Redox Reactions Test

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Redox Reactions Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Sulphur dioxide bleaches colouring matter by ___________.

A
dehydration

B
oxidation

C
decomposition

D
reduction

Solution

Sulphur dioxide bleaches coloring matter by reduction. Bleaching action of sulphur dioxide is not permanent.
$$SO_2(g) + 2H_2O \rightarrow H_2SO_4(aq) + 2[H^+]$$
Question 2
1 / -0

When an oxidising agent reacts, it consumes electrons. In the following changes, which one of the oxidising agents has consumed maximum number of electrons per mole?

A
$$Cr{O}_{4}^{2-}\longrightarrow {Cr}^{3+}$$

B
$$Mn{O}_{2}\longrightarrow {Mn}^{3+}$$

C
$$Mn{O}_{4}^{2-}\longrightarrow {Mn}^{2+}$$

D
$$Mn{O}_{4}^{-}\longrightarrow {Mn}^{2+}$$

Solution

Oxidation number of $$Mn$$ in $$Mn{O}_{4}$$ is $$+7$$
Therefore, conversion of $${Mn}^{2+}$$ from $${Mn}^{+7}$$ will have to accept $$5$$ electrons, which is highest.
Question 3
1 / -0

The formation of nitrous oxide from nitrogen and oxygen is the example for

A
Decomposition reaction

B
Chemical combination of one element and one compound

C
Chemical combination of two compounds

D
Chemical combination of two elements

Solution

The formation of nitrous oxide from nitrogen and oxygen is the example for chemical combination of two elements.
Question 4
1 / -0

Identify the pair of binary corresponds in which nitrogen exhibits the lowest and the highest oxidation state.

A
$$N{ H }_{ 3 },N{ O }_{ 2 }$$

B
$${ N }_{ 3 }H,{ N }_{ 2 }{ O }_{ 5 }$$

C
$${ N }_{ 2 },HN{ O }_{ 3 }$$

D
$${ N }_{ 2 }O,{ N }_{ 2 }{ O }_{ 5 }$$

Solution

Oxidation of nitrogen in $${ N }_{ 3 }H$$ is $$-\frac { 1 }{ 3 } $$, which is the lowest possible.
Oxidation of nitrogen in $${ N }_{ 2 }{ O }_{ 5 }$$ is $$+5$$ which is the highest possible.
Question 5
1 / -0

"Nitric oxide reacts with oxygen to form reddish brown nitrogen dioxide gas". Choose the correct equation for the above reaction

A
$$2NO + O_{2} \rightarrow 2NO_{2}$$

B
$$N_{2}O + O_{2} \rightarrow NO_{2} + NO$$

C
$$NaNO_{3} \xrightarrow {heat} 2NaNO_{2} + O_{2}$$

D
$$2NO_{3} + O_{2} \rightarrow 2NO_{2} + 2O_{2}$$

Solution

$$NO$$: Nitric oxide
$$NO_2$$: Nitrogen dioxide gas; It has a reddish brown colour.
Therefore, equation is $$2NO+O_2\longrightarrow 2NO_2$$.
Question 6
1 / -0

A compound contains $$X, Y\ and\ Z$$ atoms. The oxidation states of $$X, Y\ and\ Z$$ are $$+2, + 2\ and\ -2$$ respectively. The possible formula of the compound is:

A
$$XYZ_2$$

B
$$Y_2(XZ_3)_2$$

C
$$X_3(Y_4Z)_2$$

D
$$X_3(YZ_4)_3 $$

Solution

Sum of the oxidation numbers of atoms in it, is zero.
Therfore Answer is option(A)
Question 7
1 / -0

Heat is released in :

A
Exothermic reactions

B
Endothermic reactions

C
Both A and B

D
None of these

Solution

Heat is released in exothermic reactions.
Question 8
1 / -0

What is the oxidation state of $$Mn$$ in $$KMnO_4$$?

