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Introduction to Oxidation and Reduction

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Question

A solution of $0.2 g$ of a compound containing $C u^{2 +}$ and $C_{2} O_{4}^{2 -}$ ions on titration with $0.02 M K M n O_{4}$ in presence of $H_{2} S O_{4}$ consumes $22.6 m L$ of the oxidant. The resultant solution is neutralized with $N a_{2} C O_{3}$ , acidified with dil. acetic acid and treated with excess $K I$ .The liberated iodine requires $11.3 m L$ of $0.05 M N a_{2} S_{3} O_{3}$ solution for complete reduction. Find out the mole ratio of $C u^{2 +}$ to $C_{2} O_{4}^{2 -}$ in the compound is $a : b$ . The sum of $a + b$ is

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Solution

In the given problem, a solution containing $C u^{2 +}$ and $C_{2} O_{4}^{2 -}$ is titrated first with $K M n O_{4}$ and then with $N a_{2} S_{2} O_{3}$ in presence of $K I$ . In titration with $K M n O_{4}$ , it is the $C_{2} O_{4}^{2 -}$ ions that react with the $M n O_{4}^{-}$ ions. The concerned balanced equation may be written as below.
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$
Thus according to the above reaction
2 mol $M n O_{4}^{-}$ reacts with 5 mol of $C_{2} O_{4}^{2 -}$
However,
No. of mmol of $M n O_{4}^{-}$ and used in titration
$= 0.02 \times 22.6$
Now,
milimoles of oxalate $= \frac{0.02 \times 22.6 \times 5}{2} = 1.130$
$K I$ reacts with $C u^{2 +}$ as
$2 C u^{2 +} + 4 I^{-} \to C u_{2} I_{2} + I_{2}$
$I_{2} + 2 S_{2} O_{3}^{-} \to 2 I^{-} + S_{4} O_{6}^{2 -}$
According to the above reactions
mmoles of thiosulfate =mmoles of $C u^{2 +}$
$= 0.05 \times 11.3$
$∴,$
Mole ratio of $C u^{2 +}$ to $C_{2} O_{4}^{2 -} = \frac{0.565 m m o l}{1.130 m m o l} = 1 : 2$
Hence, $a : b = 1 : 2$
So, $a + b = 1 + 2 = 3$

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Similar questions

Q. A solution of

0.2 g

of a compound containing

C u^{2 +}

{C_{2} O_{4}}^{2 -}

ions on titration with

0.02 M K M n O_{4}

H_{2} S O_{4}

22.6 m L

oxidant. The resulting solution is neutralised by

N a_{2} C O_{3}

, acidified with dilute acetic acid and titrated with excess of

K I

. The liberated iodine required

1.3 m L

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

for complete reduction. Find out the whole ratio of

C u^{2 +}

{C_{2} O_{4}}^{2 -}

in the compound.

Q. A solution containing

C u^{2 +}

C_{2} O_{4}^{-}

ions which is titrated with

20 m l

M / 4 K M n O_{4}

solution in acidic medium. The resulting solution is treated with excess of

K I

after neutralization. The evolved

I_{2}

is then absorbs in

25 m l

N / 10

hypo solution, then the difference of the number millimoles of

C u^{2 +}

C_{2} O_{4}^{2 -}

ions in the solution is:

3.0 g

sample containing

F e_{3} O_{4}, F e_{2} O_{3}

and an inert impure substance is treated with excess of

K I

solution in presence of dilute

H_{2} S O_{4}

. The entire iron is converted to

F e^{2 +}

along with the liberation of iodine. The resulting solution is diluted to

100 m L

20 m L

of the dilute solution requires

11.0 m L

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

solution to reduce the iodine present. A

50 m L

of the diluted solution after complete extraction of iodine requires

12.8 m L

0.25 M K M n O_{4}

solution in dilute

H_{2} S O_{4}

medium for oxidation of

F e^{2 +}

. Calculate the percentage of

F e_{2} O_{3}

F e_{3} O_{4}

in the original sample.

Q. A mixture containing

0.05 m o l e

K_{2} C r_{2} O_{7}

0.02 m o l e

K M n O_{4}

was treated with excess of

K I

in acidic medium. The liberated iodine required

1.0 L

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

solution for titration. Concentration of

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

Q. A 5g sample containing

F e_{3} O_{4} (F e O + F e_{2} O_{3})

and an inert impurity is treated with excess of KI solution in the presence of dilute

H_{2} S O_{4}

. The entire iron converted to Ferrous ion along with the liberation of Iodine. The resulting solution is diluted to 100 mL. 20 mL of the diluted solution requires

10 m L o f 0.5 M N a_{2} S_{2} O_{3}

solution to reduce the iodine present. Calculate approximate percentage of

F e_{2} O_{3}

?

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