Table 3
- Discussion
Table 1: For chlorine, it can be prepared from the oxidation of chlorine by MnO2 and conc. HCl Heating will be required.
MnO2 + 4H+ + 2Cl- → Mn2+ + 2H2O + Cl2
The Cl2 is purified by washing with water to remove HCl and then with conc. H2SO4, conc. H3PO4, or anhydrous CaCl2 to remove moisture.
For bromine, it can be prepared by the action of MnO2 on KBr and conc. H2SO4. No heating is required because Br- is a stronger agent than Cl-. The Br2 is then dried with anhydrous CaCl2. The bromine liquid is purified by distillation.
NaBr + H2SO4 → HBr + NaHSO4
MnO2 + 4H+ + 2Br- → Mn2+ + 2H2O + Br2
For iodine, it is similar to Br2, using KI + conc. H2SO4. MnO2 is not required because I- is a strong reducing agent. The I2 collected is purified by sublimation.
KI + H2SO4 → HI + KHSO4
8HI + H2SO4 → H2S + 4H2O + 4I2
Table 2: Concentrated sulphric (VI) acid is often said to be an oxidizing acid as it exhibits both oxidizing agent and acidic properties. On treatment with concentrated sulphric acid (VI) acid, chlorides give hydrogen chloride. However, bromides and iodides do not give hydrogen bromide and hydrogen iodide respectively when concentrated sulphuric (VI) acid is added to them. Instead, sulphur dioxide or hydrogen sulphide is formed. Bromides and iodide do not react in the same way with concentrated sulphuric (VI) acid as chlorides. It is because the hydrogen bromide and hydrogen iodide produced are oxidized by concentrated sulphuric (VI) acid to bromine and iodine respectively. However, hydrogen chloride is not oxidized by concentrated sulphuric (VI) acid. In fact, the reaction of chlorides with concentrated sulphuric (VI) acid can be used for the preparation of hydrogen chloride in the laboratory. However, hydrogen bromide and hydrogen iodide cannot be prepared in this way.
Table 3: Phosphoric (V) acid is not an oxidizing agent. It reacts with halides to form the corresponding hydrogen halides. In fact, this is a general method to prepare hydrogen halides in the laboratory.
5. Answers to Question
5.1 Write the chemical equations in each case. Explain the differences in reactivity between concentrated sulphuric(VI) acid and concentrated phosphoric(V) acid.
Table 1:
Table 2:
KCl(s) + H2SO4 (l) →KHSO4(s) + HCl(g)
For bromides:
KBr(s) + H2SO4 (l) →KHSO4(s) + HBr(g)
2HBr (g) + H2SO4 (l) →SO2 (g) + Br2 (g) + 2H2O( l)
The chemical equation for the overall reaction is:
2KBr(s) + 3H2SO4 (l) →2KHSO4(s) + SO2 (g) + Br2 (g) + 2H2O (l)
For iodides:
KI(s) + H2SO4 (l) →KHSO4(s) + HI (g)
8HI (g) + H2SO4 (l) →H2S (g) + 4I2 (g) + H2O (l)
The chemical equation for the overall reaction is:
8KI(s) + 9H2SO4 (l) →8KHSO4(s) + H2S (g) + 4I2 (g) + 4H2O (l)
Table 3:
3KCl(s) + H3PO4 (l)→ K3PO4(s) + 3HCl (g)
3KBr(s) + H3PO4 (l)→ K3PO4(s) + 3HBr (g)
3KI(s) + H3PO4 (l)→ K3PO4(s) + 3HI (g)
5.2 What is the relationship between H-X
bond enthalpy and its tendency to undergo
further oxidation? [X=Cl/Br/I]
Electron affinity of halogens is the enthalpy change
when one mole of electrons is added to one mole of
halogen atoms or ions in the gaseous state. The electron
affinity decreases from chlorine to iodine. The general
decrease in electron affinity is due to the increases in
atomic size and number of electrons shells down the
group this leads to a decrease in effective charge.
Therefore, the tendency of the nuclei of halogen atoms
to attract additional electrons decreases.
5.3 State the function of manganese (IV) oxide added in experimental procedure (A).
Manganese (IV) oxide is a strong oxidizing agent. When manganese (IV) oxide reacts with reducing agents, it is often reduced to manganese (II) ions
6. Conclusion
The reducing ability of the halide ions increases as we go down the Group when react with an oxidizing acid.