Redox Reactions Test

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Redox Reactions Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

In the reaction between acidified $$ KMnO_{4}$$ and hot oxalic acid, the species that gains electrons is:

A
$$K^{+}$$

B
$$MnO_{4}^{-}$$

C
$$C_{2}O_{4}^{2-}$$

D
$$CO_{2}$$

Solution

The reaction between potassium permanganate and oxalic acid is a redox reaction.

During the reaction, potassium permanganate (which is a strong oxidizing agent) is reduced and oxalic acid is oxidized.

Thus, $$MnO_4^-$$ ion gains electrons.

Option B is correct.
Question 2
1 / -0

Oxidation number of sulphur in marshall's acid $$(H_{2}S_{2}O_{8})$$ is

A
$$+5$$

B
$$+8$$

C
$$+6$$

D
$$+7$$

Solution

Let $$x$$ be the oxidation number of sulphur in marshall's acid $$(H_2S_2O_8) $$. It contains a peroxide linkage. Thus, the oxidation number of two O atoms is $$-1$$ and that of the remaining O atoms is $$-2$$. The oxidation number of H is $$+1$$.
For the neutral molecule, the sum of the oxidation numbers is zero.
Therefore, $$2(+1)+2x+2(-1)+6(-2)=0$$ or, $$x=+6$$
The structure is given below:
Question 3
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$$H_{2}O_{2}+O_{3}\rightarrow H_{2}O+2O_{2}$$, in this reaction

A
$$H_{2} O_{2}$$ is bleached

B
$$H_{2} O_{2}$$ is oxidised

C
$$H_{2} O_{2}$$ is dehydrated

D
$$H_{2} O_{2}$$ is neither oxidised nor reduced

Solution

Oxidation half reactions is,
$$H_{2}O_{2}\rightarrow 2H^{+}+O_{2}+2e^{-}$$
There is a loss of electrons. Therefore, $$H_2O_2$$ is oxidized.
Question 4
1 / -0

Which of the following statement is wrong?

A
Oxidation number of oxygen is $$+1$$ in peroxides.

B
Oxidation number of oxygen is $$+2$$ in oxygen difluoride.

C
Oxidation number of oxygen is $$-\frac{1}{2}$$ in super oxides.

D
Oxidation number of oxygen is $$-2$$ in most of its compounds.

Solution

In peroxides ($$H_2O_2$$), oxygen has an oxidation state of $$-1$$.

In $$OF_2$$, oxidation state of oxygen is equal to $$+2$$.

In $$KO_2$$, oxidation state of oxygen is equal to $$-\cfrac {1} {2}$$

Oxidation number of oxygen is $$-2$$ in most of its compounds.
Question 5
1 / -0

Oxidation number of $$Cr$$ in $$KCrO_{4}^-$$ is:

A
$$+4$$

B
$$+6$$

C
$$+2$$

D
$$+8$$

Solution

Let x be the oxidation number of Cr in $$\displaystyle KCrO_{4}^- $$

$$\displaystyle +1+X+4(-2) = -1 $$

$$\displaystyle +1+X-8=-1 $$

$$\displaystyle X-7=-1 $$

$$\displaystyle X=+6 $$

Thus, the oxidation number of Cr in $$\displaystyle KCrO_{4}^- $$ is +6.

Option B is correct.
Question 6
1 / -0

$$1$$ mole of $$N_{2}H_{4}$$ lost $$10$$ moles of electrons and formed a new compound, $$Y$$, in which all hydrogens are present without any change in the oxidation state. The oxidation number of nitrogen in the new compound, $$Y$$, is:

A
$$+1$$

B
$$+2$$

C
$$+3$$

D
$$+5$$

Solution

The oxidation state of two nitrogen in hydrazine is $$-4$$ as four hydrogen atoms have $$=1$$ each.

Therefore, $$1$$ nitrogen has oxidation state of $$-2$$.

When hydrazine containing two nitrogen atoms loses $$10$$ electrons, the oxidation state of two nitrogen changes from $$-4$$ to $$+6$$.

Therefore, the oxidation state changes from $$-2 +5 = +3$$ per nitrogen atom.

Option C is correct.
Question 7
1 / -0

Which is not a redox reaction?

