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} ?

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

View Solution

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

View Solution

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.

View Solution

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}

View Solution

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?

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

View Solution

Q. In an aqueous solution of barium nitrate, the

[N O_{3}^{-}]

is 0.080 M. This solution can be labeled as

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

View Solution

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}

)

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

View Solution

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

 JEE MAIN 2017

View Solution

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

View Solution

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)

View Solution

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:

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

View Solution

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.

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

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 $H_{2}$

View Solution

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

View Solution

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

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