Redox Reactions — Page 2 | NEET Chemistry

Q11

(I) H 2 O 2 + O 3

\to

H 2 O + 2O 2 (II) H 2 O 2 + Ag 2 O

\to

2Ag + H 2 O + O 2 Role of Hydrogen peroxide in the above reactions is respectvely

A oxidizing in (I) and reducing in (II)

B reducing in (I) and oxidizing (II)

C reducing in (I) and (II)

D oxidizing in (I) and (II)

Correct Answer

Option C

Solution

So, H 2 O 2 acts as reducing agent in all those reactions in which O 2 is evolved.

Q12

In acidic medium, H 2 O 2 changes Cr 2 O 7 2- to CrO 5 which has two (

-

O

-

O

-

) bonds. Oxidation state of Cr in CrO 5 is

A +5

B +3

C +6

D

-

10

Correct Answer

Option C

Solution

CrO 5 has butterfly structure having two peroxo bonds.

Peroxo oxygen has –1 oxidation state.

Let oxidation state of Cr be ‘x’ CrO5 : x + 4(–1) + 1 (–2) = 0 $\Rightarrow$ x = +6

Q13

The pair of compounds that can exist together is

A FeCl 3 , SnCl 2

B HgCl 2 , SnCl 2

C FeCl 2 , SnCl 2

D FeCl 3 , KI

Correct Answer

Option C

Solution

Both FeCl 2 and SnCl 2 are reducing agents with low oxidation numbers.

Q14

When Cl 2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

A zero to +1 and zero to

-

5

B zero to

-

1 and zero to +5

C zero to

-

1 and zero to +3

D zero to +1 and zero to

-

3

Correct Answer

Option B

Solution

This is an example of disproportionation reaction and oxidation state of chlorine changes from 0 to –1 and +5.

Q15

A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number?

A S

B H

C Cl

D C

Correct Answer

Option C

Solution

KClO 3 + C 2 H 2 O 4 + H 2 SO 4 $\to$ K 2 SO 4 + CO 2 + KCl + H 2 O Maximum change in oxidation number of chlorine, i.e., from +5 to –1.

Q16

Oxidation numbers of P in PO

_4^{3 - }

, of S in SO

_4^{2 - }

and that of Cr in Cr 2 O

_7^{2 - }

are respectively

A +3, +6 and +5

B +5, +3 and +6

C

-

3, +6 and + 6

D +5, +6 and +6

Correct Answer

Option D

Solution

Let oxidation number of P in PO 4 3– be x. $\therefore$ x + 4(–2) = –3 $\Rightarrow$ x = +5 Let oxidation number of S in SO 4 2– be y. $\therefore$ y + 4(–2) = –2 $\Rightarrow$ y = +6 Let oxidation number of Cr in Cr 2 O 7 2– be z. $\Rightarrow$ 2z + 7(–2) = –2 $\Rightarrow$ z = +6

Q17

Number of moles of MnO

_4^{ - }

required to oxidize one mole of ferrous oxalate completely in acidic medium will be

A 7.5 moles

B 0.2 moles

C 0.6 moles

D 0.4 moles

Correct Answer

Option C

Solution

The balanced ionic equation for oxidation of ferrous oxalate by MnO 4 – in acidic medium is as follows : 3MnO 4 - + 5FeC 2 O 4 + 24H + $\to$ 3Mn 2+ + 10CO 2 + 12H 2 O + 5Fe 3+ Thus, 5 moles of FeC 2 O 4 require 3 moles of MnO 4 - .

So, 1 mole of FeC 2 O 4 requires =

{3 \over 5}

= 0.6 moles of MnO 4 -

Q18

The oxidation states of sulphur in the anions SO3

{_3^{2 - }}

, S 2 O

{_4^{2 - }}

and S 2 O

{_6^{2 - }}

follow the order

A S 2 O

{_4^{2 - }}

< SO

{_3^{2 - }}

< S 2 O

{_6^{2 - }}

B SO

{_3^{2 - }}

< S 2 O

{_4^{2 - }}

< S 2 O

{_6^{2 - }}

C S 2 O

{_4^{2 - }}

< S 2 O

{_6^{2 - }}

< SO

{_3^{2 - }}

D S 2 O

{_6^{2 - }}

< S 2 O

{_4^{2 - }}

< SO

{_3^{2 - }}

Correct Answer

Option A

Solution

SO 3 2– : oxidation state of ‘S’ is +4 S 2 O 4 2– : oxidation state of ‘S’ is +3.

S 2 O 6 2– : oxidation state of ‘S’ is +5.

So, the order is S 2 O

{_4^{2 - }}

< SO

{_3^{2 - }}

< S 2 O

{_6^{2 - }}

.

Q19

Consider the following reaction:

xMnO_4^- + yC_2O_4^{2-}

+ zH+

\to

xMn2+ + 2yCO2 +

{z \over 2}{H_2}O

The value's of x, y and z in the reaction are, respectively :

A 5, 2 and 16

B 2, 5 and 8

C 2, 5 and 16

D 5, 2 and 8

Correct Answer

Option C

Solution

After balancing the reaction we get

2MnO_4^- + 5C_2O_4^{2-}

+ 16H+ $\to$ 2Mn2+ + 10CO2 +

8{H_2}O

$\therefore$

x

= 2,

y

= 5 and z = 16

Q20

An example of a disproportionation reaction is :

A 2NaBr + Cl2

\to

2NaCl + Br2

B 2KMnO4

\to

2KMnO4 + MnO2 + O2 (3) (4)

C 2CuBr

\to

CuBr2 + Cu

D 2MnO4 + 10I– + 16H+

\to

2Mn2+ + 5I2 + 8H2O

Correct Answer

Option C

Solution

In disproportionation reaction one element undergoes both oxidation and reduction. Here disproportionation reaction is :