A
$$CaCO_{3}\rightarrow CaO+CO_{2}$$

B
$$2H_{2}+O_{2}\rightarrow 2H_{2}O$$

C
$$Na+H_{2}O\rightarrow NaOH+ \frac{1}{2}H_{2}$$

D
$$MnCl_{3}\rightarrow MnCl_{2}+ \frac{1}{2}Cl_{2}$$

Solution

$$CaCO_{3}\rightarrow CaO+CO_{2}$$

Since, the reaction does not involve oxidation and reduction, it is not a redox reaction.

In the decomposition of $$CaCO_{3}$$, there occurs no change in the oxidation state of elements.
Question 8
1 / -0

The oxidation number(s) of nitrogen in $$NH_{4} NO_{3}$$ is/are

A
$$-3$$ & $$+5$$

B
$$+3$$ & $$+5$$

C
$$+5$$

D
$$+3$$

Solution

$$NH_4NO_3$$ exists as $$\underline{N}H_{4}^{+}\underline{N}O_{3}^{-}$$
Therefore, oxidation numbers of the N are $$-3$$ and $$+5$$.
Question 9
1 / -0

What is the oxidation state of chlorine in hypochlorous acid?

A
$$+1$$

B
$$+3$$

C
$$+5$$

D
$$+7$$

Solution

Let $$x$$ be the oxidation state of chlorine in hypochlorous acid, $$HOCl$$.
The oxidation states of hydrogen and oxygen are $$+1$$ and $$-2$$ respectively.

For a neutral molecule, the sum of the oxidation states is $$0$$.

Therefore, $$1+(-2)+x=0$$ or, $$x=+1$$
Question 10
1 / -0

Oxidation state of $$N$$ is given correct in :

A
$$0$$ in $$[Co(NH_{3})_{5}Cl]Cl$$

B
$$-1$$ in $$NH_{4}OH$$

C
$$-3$$ in $$(N_{2}H_{5})_{2}SO_{4}$$

D
$$-3$$ in $$Mg_{3}N_{2}$$

Solution

The oxidation numbers of nitrogen in $$[Co(NH_{3})_{5}Cl]Cl, NH_{4}OH, (N_{2}H_{5})_{2}SO_{4}$$ and $$Mg_{3}N_{2}$$ are $$-3, \:-3,\: -2$$ and $$-3$$ respectively.
Thus, the oxidation state of $$N$$ is correctly given in $$Mg_{3}N_{2}$$.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 17

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Redox Reactions Test - 17

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Question 1

$$K^{+}$$

$$MnO_{4}^{-}$$

$$C_{2}O_{4}^{2-}$$

$$CO_{2}$$

Solution

Question 2

$$+5$$

$$+8$$

$$+6$$

$$+7$$

Solution

Question 3

$$H_{2} O_{2}$$ is bleached

$$H_{2} O_{2}$$ is oxidised

$$H_{2} O_{2}$$ is dehydrated

$$H_{2} O_{2}$$ is neither oxidised nor reduced

Solution

Question 4

Oxidation number of oxygen is $$+1$$ in peroxides.

Oxidation number of oxygen is $$+2$$ in oxygen difluoride.

Oxidation number of oxygen is $$-\frac{1}{2}$$ in super oxides.

Oxidation number of oxygen is $$-2$$ in most of its compounds.

Solution

Question 5

$$+4$$

$$+6$$

$$+2$$

$$+8$$

Solution

Question 6

$$+1$$

$$+2$$

$$+3$$

$$+5$$

Solution

Question 7

$$CaCO_{3}\rightarrow CaO+CO_{2}$$

$$2H_{2}+O_{2}\rightarrow 2H_{2}O$$

$$Na+H_{2}O\rightarrow NaOH+ \frac{1}{2}H_{2}$$

$$MnCl_{3}\rightarrow MnCl_{2}+ \frac{1}{2}Cl_{2}$$

Solution

Question 8

$$-3$$ & $$+5$$

$$+3$$ & $$+5$$

$$+5$$

$$+3$$

Solution

Question 9

$$+1$$

$$+3$$

$$+5$$

$$+7$$

Solution

Question 10

$$0$$ in $$[Co(NH_{3})_{5}Cl]Cl$$

$$-1$$ in $$NH_{4}OH$$

$$-3$$ in $$(N_{2}H_{5})_{2}SO_{4}$$

$$-3$$ in $$Mg_{3}N_{2}$$

Solution

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