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Standard XII

Chemistry

SHE

Excess of K...

Question

Excess of $K I$ reacts with $C u S O_{4}$ solution and then $N a_{2} S_{2} O_{3}$ solution is added to it.

Which of the statements is incorrect in this reaction?

Evolved

I_{2}

is reduced

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C u I_{2}

is formed

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N a_{2} S_{2} O_{3}

is oxidised

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C u_{2} I_{2}

is formed

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Solution

The correct option is B $C u I_{2}$ is formed
Reaction of process:

$2 C u S O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$
This liberated $I_{2}$ is titrated with $N a_{2} S_{2} O_{3}$ .

The reaction is given below:
$I_{2} + 2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + 2 N a l$
So, liberated $I_{2}$ is reduced and $N a_{2} S_{2} O_{3}$ is oxidised.

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Similar questions

Excess of KI reacts with $C u S O_{4}$ solution and then $N a_{2} S_{2} O_{3}$ solution is added to it. Which of the following statement is incorrect from this reaction

Q. Excess of

K I

reacts with

C u S O_{4}

solution and then

N a_{2} S_{2} O_{3}

solution is added to it. Which of the following statements is incorrect for these reactions?

C u S O_{4}

reacts with excess amount of

K I

followed by solution of

N a_{2} S_{2} O_{3}

. In this process which of the following statement is correct?

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

ml of

0.02 M

solution with excess

K I

is :

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

The color of the solution developed during the addition of

K I

C u S O_{4}

solution is :

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Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect in this reaction?

The correct option is B CuI2 is formedReaction of process:2CuSO4+4KI→Cu2I2+2K2SO4+I2This liberated I2 is titrated with Na2S2O3.The reaction is given below:I2+2Na2S2O3→Na2S4O6+2NalSo, liberated I2 is reduced and Na2S2O3 is oxidised.

Excess of $K I$ reacts with $C u S O_{4}$ solution and then $N a_{2} S_{2} O_{3}$ solution is added to it.

Which of the statements is incorrect in this reaction?