Redox Reactions Test

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Redox Reactions Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Analyze the diagram. The left beaker represents the start of the reaction and the right beaker represents the end of the reaction.
Classify this reaction.

A
Acid-Base Neutralization reaction.

B
Redox reaction.

C
Precipitation reaction.

D
More information is needed.

Solution

Redox reactions are reduction-oxidation reactions in which exchange of electrons takes place between species in which one will undergo oxidation and other result in reduction.
Question 2
1 / -0

The oxidation number of chromium in $$K_{2}CrO_{4}$$ is:

A
$$+2$$

B
$$+3$$

C
$$+6$$

D
$$+8$$

Solution

In $$K_2CrO_4$$,
Oxidation no. of $$K=+1.$$
Oxidation no. of $$O=-2.$$
Oxidation no. of $$Cr$$ be $$x$$.
$$2(1)+x+4(-2)=0,\\ 2+x-8=0,\\ x-6=0,\\x=6.$$
$$\therefore$$ Oxidation no. of $$Chromium$$ in $$K_2CrO_4$$ is $$6$$.
Question 3
1 / -0

The half-cell reaction is the one that:

A
takes place at one electrode

B
consumes half a unit of electricity

C
involves half a mole of electrolyte

D
goes half way to completion

Solution

A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
Half-reactions are often used as a method of balancing redox reactions.
Question 4
1 / -0

What is the number of coulombs required for the conversion of one mole of $$MnO_{4}^{-}$$ to one mole of $$Mn^{2+}$$?

A
$$5\times 96500$$

B
$$3\times 96500$$

C
$$96500$$

D
$$9650$$

Solution

$$MnO^-_4$$ is converted to $$Mn^{+2}$$.
In $$MnO^-_4$$, oxidation state of $$Mn$$ be $$x.$$
$$x+4(-2)=-1,\\ x-8=-1,\\x=+7.$$
Oxidation state changes from $$+7$$ to $$+2$$.
Change in oxidation state $$= 5$$.
$$1$$ Faraday ($$96500$$ coulombs) to change one oxidation state.
To change $$5$$ oxidation state, no. of coulombs required, $$ =5\times 1 \text{ Faraday }\\=5\times 96500 \text{ coulombs},$$
Question 5
1 / -0

In alkaline medium, the reaction of hydrogen peroxide with potassium permanganate produces a compound in which the oxidation state of $$Mn$$ is:

A
$$0$$

B
$$+2$$

C
$$+3$$

D
$$+4$$

Solution

$$2Mn{ O }_{4}^{-}(aq) + 3H_{2}O_{2}(aq) \rightarrow 2MnO_{2}(s) + 2H_{2}O(l) + 3O_{2}(g) + 2O{ H }^{-}_{(aq)}$$

Oxidation state of $$Mn$$ in $$MnO_{2}$$ is $$+4$$.

Hence, option $$D$$ is correct.
Question 6
1 / -0

The oxidation number of chromium in $$\displaystyle Cr{ O }_{ 5 }$$ is:

A
+6

B
+5

C
+10

D
0

Solution

The oxidation number of chromium in chromium pentaoxide is 6.
Question 7
1 / -0

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half-cell reactions and their standard potentials are given below:
$$Mn{O}_{4}^{-} (aq.) + 8H^{+}(aq.) + 5e^{-} \rightarrow Mn^{2+}(aq.) + 4H_{2}O(l); E^{\circ} = 1.51\ volt$$
$$Cr_{2}O_{7}^{2-}(aq.) + 14H^{+}(aq.) + 6e^{-} \rightarrow 2Cr^{3+}(aq.) + 7H_{2}O(l)' E^{\circ} = 1.38\ volt$$
$$Fe^{3+}(aq.) + e^{-}\rightarrow Fe^{2+}(aq.); E^{\circ} = 0.77\ volt$$
$$Cl_{2}(g) + 2e^{-} \rightarrow 2Cl^{-}(aq.); E^{\circ} = 1.40\ volt$$
Identity the only incorrect statement regarding the quantitative estimation of aqueous $$Fe(NO_{3})_{2}$$.

