Redox Reactions Test

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Redox Reactions Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

$$One\ mole$$ of $$N_2H_4$$ loses $$10\ moles$$ of electrons to form a new compound $$Y$$. Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in compound $$Y$$?
[There is no change in the oxidation state of hydrogen.]

A
$$-1$$

B
$$-3$$

C
$$+3$$

D
$$+5$$

Solution

$$\text{One} $$ $$\text{Mole}$$ of $$N_2H_4$$ loses $$10$$ $$\text{moles}$$ of $$e^-\longrightarrow Y$$

The oxidation state of $$H$$ remains unchanged and all the $$\text{Nitrogen"}$$ appears in $$Y.$$

$$\therefore$$ $$e^-$$ must be lost from $$\text{nitrogen}$$,

$$1$$ $$\text{mole}$$ of $$N_2H_4$$ contains $$2$$ $$\text{moles}$$ of $$\text{nitrogen}$$.

$$\Longrightarrow 2$$ $$\text{moles}$$ of $$N$$ loses $$10$$ $$\text{moles}$$ of $$e^-$$.

$$\therefore 1$$ $$\text{mole}$$ loses $$5$$ $$\text{moles}$$ of $$e^-$$

Initial oxidation state of $$N$$ in $$N_2H_4=-2$$

when it loses $$5$$ $$\text{moles}$$ of $$e^-$$

Final oxidation state of each $$N=-2+5=+3.$$

(since it loses $$e^-$$, therefore it must acquire positive charge).

Hence, the correct option is $$C$$
Question 2
1 / -0

Which of the following statements is correct?

A
Oxidation of a substance is followed by reduction of another

B
Reduction of a substance is followed by oxidation of another

C
Oxidation and reduction are complementary reactions

D
It is not necessary that both oxidation and reduction should take place in the same reaction

Solution

Oxidation and reduction are complementary reactions because oxidation is followed by loss of electrons and reduction is followed by gain of electrons.
So this statement is correct.
Question 3
1 / -0

Select the compound in which chlorine is assigned the oxidation number $$+5$$.

A
$$HClO_4$$

B
$$HClO_2$$

C
$$HClO_3$$

D
HCl

Solution

Oxidation state of hydrogen=+1
Oxidation state of oxygen=-2
Let oxidation state of chlorine is x,
In $$HCl{ O }_{ 4 }$$, $$1+x+4(-2)=0, x=+7$$
In $$HCl{ O }_{ 3 }$$  $$1+x+3(-2)=0, x=+5$$
In $$HCl{ O }_{2 }$$  $$1+x+2(-2)=0, x=+3$$
In $$HCl{ }_{ }$$   $$1+x=0, x=-1$$
In $$HCl{ O }_{ 4 }$$, chlorine is assigned the oxidation state of +5.
Question 4
1 / -0

Oxidation number of P in $$Mg_2P_2O_7$$ is?

A
$$+3$$

B
$$+2$$

C
$$+5$$

D
$$-3$$

Solution

Let the oxidation number of $$P$$ in $${ Mg }_{ 2 }P_{ 2 }{ O }_{ 7 }$$ be $$x$$. The oxidation states of $$Mg$$ and $$O$$ are +2 and -2 respectively. Since the atom as whole is neutral, following equation holds true:
$$2\times (+2)\quad +\quad 2\times (x)\quad +\quad 7\times (-2)\quad =\quad 0\\ \Longrightarrow \quad 4\quad +\quad 2x\quad -14\quad =\quad 0\\ \Longrightarrow \quad 2x\quad =\quad 14-4\\ \Longrightarrow \quad 2x\quad =\quad 10\\ \Longrightarrow \quad x\quad =\quad \dfrac { 10 }{ 2 } \quad =\quad 5$$

Hence, answer is C.
Question 5
1 / -0

In which of the following pairs, the oxidation states of sulphur and chromium are same?

A
$$SO^{2-}_3, CrO^{2-}_4$$

B
$$SO_3, CrO^{2-}_4$$

C
$$SO_2, CrO^{2-}_4$$

D
$$SO_2, Cr_2O^{2-}_7$$

Solution

Oxidation number of oxygen is $$-2$$. Therefore,
In $$SO_3^{-2}$$,oxidation number of sulphur is $$x+3(-2)=-2,x=+4$$
In $$CrO_4^{-2}$$,oxidation number of chromium is $$x+4(-2)=-2,x=+6$$
In $$SO_3$$,oxidation number of sulphur is $$x+3(-2)=0,x=+6$$
In $$SO_2$$,oxidation number of sulphur is $$x+2(-2)=0,x=+4$$
In $$Cr_2O_7^{-2}$$,oxidation number of chromium is $$2(x)+7(-2)=-2,x=+6$$
So B) is correct pair having same oxidation state of sulphur and chromium.
Question 6
1 / -0

The oxidation number of phosphorus in $$PO^{3-}_4, P_4O_{10}$$ and $$P_2O^{4-}_7$$ is?

