Introduction ...

Introduction to Oxidation and Reduction

Trending Questions

Q. For

109 %

labeled oleum if the number of moles of

H_{2} S O_{4}

and free

S O_{3}

be x and y respectively, then what will be the value of

\frac{x + y}{x - y} ?

$\frac{1}{3}$
$9.9$
$1$
$18$

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20 mL

0.1 M

solution of compound

N a_{2} C O_{3} \cdot N a H C O_{3} \cdot 2 H_{2} O

is titrated against

0.05 M H C l

x mL

H C l

is used when phenolphthalein is used as an indicator and

y mL

H C l

is used when methyl orange is the indicator in two separate titrations. Hence

(y - x)

is:

$40 mL$
$80 mL$
$120 mL$
None of these

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Q. Which one of the following reaction represent redox reaction ?

$Z n O (s) + C (s) \to Z n (s) + C O (g)$
$C u O (s) + H_{2} (s) \to C u (s) + H_{2} O (g)$
$2 A g_{2} O (s) \to 4 A g (s) + O_{2} (g)$
All of the above.

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Q. Calculate the percentage of free

S O_{3}

in an oleum that is labelled

118 % H_{2} S O_{4}

$18 %$
$40 %$
$80 %$
$72 %$

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Q. What is the

%

of free

S O_{3}

in an oleum that is labelled as

104.5 % H_{2} S O_{4}

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Q. Both iodometric and iodimetric titrations are carried out in:

Basic medium
Acidic medium
Both acidic and basic medium
Neutral medium

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4.5 g

water is added into an oleum sample labelled as

109 % H_{2} S O_{4} .

What is the amount of free

S O_{3}

remaining in the solution?

$10 g$
$12 g$
$20 g$
$24 g$

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Q. In an aqueous solution of barium nitrate, the

[N O_{3}^{-}]

is 0.080 M. This solution can be labeled as

$0.040 N (B a (N O_{2})_{2}$
$0.160 M B a (N O_{3})_{2}$
$0.080 N B a (N O_{3})_{2}$
$0.080 M N O_{3}^{-}$

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Q. Calculate the weight of

N a_{2} C O_{3}

of 85% purity required to prepare to neutralize 45.6 mL of 0.235 N

H_{2} S O_{4}

2.05 g
1.44 g
0.25 g
0.67 g

View Solution

20 g

of a sample of

B a (O H)_{2}

is dissolved in

10 m L

0.5 N H C l

solution. The excess of

H C l

was titrated with

0.2 N N a O H .

The volume of

N a O H

required was

10 m L

. Calculate the percentage of

B a (O H)_{2}

in the sample.
(Given molar mass of

B a

137 g m o l^{- 1}

)

$12.8 %$
$1.28 %$
$50 %$
$33.33 %$

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Q. Which of the following reactions is an example of a redox reaction?

 JEE MAIN 2017

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H_{2} O_{2} + 2 K I 40 % y i e l d ⟶ I_{2} + 2 K O H H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O

150 mL

of a

H_{2} O_{2}

sample was divided into two parts. The first part was treated with

K I

and the formed

K O H

required

200 mL

\frac{M}{2} H_{2} S O_{4}

for complete neutralisation. The other part was treated with

K M n O_{4}

yielding

6.72 L

O_{2}

1 atm

and

273 K

. Using the

%

yields indicated, find the volume strength of the

H_{2} O_{2}

sample used.

View Solution

Q. The percentage of free

S O_{3}

in an oleum is

60 %

. The given sample in terms of

% H_{2} S O_{4}

can be labelled as :

$160 %$
$140 %$
$113.5 %$
$104.5 %$

View Solution

Q. In basic medium,

C r O_{4}^{2 -}

oxidize

S_{2} O_{3}^{2 -}

to form

S O_{4}^{2 -}

and itself changes to

C r (O H)_{4}^{-} .

How many mL of 0.154 M

C r O_{4}^{2 -}

are required to react with 40 mL of 0.246 M

S_{2} O_{3}^{2 -}

200 mL
156.4 mL
170.4 mL
190.4 mL

View Solution

Q. 2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of

C a (O H)_{2}

solution is required. What is the molarity of

C a (O H)_{2}

solution?

1.0 M
0.1 M
0.05 M
0.5 M

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Q. For a cell involving one electron

E_{c e l l}^{o} = 0.59 V a t 298 K,

the equilibrium constant for cell reaction is :

[Given that \frac{2.303 R T}{F} = 0.059 V a t T = 298 K]

(NEET-2019)

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Q. In a reaction vessel, 1.184 g of NaOH is required to be added for completing the reaction. How many millilitre of 0.15 M NaOH should be added for this requirement?

197.33 ml
297.33 ml
97.33 ml
107.33 ml

View Solution

0.75 g

of oleum sample was diluted with water. If the solution required

62.5 m L

0.25 N N a O H,

the

%

of free

S O_{3}

in the sample will be:

$9.3 %$
$23.5 %$
$26.0 %$
$12.5 %$

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Q. According to the calssical concept, which of the following cannot be termed as reduction?

Removal of electropositive element
Addition of Hydrogen
Removal of Oxygen
Removal of electronegative element.

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Q. In basic medium,

C r O_{4}^{2 -}

oxidize

S_{2} O_{3}^{2 -}

to form

S O_{4}^{2 -}

and itself changes to

C r (O H)_{4}^{-} .

How many mL of 0.154 M

C r O_{4}^{2 -}

are required to react with 40 mL of 0.246 M

S_{2} O_{3}^{2 -}

200 mL
156.4 mL
170.4 mL
190.4 mL

View Solution

12 g

M g

will react completely with a monoprotic acid to give:

1 mole of $O_{2}$
1 mole of acid
5 moles of $H_{2}$
1/2 mole of $H_{2}$

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Q. What is the

%

of free

S O_{3}

in an oleum that is labelled as

104.5 % H_{2} S O_{4}

View Solution

Q. In compounds formed by the union of different elements, the more electronegative atom has a negative oxidation number.

False
True

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Q. Oxidation number of cobalt in

[C o (N H_{3})_{6}] C l_{2} B r

is -

6
zero
3
2

View Solution

Q. 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:

eq. wt. of $C u_{2} S$ is $\frac{M_{1}}{8}$
eq. wt. of $C u S$ is $\frac{M_{2}}{6}$
eq. wt. of $B a (M n O_{4})_{2}$ is $\frac{M_{3}}{5}$
$C u_{2} S$ and $C u S$ both have same no. of equivalents in mixture

View Solution