A
$$Mn{O}_{4}^{-}$$ can be used in aqueous $$HCl$$

B
$$Cr_{2}O_{7}^{2-}$$ can be used in aqueous $$HCl$$

C
$$Mn{O}_{4}^{-}$$ can be used in aqueous $$H_{2}SO_{4}$$

D
$$Cr_{2}O_{7}^{2-}$$ can be used in aqueous $$H_{2}SO_{4}$$

Solution

According to the given half cell reactions for the quantitative estimation of aqueous $$Fe(NO_3)_2$$
That $$Mn{O}_4^-$$ cannot be used in aqueous $$HCl$$ because $$Mn{O}_4^-$$ wii not oxidise $$Fe^{+2}$$ to $$Fe^{+3}$$ instead of that it also oxidise $$Cl^-$$ to $$Cl_2$$.
Whereas all other reactions are feasible.
Question 8
1 / -0

Oxidation number of nitrogen in which among the oxides of nitrogen is the lowest?

A
Nitric oxide

B
Nitrous oxide

C
Nitrogen dioxide

D
Dinitrogen trioxide

Solution

Nitrogen can exhibit +1 to +5 oxidation state due to its half-filled orbital.
The least oxidation state will be $$+1$$ which is for $$N_2O$$ ( Nitrous oxide. )

So, the correct option is $$B$$
Question 9
1 / -0

Arrange the following in the increasing order of oxidation state of $$Mn.$$
(i) $$Mn^{2+}$$
(ii) $$MnO_2$$
(iii) $$KMnO_4$$
(iv) $$K_2MnO_4$$.

A
(i) $$>$$ (ii) $$>$$ (iii) $$>$$ (iv)

B
(i) $$<$$ (ii) $$<$$ (iv) $$<$$ (iii)

C
(ii) $$<$$ (iii) $$<$$ (i) $$<$$ (iv)

D
(iii) $$<$$ (i) $$<$$ (iv) $$<$$ (ii)

Solution

Oxidation state of given compounds is as follow:
$$\underset{(+2)}{Mn^{2+}} < \underset{(+4)}{MnO_2} < \underset{(+6)}{K_2MnO_4} < \underset{(+7)}{KMnO_4}$$.
Question 10
1 / -0

Oxidation state of $$S$$ in $${SO_4}^{2-}$$ is:

A
$$+6$$

B
$$+3$$

C
$$+2$$

D
$$-2$$

Solution

Let oxidation state of $$S$$ is x.

and oxidation state of $$O$$ in $$SO^{2-}_4$$ is -2.

and the net charge on the compound is -2.

$$\therefore x+4(-2)=-2$$

$$x=+6$$.

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Redox Reactions Test - 32

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Redox Reactions Test - 32

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Question 1

Acid-Base Neutralization reaction.

Redox reaction.

Precipitation reaction.

More information is needed.

Solution

Question 2

$$+2$$

$$+3$$

$$+6$$

$$+8$$

Solution

Question 3

takes place at one electrode

consumes half a unit of electricity

involves half a mole of electrolyte

goes half way to completion

Solution

Question 4

$$5\times 96500$$

$$3\times 96500$$

$$96500$$

$$9650$$

Solution

Question 5

$$0$$

$$+2$$

$$+3$$

$$+4$$

Solution

Question 6

+6

+5

+10

0

Solution

Question 7

$$Mn{O}_{4}^{-}$$ can be used in aqueous $$HCl$$

$$Cr_{2}O_{7}^{2-}$$ can be used in aqueous $$HCl$$

$$Mn{O}_{4}^{-}$$ can be used in aqueous $$H_{2}SO_{4}$$

$$Cr_{2}O_{7}^{2-}$$ can be used in aqueous $$H_{2}SO_{4}$$

Solution

Question 8

Nitric oxide

Nitrous oxide

Nitrogen dioxide

Dinitrogen trioxide

Solution

Question 9

(i) $$>$$ (ii) $$>$$ (iii) $$>$$ (iv)

(i) $$<$$ (ii) $$<$$ (iv) $$<$$ (iii)

(ii) $$<$$ (iii) $$<$$ (i) $$<$$ (iv)

(iii) $$<$$ (i) $$<$$ (iv) $$<$$ (ii)

Solution

Question 10

$$+6$$

$$+3$$

$$+2$$

$$-2$$

Solution

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