Redox Reactions Test

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Redox Reactions Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

In which of the following species,the average oxidation state of oxygen is $$\frac{-1}{2}$$?

A
$$PbO_{2}$$

B
$$H_{2}O_{2}$$

C
$$D_{_{2}}O_{2}$$

D
$$Sr\left (O_{2} \right )_{2}$$

Solution

In $$Sr{({O}_{2})}_{2}$$ , the average oxidation of oxygen is $$\dfrac{-1}{2}$$

In $$Pb{O}_{2}$$ , $${H}_{2}{O}_{2}$$ , $${D}_{2}{O}_{2}$$ , the avg oxidation of oxygen is $$-1$$
Question 2
1 / -0

Which of the following equation depict the oxidising nature of $$H_{2}O_{2}$$?

A
$$2MnO^{-}_{4}+6H^{+}+5H_{2}O_{2}\rightarrow 2Mn^{2}+8H_{2}O+5O_{2}$$

B
$$2Fe^{3+}+2H^{+}+H_{2}O_{2}\rightarrow 2Fe^{2}+2H_{2}O+O_{2}$$

C
$$2I^{-}+2H^{+}+H_{2}O_{2}\rightarrow I_{2}+2H_{2}O$$

D
$$KIO_{4}+H_{2}O_{2}\rightarrow KIO_{3}+H_{2}O+O_{2}$$

Solution

Oxidising nature of $$H_2O_2$$ means it oxidises other substance and itself gets reduced. In such reactions $$O_2$$ is not evolved.

In (a), $$MnO_4^-$$ gets reduced from $$Mn^{7+}$$ to $$Mn^{2+}$$.

In (b), $$Fe^{3+}$$ gets reduced to $$Fe^{2+}$$

In (c), $$I^-$$ gets oxidised to $$I_2$$.

In (d), $$KIO_4$$ gets reduced from $$I^{7+}$$ to $$I^{5+}$$.

Therefore in option (C) is correct Answer.
Question 3
1 / -0

How many oxygen atoms in blue per chomate $${ CrO }_{ 5 }$$ are having $$-1$$ oxidation state?

A
1

B
2

C
4

D
5

Solution

CrO₅, also called as per chromate, has a butterfly shape and neutral charge.

Chromium lies in +6 oxidation state.

So, four oxygen will have -1 charge and one will have -2 charge, thereby satisfying the overall neutral charge.

Hence, option C is correct.
Question 4
1 / -0

Which one of the following statements is not correct?

A
Oxidation state of $$S$$ in $$(NH_4)_2 S_2O_2$$ is $$+6$$

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

C
Oxidation state of $$S$$ in $$H_2SO_2$$ is $$+8$$

D
Oxidation number of $$O$$ in $$KO_2$$ is $$-\dfrac { 1 }{ 2 } $$
Question 5
1 / -0

Given that $$ { Cl }_{ 2\left( g \right) }+{ 2e }^{ - }\longrightarrow { 2Cl }^{ - }\left( aq \right)$$ is the reduction half-reaction for the overall reaction $$ 2Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) \longrightarrow 2AgCl\left( s \right)$$.
What is the oxidation half reaction?

A
$$ Ag\left( s \right) \longrightarrow Ag^{ + }\left( aq \right) +e^{ - }$$

B
$$ Ag\left( s \right) +{ Cl }^{ - }\left( aq \right) \longrightarrow Ag\left( s \right) +e^{ - }$$

C
$$ Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) +e^{ - }\longrightarrow AgCl\left( s \right) +Cl^{ - }\left( aq \right)$$

D
$$ { 2Cl }^{ - }\left( aq \right) \longrightarrow { Cl }_{ 2 }\left( g \right) +2e^{ - }$$

Solution
Question 6
1 / -0

The oxidation state of Fe in $$ \left[ Fe\left( { H }_{ 2 }{ O } \right) _{ 5 }NO \right] { SO }_{ 4 }$$ is :

A
$$+1$$

B
$$+2$$

C
$$+3$$

D
$$+4$$

Solution

$$NO$$ in brown ring complex has $+ 1$ oxidation state.

Let oxidation state of Fe be $x$ .

Therefore, $$x + 5(0)+(1)+(-2)=0$$ = $$ x+ 1-2= 0$$

$$x= +1$$

Hence, the oxidation state of iron in the brown ring complex is $+ 1$ .

Thus option A is correct.
Question 7
1 / -0

The oxidation state(s) of $$Cl$$ in $$CaO_{2}Cl_{2}$$ (bleaching powder) is/are:

A
+1 only

B
-1 only

C
+1 and -1

D
none of these

Solution

In bleaching powder, $$Ca(OCl)Cl$$, the two $$Cl$$ atoms are in different oxidation states i.e., one $$Cl–$$ having oxidation number of $$–1$$ and the other as $$OCl–$$ having oxidation number of $$+1.$$

Hence, option C is correct.
Question 8
1 / -0

Which of the following metals form complex in zero oxidation state?

A
Cu

B
Ni

C
Mn

D
V

Solution

(B) Ni
Because in zero oxidation stalt it forms metal
carbonyles and carbonyly is Pi acid ligand. and
out of four options Ni form carbonyls as $$Ni(CO)_{4}$$
Question 9
1 / -0

Covalency and oxidation number of nitrogen in $${ N }_{ 2 }{ O }_{ 5 }$$ is respectively :

A
5, +5

B
4, +5

C
3, +3

D
3 , +5

Solution
Question 10
1 / -0

The oxidation number of 'N' in NOClO:

A
+3

B
+1

C
+2

D
+7

Solution

1)$$NOClO$$
$$x+(-1)+(-1)+(-1)=0$$
$$x-1-1-1=0$$
$$X=+3$$
2)Thus oxidation state of N in given compound is $$+3$$.