43.

(i) How is iodine manufactured from sea weeds ?   

(ii) How is hydrogen peroxide prepared in the laboratory ?

(iii) Give a balanced equation for a reaction in which hydrogen peroxide acts as a reducing agent and one in which it acts as an oxidizing agent.

(i) Sea weeds of laminaria veriety contain iodine in the form of iodides. The extraction of iodine from them is done in three steps :

1. Preparation of kelp or ash from sea weeds. The dried sea weeds are burnt in shallow pit and ash thus obtained is called kelp. It has 0.4 to 1.3% of iodine in the form of iodides, rest are chlorides and sulphates.

2. Extraction of kelp with water. The kelp is dissolved in water and the solution is filtered to remove insoluble impurities. The filtrate is now concentrated when sulphates and chlorides separate out, more soluble iodides remain in the mother liquor.

3. Liberaton of iodine. The mother liquor is mixed with conc H₂SO₄ and MnO₂ and is now heated in cast iron retorts with lead covers and connected to series of earthen ware condensers called aludels. The iodine vapours are formed and they condense on walls of aludels. When reaction is over aludels are broken to get iodine.

2NaI + 3H₂SO₄ + MnO₂ → 2NaHSO₄ + MnSO4 + 2 H₂O +I₂

(ii) In laboratory H₂O₂ is prepared by either Na₂O₂ or BaO₂.

From sodium peroxide. H₂O₂ is obtained by adding calculated quantity of Na₂O₂ to 20 % ice cold H₂SO₄ solution in small lots and constant stirring.

Na₂O₂., + H₂,SO₄ → Na₂SO₄ + H₂O

When the resulting solution is cooled is 271 K crystals of Na₂SO₄.10H₂O separate and resulting solution is 30% H₂,O₂ solution.

iii)