IIT-JEE 2012 · Paper 2 · Chemistry Q37

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ChemistryMultiple correct+4Question 37 of 60

For the given aqueous reactions, which of the statement(s) is (are) true?

Excess KI + K₃[Fe(CN)₆] with dilute H₂SO₄ gives a brownish-yellow solution; adding ZnSO₄ gives a white precipitate + brownish-yellow filtrate; the filtrate with Na₂S₂O₃ gives a colourless solution.

The first reaction is a redox reaction.

White precipitate is

\ce{Zn3[Fe(CN)6]2}

Addition of filtrate to starch solution gives blue colour.

White precipitate is soluble in NaOH solution.

Answer and worked solution

Official answer: A, C, D

First step. Ferricyanide oxidises iodide:
$\ce{2[Fe(CN)6]^{3-} + 2I- -> 2[Fe(CN)6]^{4-} + I2},$
and the iodine (as $\ce{I3-}$ with excess KI) colours the solution brownish-yellow. (A) true.

Zinc sulphate removes the ferrocyanide as white $\ce{K2Zn3[Fe(CN)6]2}$ (zinc ferrocyanide), which also pulls the first equilibrium to completion. It is not $\ce{Zn3[Fe(CN)6]2}$ , the ferricyanide. (B) false.

Filtrate. It contains the iodine, so it turns starch blue (and thiosulphate decolourises it). (C) true.

Precipitate in NaOH. Zinc is amphoteric; the precipitate dissolves as zincate. (D) true.

(A), (C), (D)

Topics:Cation analysis Stability and applications Oxidation number

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IIT-JEE 2012 · Paper 2, Chemistry question 37