A
$$+5$$

B
$$+3$$

C
$$-3$$

D
$$+2$$

Solution

Oxidation number of oxygen in these compounds is -2. So we can find oxidation number of phosphorus.
In $${ PO }_{ 4 }^{ 3- }$$,$$x+4(-2)=-3,x=+5$$
In $${ P }_{ 4 }{ O }_{ 10 }$$,$$4(x)+10(-2)=0,x=+5$$
In $${ { P }_{ 2 }O }_{ 7 }^{ -2 }$$,$$2(x)+7(-2)=-4,x=+5$$
Question 7
1 / -0

Which one of the following statements is not correct?

A
Oxidation number of S in $$(NH_4)_2S_2O_8$$ is $$+6$$

B
Oxidation number of Os in $$OsO_4$$ is $$+8$$

C
Oxidation number of S in $$H_2SO_5$$ is $$+8$$

D
Oxidation number of O in $$KO_2$$ is $$-\displaystyle\frac{1}{2}$$

Solution

When $${ (N{ H }_{ 4 }) }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$$ dissolved in water, it dissociates to give $$NH_4^{+}$$ and $$S_2O_8^{-2}$$.As oxidation number of oxygen is -2 so oxidation number of sulphur is $$(x+8(-2)=-2,x=+7) is +7$$.
Oxidation number of $$Os$$ in $$OsO_4$$ is $$x+4(-2)=0,x=+8$$
Oxidation number of $$S$$ in $$H_2SO_5$$ is $$2(+1)+x+5(-2)=0,x=+8$$
Oxidation number of $$O$$ in $$KO_2$$ is $$-\frac { 1 }{ 2 } $$ (when oxygen reacts with alkali metal,alkali metals have +1 oxidation state)
So A is not correct statement.
Question 8
1 / -0

It is found that V forms a double salt isomorphous with Mohr's salt. The oxidation number of V in this compound is_________.

A
$$+3$$

B
$$+2$$

C
$$+4$$

D
$$-4$$

Solution

Double salt of $$V$$ isomorphous with Mohr's salt is $${ (N{ H }_{ 4 }) }_{ 2 }V{ { (SO }_{ 4 } })_{ 2 }.6{ H }_{ 2 }O$$. When double salt dissolves in water it dissociates to give $$NH_4^+$$,$$V^{+2}$$ and $$SO_4^{-2}$$.So oxidation state of vanadium is +2.
Question 9
1 / -0

The oxidation number of phosphorus in $$Ba(H_2PO_2)_2$$ is:

A
$$+3$$

B
$$+2$$

C
$$+1$$

D
$$-1$$

Solution

$$Ba{ { (H }_{ 2 }{ PO }_{ 2 }) }_{ 2 }$$ can be written as $${ Ba }^{ +2 }{ { (H }_{ 2 }{ PO }_{ 2 }) }^{ -1 }$$

Let the oxidation number of $$P$$ in $${ { (H }_{ 2 }{ PO }_{ 2 }) }^{ -1 }$$ be $$x$$. The oxidation states of $$H$$ and $$O$$ are $$+1$$ and $$-2$$ respectively. Since the atom as whole is single negatively charged, following equation holds true:

$$2\times (+1) + 1\times (x) + 2\times (-2) = -1\\ \Longrightarrow 2 + x - 4 =-1\\ \Longrightarrow x = 4-2-1\\ \Longrightarrow x = 4-3\\ \Longrightarrow x = +1$$

Hence, an answer is $$C$$.
Question 10
1 / -0

The oxidation number of phosphorus in $$Ba(H_2PO_2)_2$$ is:

A
$$+3$$

B
$$+2$$

C
$$+1$$

D
$$-1$$

Solution

In this compound,oxidation number of $$Ba$$ is +2, $$H$$ is +1 and $$O$$ is -2. Now we can oxidation number of $$P$$:
Let oxidation number of phosphorus is $$x$$, so
$$2+2(2(+1)+x+2(-2))=0$$
$$2+2(x-2)=0$$
$$x=+1$$

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Redox Reactions Test - 34

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Redox Reactions Test - 34

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Question 1

$$-1$$

$$-3$$

$$+3$$

$$+5$$

Solution

Question 2

Oxidation of a substance is followed by reduction of another

Reduction of a substance is followed by oxidation of another

Oxidation and reduction are complementary reactions

It is not necessary that both oxidation and reduction should take place in the same reaction

Solution

Question 3

$$HClO_4$$

$$HClO_2$$

$$HClO_3$$

HCl

Solution

Question 4

$$+3$$

$$+2$$

$$+5$$

$$-3$$

Solution

Question 5

$$SO^{2-}_3, CrO^{2-}_4$$

$$SO_3, CrO^{2-}_4$$

$$SO_2, CrO^{2-}_4$$

$$SO_2, Cr_2O^{2-}_7$$

Solution

Question 6

$$+5$$

$$+3$$

$$-3$$

$$+2$$

Solution

Question 7

Oxidation number of S in $$(NH_4)_2S_2O_8$$ is $$+6$$

Oxidation number of Os in $$OsO_4$$ is $$+8$$

Oxidation number of S in $$H_2SO_5$$ is $$+8$$

Oxidation number of O in $$KO_2$$ is $$-\displaystyle\frac{1}{2}$$

Solution

Question 8

$$+3$$

$$+2$$

$$+4$$

$$-4$$

Solution

Question 9

$$+3$$

$$+2$$

$$+1$$

$$-1$$

Solution

Question 10

$$+3$$

$$+2$$

$$+1$$

$$-1$$

Solution

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