Redox Reactions Test

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Redox Reactions Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

In acid medium, the standard reduction potential of $$NO$$ converted to $${ N }_{ 2 }O$$ is $$1.59 V$$. Its standard potential in alkaline medium would be:

A
$$-1.59 V$$

B
$$-0.764 V$$

C
$$0.764 V$$

D
$$0.062 V$$

Solution

$$NO \longrightarrow N_2O$$ $$E^o= 1.59V$$
The reaction in acidic medium,
$$2NO+2H^+ \longrightarrow N_2O+H_2O$$; $$E^o=1.59$$
In basic medium,
$$2NO+H_2O \longrightarrow N_2O+2OH^-$$; Let $$E^o=x$$
Now, in alkaline medium,
$$H_2O \rightleftharpoons H^++OH^-$$
$$\Rightarrow K=10^{-14}$$
or, $$2H_2O \rightleftharpoons 2H^++2OH^-$$, $$K= 10^{-28}$$
We know, $$E^o= \cfrac {0.059}{n}\log K$$
$$=\cfrac {0.059}{2}\log 10^{-28}= 0.826$$
For standard potential,
$$E^o_{acidic}+E^o_{basic}=E^o_{cell}$$
$$\Rightarrow 1.59+x= 0.826$$
$$\Rightarrow x= 0.764$$
Question 2
1 / -0

Write balanced equations for the half reactions and calculate the reduction potentials at $$25^0C$$ for the following half cells :
$$Cl^-(1.2M) | Cl_2(g, 3.6 atm) | Pt E^o = 1.36 V$$

A
$$1.372 V$$

B
$$1.363 V$$

C
$$1.358 V$$

D
$$1.342 V$$

Solution
Question 3
1 / -0

Alizarin belongs to the class of

A
Vat dyes

B
Mordant dye

C
Basic dye

D
Reactive dye

Solution
Question 4
1 / -0

Oxidation number of Cl in $$KClO_3$$ :

A
-1

B
-5

C
+1

D
+5

Solution
Question 5
1 / -0

The oxidation number of $$Co$$ in the complex ion

A
$$+2 $$

B
$$+3 $$

C
$$+4 $$

D
$$+6 $$
Question 6
1 / -0

The oxidation state of A,B and C in compaund are -2, +5, and -2, respectively. the compound is ?

A
$$A_3( BC)_{2}$$

B
$$A_2( BC )_{3}$$

C
$$A_2 ( BC )_{4}$$

D
$$A_2 ( BC )_{5}$$
Question 7
1 / -0

Justify that the following reactions are redox reactions.

A
$$CuO\left( s \right) +{ H }_{ 2 }\left( g \right) \rightarrow Cu\left( s \right) +{ H }_{ 2 }O\left( g \right) $$

B
$${ Fe }_{ 2 }{ O }_{ 3 }\left( s \right) +3CO\left( g \right) \rightarrow 2Fe\left( s \right) +3{ CO }_{ 2 }\left( g \right) $$

C
$$4{BCl }_{ 3 }\left( g \right) +3Li{AlH }_{ 4 }\left( s \right) \rightarrow 2{ B }_{ 2 }{ H }_{ 6 }\left( s \right) +3LiCl\left( s \right)+3{AlCl }_{ 3 }\left( s \right) $$

D
$$2K\left( s \right) +{ F }_{ 2 }\left( g \right) \rightarrow { 2K }^{ + }{ F }^{ - }\left( s \right)$$

Solution

$$(A)$$  $$\overset { +2 }{ Cu } O\left( s \right) +{ \overset { 0 }{ H } }_{ 2 }\left( g \right) \longrightarrow \overset { 0 }{ Cu } \left( s \right) +{ \overset { +1 }{ H } }_{ 2 }O\left( g \right) $$
Here, reduction of $$\overset { +2 }{ Cu } O$$ to $$\overset { 0 }{ Cu } $$ and oxidation of $${ \overset { 0 }{ H } }_{ 2 }$$ to $${ \overset { +1 }{ H } }_{ 2 }O$$ takes place. Therefore, it is a redox reaction.
$$(B)$$  $${ \overset { +3 }{ Fe } }_{ 2 }{ O }_{ 3 }\left( s \right) +3\overset { +2 }{ CO } \left( g \right) \longrightarrow 2\overset { 0 }{ Fe } \left( s \right) +3{ \overset { +4 }{ CO } }_{ 2 }\left( g \right) $$
Here, reduction of $${ \overset { +3 }{ Fe } }_{ 2 }{ O }_{ 3 }$$ to $$\overset { 0 }{ Fe } $$ and oxidation of $$\overset { +2 }{ CO } $$ takes place. Therefore, it is a redox reaction.
$$(C)$$  $$4\overset { +3 }{ B } { Cl }_{ 3 }\left( g \right) +3Li\overset { +3 }{ AI } { H }_{ 4 }\left( s \right) \longrightarrow 3{ \overset { +3 }{ B } }_{ 2 }{ H }_{ 6 }\left( s \right) +3LiCl\left( s \right) +3\overset { +3 }{ AI } { Cl }_{ 3 }\left( s \right) $$
Here, no change of oxidation states take place. Therefore, it is not a redox reaction.
$$(D)$$  $$2\overset { 0 }{ K } \left( s \right) +{ \overset { 0 }{ F } }_{ 2 }\left( g \right) \longrightarrow 2{ K }^{ + }{ F }^{ - }\left( s \right) $$
Here, $$\overset { 0 }{ K } $$ is oxidized to $${ K }^{ +1 }$$ and $${ \overset { 0 }{ F } }_{ 2 }$$ is reduced to $${ F }^{ -1 }$$. So, it is a redox reaction.
Question 8
1 / -0

In the redox reaction, $$xKMnO_4$$ + $$NH_3$$ $$\rightarrow$$ $$yKNO_3$$ + $$MnO_2$$ +$$KOH$$ +$$H_2O$$
The $$x$$ and $$y$$ are:

A
$$x=4, y=6$$

B
$$x=8, y=3$$

C
$$x=2, y=5$$

D
$$x=7, y=10$$
Question 9
1 / -0

In the balanced equation
$${MnO_{4}}^{-}+H^{+}+{C_{2}O_{4}}^{2-}\rightarrow Mn^{2+}+CO_{2}+H_{2}O$$, the moles of $$CO_{2}$$ formed are :-

A
2

B
4

C
5

D
10

Solution
Question 10
1 / -0

In the balanced equation, coefficient of $${{\text{N}}{{\text{H}}_{\text{4}}}^ + }$$ will be:-
$$Zn + H^ + + NO_3^{−}→NH_4^+ + Zn^{ 2+} + H_2O$$

A
4

B
3

C
2

D
1