Redox Reactions Test

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Redox Reactions Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

Consider the following out of $$\underset { \left( I \right) }{ H_{ 2 }SO_{ 4 } } \underset { \left( II \right) }{ ,H_{ 2 }S_{ 2 }O_{ 8 }, } { \underset{III}{H_{ 2 }SO_{ 5 }},\underset{IV}{H_{ 2 }S_{ 2 }O_{ 3 } }} $$ and $$\underset { \left( V \right) }{ { H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 } } $$
Choose the correct option.

A
Oxidation number of Sulphur is $$+6$$ in all except $$IV$$

B
Peroxy linkage is in $$II$$ and $$III$$

C
Both statements are correct

D
None of the above is correct

Solution

The oxidation state of $$S$$ in $${ H }_{ 2 }{ SO }_{ 4 }$$ is +6
Because
$$x+2-8=0$$
$$x=6$$
The oxidation state of $$S$$ in $${ H }_{ 2 }{ { S }_{ 2 }O }_{ 8 }$$ is +7
The oxidation state of $$S$$ in $${ H }_{ 2 }{ { S }O }_{ 5 }$$ is +8
The oxidation state of $$S$$ in $${ H }_{ 2 }{ { { S }_{ 2 } }O }_{ 3 }$$ is +2
The oxidation state of $$S$$ in $${ H }_{ 2 }{ { { S }_{ 2 } }O }_{ 7 }$$ is +6
Peroxy linkage is the covalent $$O-O$$ bond.it is found in $$H_2S_2O_8$$ & $$H_2SO_5$$
Question 2
1 / -0

Which of the following show non-zero multiple oxidation state?

A
$$S$$

B
$$O$$

C
$$Zn$$

D
$$H$$

Solution

The possible oxidation states of elements are :
$$S\rightarrow -2,-1,0,+1,+2,+3,+4,+5,+6$$
$$O\rightarrow -2,-1,+1,+2$$
$$Zn\rightarrow -2,+1,+2$$
$$H\rightarrow -1,+1$$
Out of the above elements, only S shows non-zero multiple oxidation states.
Question 3
1 / -0

Sugarcane on reaction with nitric acid gives?

A
$${ CO }_{ 2\quad }and\quad { SO }_{ 2 }\quad $$

B
$$(COOH)_{ 2 }$$

C
$$2\ HCOOH$$

D
$$No\ reaction\ takes\ place$$

Solution
Question 4
1 / -0

Which of the following pairs of elements show similar set of oxidation state?

A
$$O^{16}, O^{18}$$

B
$$Na, K$$

C
$$C, Be$$

D
$$Zn, Rb$$

Solution

Oxidation state is decided by number of electrons an element loose or gain.In isotopes number of electrons are same but mass number is different. In $${ O }^{ 16 } $$ and $${ O }^{ 18 } $$ mass number is differnet 16 and 18 respectively and both have same oxidation state 2.
Question 5
1 / -0

Four $$Cl_{2}$$ molecules undergo a loss and gain of $$6$$ mole of electrons to form two oxidation states of $$Cl$$ in auto redox change. What are the $$+ve$$ and $$-ve$$ oxidation states of $$Cl$$ in the change:

A
$$Cl^{5+},Cl^{o}$$

B
$$Cl^{7+},Cl^{-}$$

C
$$Cl^{3+},Cl^{o}$$

D
$$Cl^{3+},Cl^{-}$$

Solution

The overall reaction is:
$$4{ Cl }_{ 2 }\rightarrow 2{ Cl }^{ +3 }+6{ Cl }^{ - }$$
Here total number of atom is balanced and total charge is balance.
Change of oxidation state is +3 and -1.
Question 6
1 / -0

The $$O.S.$$ of $$Fe$$ in $$Na_{2}[Fe(CN)_{5}NO]$$ and $$Na_{4}[Fe(CN)_{5}NOS]$$ is:

A
$$+2, +2$$

B
$$+2, +4$$

C
$$+4, +4$$

D
$$+3, +3$$

Solution

$${ Na }_{ 2 }\left[ Fe{ \left( CN \right) }_{ 5 }NO \right] \Rightarrow { Na }^{ + }{ \left[ Fe{ \left( CN \right) }_{ 5 }NO \right] }^{ -2 }$$
$$\therefore \quad x-5+1=-2$$
$$\therefore \quad x-4=-2$$
$$\therefore \quad \boxed { x=+2 } $$
$$\therefore $$ oxidation state of $$Fe$$ is $$+2$$
$${ Na }_{ 4 }\left[ Fe{ \left( CN \right) }_{ 5 }NOS \right] \Rightarrow { Na }^{ + }{ \left[ Fe{ \left( CN \right) }_{ 5 }NOS \right] }^{ 4- }$$
$$\therefore \quad x-5+1=-2$$
$$\therefore \quad \boxed { x=+2 } $$
The O.S of $$Fe$$ in both the molecules in $$+2$$.
Question 7
1 / -0

