1 g of a moist sample of a mixture of KClO 3 and KCl was dissolved in water and made upto 250 mL . 25 mL of this solution was treated with SO 2 to reduced chlorate into chloride and the excess SO 2 was boiled off. When the total chloride was precipitated, 0 . 1435 g of AgCl was obtained. In another experiment 25 mL of the original solution was treated with 30 mL of 0 . 2 N solution of FeSO 4 and unreacted FeSO 4 required 37 . 5 mL of 0 . 08 N solution of an oxidizing agent for complete oxidation. Calculate the molar ratio of chlorate and chloride in the given mixture. Fe 2 + reacts with ClO 3 - according to equation: ClO 3 - + 6 Fe 2 + + 6 H + → Cl - + 6 Fe 3 + + 3 H 2 O Also calculate the mass percent of moisture present in the moist sample.

Let the mass of KClO 3 be x gm Let mass of KCl be y gm mole = mass atomic mass n KClO 3 ( moles ) = x 122 . 5 n KCl = y 74 . 5 For Cl 6 e - + 6 H + + Cl + 5 O 3 - ⟶ Cl - + 3 H 2 O Valence factor = 5 - - 1 = 6 It is given that When the total chloride was precipitated, 0 . 1435 gm AgCl was obtained ⇒ n KClO 3 + n KCl = n AgCl ⇒ x 122 . 5 + y 74 . 5 = 0 . 1435 143 . 5 = 0 . 001 . . . i for complete oxidation of an oxidizing agent Milliequivalents of KClO 3 or KCl = Milliequivalents of ( reacted FeSO 4 solution - unreacted FeSO 4 solution ) gm Eq . = moles × valence factor M . eq . = Normality × Volume = N 1 V 1 - N 2 V 2 = 30 × 0 . 2 - 37 . 5 × 0 . 08 = 3 gm Eq . of KClO 3 = x ( 6 ) 122 . 5 = 0 . 003 x 122 . 5 = 0 . 0005 ( n KClO 3 ) gm Eq . of KCl = y ( 6 ) 74 . 5 = 0 . 003 y 74 . 5 = 0 . 0005 ( n KCl ) Molar ratio n KClO 3 n KCl = 1 1 moisture = 1 - mass ( KClO 3 + KCl ) = 1 - ( 122 . 5 + 74 . 5 ) × 0 . 0005 = 0 . 015 g Mass % of moisture = 0 . 015 1 × 100 = 1 . 5 % Hence, the molar ratio of KClO 3 and KCl is 1 : 1 and % of moisture is 1 . 5 %

Consider the reaction H + + IO 4 - + I - → I 2 + H 2 O . Find the ratio of coefficients of IO 4 - , I - and I 2 .

H + + IO 4 - + I - → I 2 + H 2 O Now, balance the iodine atom. The oxidation number one iodine changes from the + 7 to zero and another iodine changes from the - 1 to zero. So, multiply the iodide ion by the seven and iodine by the four for balance the iodine atoms. H + + IO 4 - + 7 I - → 4 I 2 + H 2 O Now, balance the oxygen atom by adding the water molecule on the right-hand side. H + + IO 4 - + 7 I - → 4 I 2 + 4 H 2 O Now, balance the hydrogen atoms by adding the hydrogen on left-hand side. 8 H + + IO 4 - + 7 I - → 4 I 2 + 4 H 2 O So, the balanced chemical equation is 8 H + + IO 4 - + 7 I - → 4 I 2 + 4 H 2 O . Hence, the ratio of coefficients of IO 4 - , I - and I 2 is 1 : 7 : 4 .

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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-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?

HARD

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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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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 :

HARD

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 8

1 g

of a moist sample of a mixture of

{KClO}_{3}

and

KCl

was dissolved in water and made upto

250 mL . 25 mL

of this solution was treated with

{SO}_{2}

to reduced chlorate into chloride and the excess

{SO}_{2}

was boiled off. When the total chloride was precipitated,

0.1435 g

AgCl

was obtained. In another experiment

25 mL

of the original solution was treated with

30 mL

0.2 N

solution of

{FeSO}_{4}

and unreacted

{FeSO}_{4}

required

37.5 mL

0.08 N

solution of an oxidizing agent for complete oxidation. Calculate the molar ratio of chlorate and chloride in the given mixture.

{Fe}^{2 +}

reacts with

{ClO}_{3}^{-}

according to equation:

{ClO}_{3}^{-} + 6 {Fe}^{2 +} + 6 H^{+} \to {Cl}^{-} + 6 {Fe}^{3 +} + 3 H_{2} O

Also calculate the mass percent of moisture present in the moist sample.

HARD

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 9

A steel sample is to be analysed for

Cr

and

Mn

simultaneously. By suitable treatment the

Cr

is oxidized to

{Cr}_{2} O_{7}^{2 -}

and the

Mn

{MnO}_{4}^{-} . A 10.00 g

sample of steel is used to produce

250.0 mL

of a solution containing

{Cr}_{2} O_{7}^{2 -}

and

{MnO}_{4}^{-} . A 10.00 mL

portion of this solution is added to

{aBaCl}_{2}

solution and by proper adjustment of the acidity, the chromium is completely precipitated as

{BaCrO}_{4}; 0.0549 g

is obtained. A second

10.00 mL

portion of this solution requires exactly

15.95 mL

0.0750 M

standard

{Fe}^{2 +}

solution for its titration (in acid solution). Calculate the

%

Mn

and

%

Cr

in the steel sample.

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 10

1.5 g

sample containing oxalic acid and some inert impurity was dissolved in enough water and volume made up to

250 mL

. A

20 mL

portion of this solution was then mixed with

30 mL

of an alkali solution. The resulting solution was then treated with stoichiometric amount of

{CaCl}_{2}

just needed for precipitation of oxalate as

{CaC}_{2} O_{4}

. Solution was filtered off and filtrate was finally titrated against

0.1 M HCl

solution.

8.0 mL

of acid was required to reach the equivalence point. At last, the above neutral solution was treated with excess of

{AgNO}_{3}

solution and

AgCl

obtained was washed, dried and weighed to be

0.4305 g . De -

termine mass percentage of oxalic acid in the original sample:

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 13

Write balanced net ionic equations for the following reactions in basic solution :

(a) ${Tl}_{2} O_{3} (s) + {NH}_{2} OH (aq) ⟶ TlOH (s) + N_{2} (g)$

(b) $Cu {({NH}_{3})}_{4}^{2 +} (aq) + S_{2} O_{4}^{2 -} (aq) ⟶ {SO}_{3}^{2 -} (aq) + Cu (s) + {NH}_{3} (aq)$

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 15

Consider the reaction

H^{+} + {IO}_{4}^{-} + I^{-} \to I_{2} + H_{2} O

. Find the ratio of coefficients of

{IO}_{4}^{-}, I^{-}

and

I_{2}

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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