Redox Reactions(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

REDOX REACTIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

A reducing agent is a substance which can

Accept electron
Donate electrons
Accept protons
Donate protons

Discuss Question

Answer: Option B. -> Donate electrons
:
B

A substance which is capable of reducing other substances by donating electrons is called a reducing agent or reductant or reducer.

Question 2.

Which of the following behaves as both oxidising and reducing agents

$H_{2} S O_{4}$
$S O_{2}$
$H_{2} S$
$H N O_{3}$

Discuss Question

Answer: Option B. ->

S O_{2}

:
B

In $S O_{2}$ , sulphur has a +4 oxidation state. In $H_{2} S O_{4}$ , Sulphur exhibits its highest oxidation state of +6.
On the other hand - in $H_{2} S$ , sulphur exists in its lowest -2 oxidation state. Thus, we can surmise
that $S O_{2}$ could act as both an oxidising agent as well as a reducing agent. Nitrogen exhibits its highest oxidation
state of +5 in $H N O_{3}$ .

Question 3.

Which of the following cannot work as oxidising agent?

$O_{2}$
$K M n O_{4}$
$I_{2}$
None of these

Discuss Question

Answer: Option C. ->

I_{2}

:
C

Both $K M n O_{4}$ and $O_{2}$ are well known oxidizing agents. Of the given options $I_{2}$ is a reducing agent.

Question 4.

Which of the following is the most powerful oxidizing agent

$F_{2}$
$C l_{2}$
$B r_{2}$
$I_{2}$

Discuss Question

Answer: Option A. ->

F_{2}

:
A

Fluorine is the most powerful oxidising agent as reflected by a very high reduction potential of nearly 2.87.

Question 5.

Several blocks of magnesium are fixed to the bottom of a ship to

Keep away the sharks
Make the ship lighter
Prevent action of water and salt
Prevent puncturing by under-sea rocks

Discuss Question

Answer: Option C. -> Prevent action of water and salt
:
C

Magnesium provides cathodic protection and prevents rusting or corrosion thus preventing the action of water and salt.

Question 6.

In the reaction, $8 A l + 3 F e_{3} O_{4} \to 4 A l_{2} O_{3} + 9 F e$ , the number of electrons transferred from reductant to oxidant is:

Discuss Question

Answer: Option D. -> 24
:
D
Elemental metallic Aluminium carries an oxidation number of 0.
From the products, we see that

0 A l

becomes

3 + A l^{3 +}

.
Clearly, Aluminium is the reducing agent.
There are 8 Aluminium atoms on the left and they all become

A l^{3 +}

.
So overall, there are 24 electrons transferred.

Question 7.

When $K M n O_{4}$ acts as an oxidising agent and ultimately forms $[M n O_{4}]^{- 2}, M n O_{2}, M n_{2} O_{3}, M n^{+ 2}$ then the number of electrons transferred in each case respectively is

4, 3, 1, 5
1, 5, 3, 7
1, 3, 4, 5
3, 5, 7, 1

Discuss Question

Answer: Option C. -> 1, 3, 4, 5
:
C

This has to be answered based on the oxidation number of the central metal atom in each case.
To start with Mn has +7 oxidation number in $K M n O_{4}$ . Likewise, we have
$+ 6 M n O_{4}^{2 -}$
$+ 4 M n O_{2}$
$+ 3 M n_{2} O_{3}$
$+ 2 M n^{2 +}$
The number of electrons transferred in each case is 1, 3, 4, 5 respectively.

Question 8.

Strongest reducing agent is

$F^{-}$
$C l^{-}$
$B r^{-}$
$I^{-}$

Discuss Question

Answer: Option D. ->

I^{-}

:
D

$I^{-}$ act as a more reducing agent than other ions.

Question 9.

An element A in a compound ABD has oxidation number A ^n-. It is oxidised by $C r_{2} O_{7}^{- 2}$ in acidic medium. In the experiment 1.68 $\times$ $10^{- 3}$ x moles of $K_{2} C r_{2} O_{7}$ were used for 3.26 $\times$ $10^{- 3}$ moles of ABD. The new oxidation number of A after oxidation is:

Discuss Question

Answer: Option B. -> 3-n
:
B

First - let us write down the reaction:
$C r_{2} O_{7}^{2 -} + A^{n -} B D ⟶ C r^{3 +} + A^{- n + x}$
From stoichiometry, we have:
Equivalents of $C r_{2} O_{7}^{2 -}$ = equivalents of ABD
$1.68 \times 10^{- 3} = 3.26 \times 10^{- 3} \times x$
Solving, we get $x = 3$ .
Hence, the new oxidation number is: 3-n