Redox Reactions Test

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Redox Reactions Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

$$H_2SO_5 + H_2O \rightarrow H_2SO_4 + H_2O_2$$
Oxidation number of sulphur in $$H_2SO_5$$ in the above reaction is :

A
$$6$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

Formation of $$H_2O_2$$ indicates that there is one peroxy linkage in $$H_2SO_5$$. The oxidation state of H = $$+1$$.
Therefore, $$x+2-2-6 = 0$$ or, $$x=+6$$
Hence, $$x= 6$$.
Thus, oxidation number of sulphur = $$+6$$.
The structure is given below:
Question 2
1 / -0

The difference in the oxidation numbers of the two types of sulphur atoms in $$Na_2S_4O_6$$ is :

A
5

B
4

C
3

D
6

Solution

In $$Na_2S_4O_6$$, the oxidation number of end sulphur atoms is $$+5$$ each and the oxidation number of middle sulphur atoms is $$0$$ each.
The difference in the oxidation numbers of the two types of sulphur atoms is $$5-0=0$$.
The structure of the compound is given below:
Question 3
1 / -0

In the following reaction:
$$Cr(OH)_3 + (OH)^- + {IO_3}^- \rightarrow {CrO_4}^{2-} + H_2O + I^-$$

A
$${IO_3}^- $$ is oxidising agent

B
$$Cr(OH)_3$$ is oxidised

C
$$6e^-$$ are being taken per $$I$$ atom

D
All of the above are correct

Solution

The redox reaction is shown below:
$$Cr(OH)_3 + (OH)^- + {IO_3}^- \rightarrow {CrO_4}^{2-} + H_2O + I^-$$
The oxidation number of $$I$$ changes from $$+5$$ to $$-1$$.
Thus, each iodine takes $$6$$ electrons and is reduced.
Hence, $${IO_3}^-$$ is the oxidizing agent.
The oxidation number of chromium changes from $$+3$$ to $$+6$$.
Thus, $$Cr(OH)_3$$ is oxidized.
Question 4
1 / -0

What is the oxidation state of Mn in the compound $$K_2MnO_4$$?

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$
Question 5
1 / -0

Which among the following shows maximum oxidation state?

A
V

B
Fe

C
Mn

D
Cr

Solution

Metal Maximum oxidation state
V +3
Cr +6
Fe +3
Mn +7
Hence, $$Mn$$ shows the maximum oxidation state.
Question 6
1 / -0

Oxidation number of carbon in "$$C_3O_2$$" and "$$Mg_2C_3$$" are respectively :

A
$$-\frac{4}{3}, + \frac{4}{3}$$

B
$$+\frac{4}{3}, -\frac{4}{3}$$

C
$$-\frac{2}{3}, + \frac{2}{3}$$

D
$$+\frac{2}{3}, -\frac{2}{3}$$

Solution

Let oxidation state of $$C$$ in $$C_3O_2$$ be $$x$$.
Since the compound is neutral, sum of oxidation states of all elements = 0
$$ 3x+2(-2)=0$$
or, $$ x = \frac{4}{3} $$
Oxidation state of $$C$$ in $$Mg_2C_3$$ is $$y$$.
Since the compound is neutral, sum of oxidation states of all elements = 0
$$2\times2+3\times y = 0$$
or, $$ y = -\frac{4}{3} $$
Question 7
1 / -0

In the reaction $$2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 NaI$$, the oxidation state of sulphur is :

A
decreased

B
increased

C
unchanged

D
none of these

Solution

Given the reaction:
$$2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 NaI$$
+2 +2.5
We can see that the oxidation state of S increases.
Question 8
1 / -0

The oxidation number and covalency of sulphur in the sulphur molecule ($$S_8$$) are, respectively :

A
0 and 2

B
+6 and 8

C
0 and 8

D
+6 and 2

Solution

From the figure, it can be seen that oxidation numberof S is zero (elemental form) and its co-valency is 2 as each atom is linked to two other atoms.
Question 9
1 / -0

The violent reaction between sodium and water is an example of :

A
reduction

B
oxidation

C
redox reaction

D
neutralisation reaction

Solution

Reaction:
$$Na + H_2O \rightarrow NaOH + (1/2)H_2$$
0 +1 +1 0
The oxidation state of $$Na$$ changes from $$0$$ to $$+1$$ while that of $$H$$ from $$+1$$ to $$0$$.
Here, oxidation state of reactants changes so, it is a redox reaction.
Question 10
1 / -0

Arrange the following in correct order with respect to oxidation state of S.
$$ SO_{3}^{2-}, S_2 O_{4}^{2-}$$ and $$S_2 O_{6}^{2-}$$

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
$$S_2 O_{6}^{2-}< S_2 O_{4}^{2-} < SO_{3}^{2-} $$

Solution

Oxidation state of S in
$$S_2 O_{4}^{2-} = +3$$
$$SO_{3}^{2-} = +4$$
$$S_2 O_{6}^{2-} = +5$$
Hence, the correct order is $$ S_2 O_{4}^{2-} < SO_{3}^{2-} < S_2 O_{6}^{2-}$$.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 21

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Redox Reactions Test - 21

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Question 1

$$6$$

$$4$$

$$5$$

$$2$$

Solution

Question 2

5

4

3

6

Solution

Question 3

$${IO_3}^- $$ is oxidising agent

$$Cr(OH)_3$$ is oxidised

$$6e^-$$ are being taken per $$I$$ atom

All of the above are correct

Solution

Question 4

$$3$$

$$4$$

$$5$$

$$6$$

Question 5

V

Fe

Mn

Cr

Solution

Question 6

$$-\frac{4}{3}, + \frac{4}{3}$$

$$+\frac{4}{3}, -\frac{4}{3}$$

$$-\frac{2}{3}, + \frac{2}{3}$$

$$+\frac{2}{3}, -\frac{2}{3}$$

Solution

Question 7

decreased

increased

unchanged

none of these

Solution

Question 8

0 and 2

+6 and 8

0 and 8

+6 and 2

Solution

Question 9

reduction

oxidation

redox reaction

neutralisation reaction

Solution

Question 10

$$ S_2 O_{4}^{2-} < SO_{3}^{2-} < S_2 O_{6}^{2-}$$

$$SO_{3}^{2-} < S_2 O_{4}^{2-} < S_2 O_{6}^{2-}$$

$$S_2 O_{4}^{2-} < S_2 O_{6}^{2-} < SO_{3}^{2-}$$

$$S_2 O_{6}^{2-}< S_2 O_{4}^{2-} < SO_{3}^{2-} $$

Solution

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