When copper sulphate solution is treated with potassium iodide and excess of hypo solution is added in resulting solution, a white precipitate is formed. The white ppt. is due to

Engineering

Chemistry

Classification of Chemical Reaction

Balancing of Redox Reaction

Dichromate and Permanganate Ion Reactions

Question

When copper sulphate solution is treated with potassium iodide and excess of hypo solution is added in resulting solution, a white precipitate is formed. The white ppt. is due to

Na₂S₄O₆

CuI

CuI₂

NaI

When copper sulphate (CuSO₄) reacts with potassium iodide (KI), a redox reaction occurs. Iodide ions reduce Cu²⁺ to Cu⁺, forming copper(I) iodide (CuI), which is a white precipitate, and iodine (I₂). The reaction is:

$2 Cu^{2 +} + 4 I^{-} \to 2 CuI + I_{2}$

Adding excess hypo (sodium thiosulphate, Na₂S₂O₃) dissolves the iodine but does not affect the CuI precipitate. Thus, the white precipitate is CuI.

Final Answer: CuI

When copper sulphate (CuSO₄) solution is treated with potassium iodide (KI), a redox reaction occurs. Copper(II) ions oxidize iodide ions to iodine, and are themselves reduced to copper(I) iodide, which precipitates as a white solid. The reaction is:

$2 Cu^{2 +} + 4 I^{-} \to Cu I_{2} + I_{2}$

However, CuI₂ is unstable and decomposes to copper(I) iodide (CuI) and iodine:

$Cu I_{2} \to 2 CuI + I_{2}$

So, the overall reaction is:

$2 Cu^{2 +} + 4 I^{-} \to 2 CuI + I_{2}$

This produces a white precipitate of CuI and brown iodine solution.

When excess sodium thiosulphate (hypo, Na₂S₂O₃) is added, it reacts with the iodine to form sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI), which are soluble:

$I_{2} + 2 {Na}_{2} S_{2} O_{3} \to {Na}_{2} S_{4} O_{6} + 2 NaI$

After this reaction, the iodine is consumed, but the white precipitate of CuI remains unchanged because CuI is insoluble and does not react with hypo. Therefore, the white precipitate observed is due to CuI (copper(I) iodide).

The other options are incorrect:

Na₂S₄O₆ is soluble and does not form a precipitate.
CuI₂ is unstable and decomposes to CuI and I₂.
NaI is soluble and does not precipitate.

Formulae

The key chemical equations involved are:

1. $2 Cu^{2 +} + 4 I^{-} \to 2 CuI + I_{2}$

2. $I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-}$

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Formulae

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