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Introduction to Oxidation and Reduction

When an equim...

Question

When an equimolar mixture of $C u_{2} S$ and $C u S$ is titrated with $B a (M n O_{4})_{2}$ in acidic medium, the final product contains $C u^{2 +}$ , $S O_{2}$ and $M n^{2 +}$ . If the molecular weight of $C u_{2} S$ , $C u S$ and $B a (M n O_{4})_{2}$ are $M_{1}$ , $M_{2}$ and $M_{3}$ respectively then:

A

eq. wt. of

C u_{2} S

is

\frac{M_{1}}{8}

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B

eq. wt. of

C u S

is

\frac{M_{2}}{6}

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C

eq. wt. of

B a (M n O_{4})_{2}

is

\frac{M_{3}}{5}

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D

C u_{2} S

and

C u S

both have same no. of equivalents in mixture

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Solution

The correct option is B eq. wt. of $C u S$ is $\frac{M_{2}}{6}$
$+ 1 C u_{2} - 2 S \to C u^{+ 2} + + 4 S O_{2}$
So, n-factor of $C u_{2} S$ = 8
Hence, eq. wt. of $C u_{2} S$ is $\frac{M_{1}}{8}$

$+ 2 C u - 2 S \to C u^{+ 2} + + 4 S O_{2}$
So, n-factor = 6
Hence, eq. wt. of $C u S$ is $\frac{M_{2}}{6}$

$B a (+ 7 M n O_{4})_{2} \to M n^{+ 2}$
So, n-factor = 10
eq. wt. of $B a (M n O_{4})_{2}$ is $\frac{M_{3}}{10}$

Clearly, as number of moles of $C u S a n d C u_{2} S$ is same but n-factor is different. So, equivalents will be different.

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Similar questions

Q. When an equimolar mixture of

C u_{2} S

C u S

is titrated with

B a (M n O_{4})_{2}

in acidic medium, the final product contains

C u^{2 +}

S O_{2}

M n^{2 +}

. If the molecular weight of

C u_{2} S

C u S

B a (M n O_{4})_{2}

M_{1}

M_{2}

M_{3}

respectively then:

Q. The mixture of

C u S

(molar weight =

M_{1}

C u_{2} S

(molecular weight =

M_{2}

) is oxidised by

K M n O_{4}

(molecular weight =

M_{4}

) in acidic medium, the product obtained are

C u^{2 +}

S O_{2 +}

. The equivalent weights of

C u S, C u_{2} S

K M n O_{4}

respectively, are :

10

C u_{2} S

C u S

was treated with

200

0.75 M M n O_{4}^{-}

in acid solution producing

S O_{2}

C u^{2 +}

M n^{2 +}

S O_{2}

was boiled off and the excess of

M n O_{4}^{-}

was titrated with

175

1 M F e^{2 +}

solution. Calculate the percentage of

C u S

in the original mixture (to nearest integer).

C u_{2} S

is converted into

{C u}^{2 +}

{S O}_{2}

, then equivalent weight of

{C u}_{2} S

[M = molecular weight of

{C u}_{2} S

]

Q. Find the number of moles of

K M n O_{4}

needed to oxidise one mole

C u_{2} S

in acidic medium. The reaction is

K M n O_{4} + C u_{2} S \to M n^{2 +} + C u^{2 +} + S O_{2}

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