A
-7

B
-3

C
0

D
+3

E
+7

Solution

Option $$E$$ is the correct answer.
$$ {KMnO}_{4} $$ dissociates as $$ {K}^{+} $$ and $$ {MnO}_{4}^{-} $$ . Thus we will now calculate the oxidation number of $$ {Mn} $$
Let us suppose it has an oxidation state of X. Thus the equation is-
$$ {+1} $$$$ {+} $$ $$ {X} $$ $$ {-} $$ $$ {8} $$ $$ {=} $$ $${0}$$
Thus the value of X is $$ {+7} $$
The equation has been written in that manner because $$ {KMnO}_{4} $$ has an overall oxidation number of 0. $$ {K} $$ has an oxidation number of $$ {+1} $$ and $$ {O} $$ has an oxidation number of $$ {-2} $$. Thus four oxygen atoms will have an oxidation number of $$ {-8} $$
Oxidation numbers on both the sides should be balanced and hence we get the value to be +7.
Question 9
1 / -0

Assign oxidation numbers to each atom in $$NH_{4}NO_{2}$$:

A
$$N = 4, -4; H = 1; O = -2$$

B
$$N = -3; H = 1; O = 3$$

C
$$N = -1; H = 1; O = -2$$

D
$$N = 3, -3; H = 1; O = -2$$

Solution

$$NH_4NO_2$$
Oxidation number of $$N:x+4-1=0$$
$$\Rightarrow x+3=0$$
$$\Rightarrow x=-3$$.
Oxidation number of $$H: -3+4x-1=0$$
$$\Rightarrow 4x=3+1$$
$$\Rightarrow x=1$$
Oxidation number of $$O: -2$$
Question 10
1 / -0

What is the oxidation state of bromine in $$Al{Br}_{3}$$?

A
$$-1$$

B
$$+3$$

C
$$0$$

D
$$-3$$

Solution

Oxidation state of $$Br$$ in $$AlBr_3: +3+3x=0$$
$$\Rightarrow 3x=-3$$
$$\Rightarrow x=-1$$.

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Redox Reactions Test - 29

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Redox Reactions Test - 29

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Question 1

dehydration

oxidation

decomposition

reduction

Solution

Question 2

$$Cr{O}_{4}^{2-}\longrightarrow {Cr}^{3+}$$

$$Mn{O}_{2}\longrightarrow {Mn}^{3+}$$

$$Mn{O}_{4}^{2-}\longrightarrow {Mn}^{2+}$$

$$Mn{O}_{4}^{-}\longrightarrow {Mn}^{2+}$$

Solution

Question 3

Decomposition reaction

Chemical combination of one element and one compound

Chemical combination of two compounds

Chemical combination of two elements

Solution

Question 4

$$N{ H }_{ 3 },N{ O }_{ 2 }$$

$${ N }_{ 3 }H,{ N }_{ 2 }{ O }_{ 5 }$$

$${ N }_{ 2 },HN{ O }_{ 3 }$$

$${ N }_{ 2 }O,{ N }_{ 2 }{ O }_{ 5 }$$

Solution

Question 5

$$2NO + O_{2} \rightarrow 2NO_{2}$$

$$N_{2}O + O_{2} \rightarrow NO_{2} + NO$$

$$NaNO_{3} \xrightarrow {heat} 2NaNO_{2} + O_{2}$$

$$2NO_{3} + O_{2} \rightarrow 2NO_{2} + 2O_{2}$$

Solution

Question 6

$$XYZ_2$$

$$Y_2(XZ_3)_2$$

$$X_3(Y_4Z)_2$$

$$X_3(YZ_4)_3 $$

Solution

Question 7

Exothermic reactions

Endothermic reactions

Both A and B

None of these

Solution

Question 8

-7

-3

0

+3

+7

Solution

Question 9

$$N = 4, -4; H = 1; O = -2$$

$$N = -3; H = 1; O = 3$$

$$N = -1; H = 1; O = -2$$

$$N = 3, -3; H = 1; O = -2$$

Solution

Question 10

$$-1$$

$$+3$$

$$0$$

$$-3$$

Solution

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