A solution contains $${Fe}^{+2},{Fe}^{+3}$$ and $${I}^{-}$$ ions. This solution was treated with iodine at $${35}^{o}C$$. Then the favourable redox reaction is:
(Given that $${ E }_{ { Fe }^{ +3 }/{ Fe }^{ +2 } }^{ o }=+0.77V;\quad { E }_{ { I }_{ 2 }/{ I }^{ - } }^{ o }=0.536V$$)

A
$${I}_{2}$$ will be reduced to $${I}^{-}$$

B
there will be no redox reaction

C
$${I}^{-}$$ will be oxidised to $${I}_{2}$$

D
$${Fe}^{+}$$ will be oxidised to $${Fe}^{+3}$$

Solution

$${ 2I }^{ - }\rightarrow { I }_{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$$

$${ E }_{ Oxidation }^{ 0 }=-0.536V$$

$${ Fe }^{ 3+ }+e^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$$

$$E^{ 0 }_{ Reduction }=-0.77V$$
----------------------------------------------------------------------------------------------------------------------
$${ 2Fe }^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I }_{ 2 }$$

$$E^{ 0 }={ E }_{ oxidation }^{ 0 }+E_{ Reduction }^{ 0 },+ve$$

Hence the reaction will take place.

$${ 2I }^{ - }\rightarrow { I }_{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$$

$${ E }^{ 0 }_{ Oxidation }=-0.536V$$

$${ Fe }^{ 3+ }+{ e }^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$$

$$E_{ Reduction }^{ 0 }=-0.77V$$
------------------------------------------------------------------------------------------------------------------------
$$2Fe^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I }_{ 2 };$$

$${ E }^{ 0 }={ E }_{ oxidation }^{ 0 }+{ E }_{ reduction }^{ 0 };+ve$$
Question 8
1 / -0

Given reaction $$HCHO \rightarrow CH_{3}OH+HCOO^{-}$$ is an example of:

A
intermolecular redox change

B
intramolecular redox change

C
disproportionation or auto redox

D
none of these

Solution

$$HCHO→CH_{ 3 }OH+HCOO^{ - }$$
A B C
The above process is disproportionation reaction because A undergoes oxidation and reduction simultaneously and forms B and C.
The oxidation state of Carbon in A is 0
The oxidation state of Carbon in B is -2
The oxidation state of Carbon in C is 3
Question 9
1 / -0

In the following structure, the oxidation number on carbon will be: $$H-N=C$$:-

A
$$0$$

B
$$-2$$

C
$$+2$$

D
$$+1$$ or $$-1$$

Solution

$$\left[ \underset { +1 }{ H } -\overset { -2 }{ \underset { -1 }{ N } = } \overset { -2 }{ \underset { +2 }{ C } } \right] \quad \Rightarrow $$ Oxidation number of carbon $$=-2+2=0$$
Question 10
1 / -0

How many electrons should $${ X }_{ 2 }{ H }_{ 4 }$$ liberate so that in the new compound X shows oxidation number of -1/2? $$\left( E,N.X>H \right) $$

A
10

B
4

C
3

D
2

Solution

$$X_2H_4 $$ En of $$X > H$$
$$\therefore H $$ is in +1
and X in $$-\dfrac{1}{2}$$
$$-\dfrac{1}{2} \times 2 + (+1) \times 4 = -1 + 4 = 3$$
$$\therefore 3 \, e^-$$ must be liberated.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 50

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Redox Reactions Test - 50

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SHARING IS CARING

Question 1

Oxidation number of Sulphur is $$+6$$ in all except $$IV$$

Peroxy linkage is in $$II$$ and $$III$$

Both statements are correct

None of the above is correct

Solution

Question 2

$$S$$

$$O$$

$$Zn$$

$$H$$

Solution

Question 3

$${ CO }_{ 2\quad }and\quad { SO }_{ 2 }\quad $$

$$(COOH)_{ 2 }$$

$$2\ HCOOH$$

$$No\ reaction\ takes\ place$$

Solution

Question 4

$$O^{16}, O^{18}$$

$$Na, K$$

$$C, Be$$

$$Zn, Rb$$

Solution

Question 5

$$Cl^{5+},Cl^{o}$$

$$Cl^{7+},Cl^{-}$$

$$Cl^{3+},Cl^{o}$$

$$Cl^{3+},Cl^{-}$$

Solution

Question 6

$$+2, +2$$

$$+2, +4$$

$$+4, +4$$

$$+3, +3$$

Solution

Question 7

$${I}_{2}$$ will be reduced to $${I}^{-}$$

there will be no redox reaction

$${I}^{-}$$ will be oxidised to $${I}_{2}$$

$${Fe}^{+}$$ will be oxidised to $${Fe}^{+3}$$

Solution

Question 8

intermolecular redox change

intramolecular redox change

disproportionation or auto redox

none of these

Solution

Question 9

$$0$$

$$-2$$

$$+2$$

$$+1$$ or $$-1$$

Solution

Question 10

10

4

3

2

Solution

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