1 . 2475 g of crystalline copper sulphate was dissolved in water and excess of KI was added. The liberated iodine consumed 50 mL N 10 Na 2 S 2 O 3 solution to reach the end point of the titration. Calculate the number of water molecules of hydration in crystalline copper sulphate salt.

4 I - + 2 Cu 2 + → Cu 2 I 2 + I 2 I 2 + 2 S 2 O 3 2 - → 2 I - + S 4 O 6 2 - meq. of Hypo = meq. of I 2 mili-equivalent of I 2 = 50 × 1 10 = 5 meq. of I 2 = meq.of hydrated CuSO 4 Number of moles of Hydrated CuSO 4 = 5 × 1 × 10 - 3 = 0 . 005 moles Mass of Hydrate Copper Sulphate = 1 . 2475 g Molar Mass of Hydrated Copper Sulphate = 1 . 2475 0 . 005 = 249 . 5 g / mol Molar Mass of Anhydrous Copper Sulphate = 159 . 5 g / mol Molar Mass of water = 18 g / mol Number of water molecules of hydration in crystalline copper sulphate = 249 . 5 - 159 . 5 18 = 90 18 = 5 .

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${KMnO}_{4}$ oxidizes $X^{n +}$ ion to ${XO}_{3}^{-},$ itself changing to ${Mn}^{2 +}$ in acid medium. $2.68 \times 10^{- 3}$ $mole$ of $X^{n +}$ requires $1.61 \times 10^{- 3}$ $mole$ of ${MnO}_{4}^{-},$ if the weight of $1 g$ equivalent of ${XCl}_{n}$ is $56 .$ Report the value of $\frac{X}{n},$ where $X$ is the molar mass of $X^{n +} .$

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Important Questions on Redox Reactions

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-3>Q 23

Nitrous acid can act both as an oxidising as well as reducing agent.

EASY

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 1

1.2475 g

of crystalline copper sulphate was dissolved in water and excess of

KI

was added. The liberated iodine consumed

\frac{50 mL N}{10 {Na}_{2} S_{2} O_{3}}

solution to reach the end point of the titration. Calculate the number of water molecules of hydration in crystalline copper sulphate salt.

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 2

1 g

sample of

K_{2} {Cr}_{2} O_{7}

containing some inert material was entirely reduced with conc.

HCl

. The chlorine, liberated was passed through hot solution of

NaOH

80 ° C

, and it completely diproportionates to form

{ClO}_{3}^{-}

and

{Cl}^{-}

. This

{NaClO}_{3}

was isolated and its reduction with

KI

(aq) liberated iodine, giving

Cl

. The iodine, thus liberated required

100 mL

of decinormal hypo solution for complete titration. What is the percentage, purity of the dichromate sample?

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 3

2.5 g

of mixture of crystalline oxalic acid

(H_{2} C_{2} O_{4} . 2 H_{2} O)

and sodium oxalate

({Na}_{2} C_{2} O_{4})

was dissolved in

100 mL

of water.

50 mL

of this solution was titrated against

\frac{N}{10 NaOH}

solution when

119.05 mL

of the base was found necessary to reach the end point with phenolphthalein as the indicator.

1 g

of the mixture was dissolved in water and the solution titrated against

\frac{N}{10 {KMnO}_{4}}

in the presence of dil.

H_{2} {SO}_{4}

. What is the volume of

{KMnO}_{4}

needed for getting the end point with

0.5 g

of the mixture?

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 4

0.6213 g

of sample contains an unknown amount of

{As}_{2} O_{3}

. The sample was treated with

HCl

resulting information of

{AsCl}_{3} (g)

which was distilled into a beaker of water. The hydrolysis reaction is as follows:

{AsCl}_{3} + 2 H_{2} O \to {HAsO}_{2} + 3 H^{+} + 3 {Cl}^{-}

The amount of

{HAsO}_{2}

was determined by titration with

0.04134 {MI}_{2}

, requiring

23.04 mL

to reach the equivalence point. The redox products in the titration were

H_{3} {AsO}_{4}

and

I

. Find the amount of

{KMnO}_{4}

needed to oxidize

As

{As}_{2} O_{3}

to its maximum possible oxidation state in acidic medium.

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 5

A sample of steel weighing

0.6 g

and containing

S

as an impurity was burnt in a stream of

O_{2}

, when

S

was converted to its oxide

{SO}_{2} \cdot {SO}_{2}

was then oxidized to

{SO}_{4}^{- -}

by using

H_{2} O_{2}

solution containing

30 mL

0.04 M NaOH . 22.48 mL

0.024 M HCl

was required to neutralize the base remaining after oxidation. Calculate the

%

S

in the sample:

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Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 6

In the presence of fluoride ion,

{Mn}^{2 +}

can be titrated with

{MnO}_{4}^{-}

, both reactants being converted to a complex of

Mn

(III). A

0.545 g

sample containing

{Mn}_{3} O_{4}

was dissolved and all manganese was converted to

{Mn}^{2 +}

. Titration in the presence of fluoride ion consumed

31.1 mL

{KMnO}_{4}

that was

0.177 N

against oxalate.
(a) write a balanced chemical equation for the reaction, assuming that the complex is

{MnF}_{4}^{-}

.
(b) what was the

%

{Mn}_{3} O_{4}

in the sample?

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 7

A mixture of two gases,

H_{2} S

and

{SO}_{2}

is passed through three beakers successively. The first beaker contains

{Pb}^{2 +}

ions, which absorbs

S^{2 -}

forming

PbS

. The second beaker contains

25 mL

0.0396 {NI}_{2}

to oxidize

{SO}_{2}

{SO}_{4}^{2 -}

. The third contains

10 mL

0.0345 N

thiosulphate solution to retain any

I_{2}

carried over from the second absorber. A

25 L

gas sample was passed through the apparatus followed by an additional amount of

N_{2}

to sweep last traces of

{SO}_{2}

from first and second absorber. The solution from the first absorber was made acidic and treated with

20 mL

0.0066 M K_{2} {Cr}_{2} O_{7}

which converted

S^{2 -}

{SO}_{2}

. The excess dichromate was reacted with solid

KI

and the liberated iodine required

7.45 mL

0.0345 N {Na}_{2} S_{2} O_{3}

solution. The solutions in the second and third absorbers were combined and the resultant iodine was titrated with

2.44 mL

of the same thiosulphate solution. Calculate the concentrations of

{SO}_{2}

and

H_{2} S

\frac{mg}{L}

of the sample :

KMnO4 oxidizes Xn+ ion to XO3-, itself changing to Mn2+ in acid medium. 2.68×10-3 mole of Xn+ requires 1.61×10-3 mole of MnO4-, if the weight of 1 g equivalent of XCln is 56. Report the value of Xn, where X is the molar mass of Xn+.

Important Questions on Redox Reactions

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