ce{Cl2 + 4OH- -> 2ClO- + 2H2O + 2e-}

Adding them gives 2 ClX2+4 OHX−→2 ClX−+2 ClOX−+2 HX2O\ce{2Cl2 + 4OH- -> 2Cl- + 2ClO- + 2H2O}, which is twice the equation above.

Hot alkali: chlorate(V) 8.268.26(iv)

With hot sodium hydroxide, chlorine is oxidised further, to chlorate(V), ClO₃⁻.

3 ClX2(g)+6 NaOH(aq)→5 NaCl(aq)+NaClOX3(aq)+3 HX2O(l) \ce{3Cl2(g) + 6NaOH(aq) -> 5NaCl(aq) + NaClO3(aq) + 3H2O(l)}

The ionic equation:

3 ClX2(g)+6 OHX−(aq)→5 ClX−(aq)+ClOX3X−(aq)+3 HX2O(l) \ce{3Cl2(g) + 6OH-(aq) -> 5Cl-(aq) + ClO3-(aq) + 3H2O(l)}

Here chlorine goes 0 → −1 in five chloride ions and 0 → +5 in one chlorate(V) ion. The electrons balance: the one atom oxidised loses 5 electrons, and the five atoms reduced gain 1 each.

Worked exampleChecking the hot-alkali equation

Show that 3 ClX2+6 OHX−→5 ClX−+ClOX3X−+3 HX2O\ce{3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O} is balanced.

  1. Cl: 6 on the left (3 × 2); 5 + 1 = 6 on the right. ✓
  2. O: 6 on the left; 3 + 3 = 6 on the right. ✓
  3. H: 6 on the left; 3 × 2 = 6 on the right. ✓
  4. Charge: −6 on the left; −5 − 1 = −6 on the right. ✓
  5. Electrons: Cl 0 → +5 loses 5 electrons; 5 Cl 0 → −1 gain 5 electrons. ✓

Analogous reactions 8.268.26(v)

Bromine and iodine have the same pattern, because the halogens are in the same group. You must be able to predict an equation by analogy.

Reaction Equation Oxidation numbers
Br₂ with cold dilute NaOH BrX2+2 NaOH→NaBr+NaBrO+HX2O\ce{Br2 + 2NaOH -> NaBr + NaBrO + H2O} Br 0 → −1 and 0 → +1 (bromate(I))
Br₂ with hot NaOH 3 BrX2+6 NaOH→5 NaBr+NaBrOX3+3 HX2O\ce{3Br2 + 6NaOH -> 5NaBr + NaBrO3 + 3H2O} Br 0 → −1 and 0 → +5 (bromate(V))
I₂ with hot NaOH 3 IX2+6 NaOH→5 NaI+NaIOX3+3 HX2O\ce{3I2 + 6NaOH -> 5NaI + NaIO3 + 3H2O} I 0 → −1 and 0 → +5 (iodate(V))
Cl₂ with hot KOH 3 ClX2+6 KOH→5 KCl+KClOX3+3 HX2O\ce{3Cl2 + 6KOH -> 5KCl + KClO3 + 3H2O} Cl 0 → −1 and 0 → +5

Worked examplePredicting by analogy

Write the equation for iodine with hot sodium hydroxide solution, and state the oxidation numbers of iodine.

  1. Copy the chlorine equation with I in place of Cl and the same coefficients: 3 ClX2+6 NaOH→5 NaCl+NaClOX3+3 HX2O\ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O} becomes 3 IX2+6 NaOH→5 NaI+NaIOX3+3 HX2O\ce{3I2 + 6NaOH -> 5NaI + NaIO3 + 3H2O}.
  2. Check the oxidation numbers: I₂ is 0. NaI: I is −1. NaIO₃: Na +1, O −2, so I = +5.
  3. One element in one species goes both down and up: disproportionation.

Check yourself

  1. Write the equation for chlorine with cold, dilute sodium hydroxide, and state the name and use of the product mixture.
  2. State the oxidation numbers of chlorine in Cl₂, NaCl and NaClO, and name the type of reaction.
  3. Write the ionic equation for chlorine with hot sodium hydroxide, and state the oxidation numbers of chlorine in the products.
  4. Predict the equation for bromine with hot concentrated sodium hydroxide.
  5. Explain what is meant by "analogous reaction" and use it to write the equation for iodine with hot potassium hydroxide.
  6. Show that the equation ClX2+2 OHX−→ClX−+ClOX−+HX2O\ce{Cl2 + 2OH- -> Cl- + ClO- + H2O} is balanced for atoms and charge.

Answers cover these until you have tried the questions

  1. ClX2+2 NaOH→NaCl+NaClO+HX2O\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}. The mixture of NaCl and sodium chlorate(I), NaClO, is bleach, used for disinfecting, cleaning and bleaching textiles.
  2. Cl₂ 0; NaCl −1; NaClO +1. It is a disproportionation: chlorine is both reduced and oxidised.
  3. 3 ClX2+6 OHX−→5 ClX−+ClOX3X−+3 HX2O\ce{3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O}. In Cl⁻ chlorine is −1; in ClO₃⁻ it is +5.
  4. 3 BrX2+6 NaOH→5 NaBr+NaBrOX3+3 HX2O\ce{3Br2 + 6NaOH -> 5NaBr + NaBrO3 + 3H2O} (Br 0 → −1 and 0 → +5).
  5. An analogous reaction is one in which a similar element, from the same group, reacts in the same way. 3 IX2+6 KOH→5 KI+KIOX3+3 HX2O\ce{3I2 + 6KOH -> 5KI + KIO3 + 3H2O}
  6. Atoms: Cl 2 each side; O 2 each side (2 OH⁻; ClO⁻ and H₂O); H 2 each side. Charge: left −2; right −1 −1 + 0 = −2. ✓