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Byju's Answer

Standard XII

Chemistry

SHE

CuSO4 reacts ...

Question

$C u S O_{4}$ 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?

C u_{2} I_{2}

will be formed

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Evolved

I_{2}

will be reduce

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

will be oxidised

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C u I_{2}

will be formed

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Solution

The correct option is C $C u_{2} I_{2}$ will be formed
$2 C u S O_{4} + - 1 4 K I ⟶_{2} + C u_{2} I_{2} + 2 K_{2} S O_{4}$
Thus, in the given reaction, $C u_{2} I_{2}$ is formed
$_{2} + + 2 2 N a_{2} S_{2} O_{3} ⟶ + 2.5 N a_{2} S_{4} O_{6} + - 1 2 N a I$
Here, the evolved $I_{2}$ is oxidized.

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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 this reaction?

(a)

C u S O_{4}

reacts with

K I

in acidic medium to liberate

I_{2}

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

(b)

Mercuric perisdate,

H g_{5} (I O_{6})_{2}

reacts with a mixture of

K I

and

H C l

according to the following the equation:

H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O

(c)

The liberated iodine is titrated against

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

solution, one mL of which is equivalent to

0.0499

g of

C u S O_{4} .5 H_{2} O

What volume in mL of

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

solution will be required to react with

I_{2}

liberated from

0.7245

g of

H g_{5} (I O_{6})_{2}

? (Molar mass of

C u S O_{4} .5 H_{2} O = 249.5

g/mol and of

H g_{5} (I O_{6})_{2} = 1448.5

g/mol)

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 :

Q. What volume of

0.40 M - N a_{2} S_{2} O_{3}

would be required to react with the

I_{2}

liberated by adding the excess KI to 50 mL of

0.20 M - C u S O_{4}

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CuSO4 reacts with excess amount of KI followed by solution of Na2S2O3. In this process which of the following statement is correct?

The correct option is C Cu2I2 will be formed2CuSO4+−14KI⟶0I2+Cu2I2+2K2SO4Thus, in the given reaction, Cu2I2 is formed0I2++22Na2S2O3⟶+2.5Na2S4O6+−12NaIHere, the evolved I2 is oxidized.

$C u S O_{4}$ 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?

The correct option is C $C u_{2} I_{2}$ will be formed
$2 C u S O_{4} + - 1 4 K I ⟶_{2} + C u_{2} I_{2} + 2 K_{2} S O_{4}$
Thus, in the given reaction, $C u_{2} I_{2}$ is formed
$_{2} + + 2 2 N a_{2} S_{2} O_{3} ⟶ + 2.5 N a_{2} S_{4} O_{6} + - 1 2 N a I$
Here, the evolved $I_{2}$ is oxidized.