JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Redox Reactions and Electrochemistry

When

Cu^{2 +}

ion is treated with KI , a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are

(A) $X = Cu_{2} I_{2} Y = Na_{2} S_{4} O_{5}$
(B) $X = Cu_{2} I_{2} Y = Na_{2} S_{4} O_{6}$ ✓ Correct
(C) $X = CuI_{2} Y = Na_{2} S_{4} O_{3}$
(D) $X = CuI_{2} Y = Na_{2} S_{4} O_{6}$

Answer: (B)

Solution

Iodide ion reduces copper(II) to copper(I) and is itself oxidised to iodine, the white precipitate formed being cuprous iodide,

Cu X_{2} I X_{2}

, which is X. The liberated iodine is titrated with sodium thiosulphate, which is oxidised to sodium tetrathionate,

Na X_{2} S X_{4} O X_{6}

, which is Y. Thus X is

Cu X_{2} I X_{2}

and Y is

Na X_{2} S X_{4} O X_{6}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Redox Reactions and Electrochemistry →

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Redox Reactions and Electrochemistry

Solution

More JEE Main questions on Redox Reactions and Electrochemistry