Redox Reactions Test

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Redox Reactions Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation number of $$C$$ in $${CO}_{2}$$ is

A
$$-2$$

B
$$+2$$

C
$$-4$$

D
$$+4$$

Solution

$$\overset { \ast }{ C } { O }_{ 2 }$$

Let, $$x$$ be the oxidation number of $$C$$ in $${CO}_{2}.$$

Now,

$$x+2\times (-2)=0;\ \Rightarrow x-4=0;\ \Rightarrow x=+4.$$
Question 2
1 / -0

The oxidation number of $$Fe$$ and $$S$$ in iron pyrites are

A
$$4$$, $$-2$$

B
$$2$$, $$-1$$

C
$$3$$, $$-1.5$$

D
$$3$$, $$-1$$

Solution

$$\overset { \ast }{ F }e{S}_{2} $$ $$Fe\overset { \ast }{ {S}_{2} }$$
$$x-4=0;4+2x=0$$
$$x=\cfrac{-4}{2}=-2$$
Question 3
1 / -0

The oxidation number of $$Ba$$ in barium peroxide is

A
$$+6$$

B
$$+2$$

C
$$1$$

D
$$+4$$

Solution

Barium peroxide is $$BaO_2.$$

The oxidation number of $$Ba$$ in barium peroxide is $$+2$$ as $$Ba$$ is second group element.
Question 4
1 / -0

$${Sn}^{2+}$$ loses two electrons in a reaction. What will be the oxidation number of tin after the reaction?

A
$$+2$$

B
zero

C
$$+4$$

D
$$-2$$

Solution

$${Sn}^{2+}\rightarrow {Sn}^{4+}+2{e}^{-}$$

$${Sn}^{2+}$$ loses two electrons in the reaction to form $$Sn^{4+}$$ and its oxidation number changes from $$+2$$ to $$+4.$$
Question 5
1 / -0

The sum of the oxidation numbers of all the carbons in $${C}_{6}{H}_{5}CHO$$ is

A
$$+2$$

B
$$0$$

C
$$+4$$

D
$$-4$$

Solution

Let, $$x$$ be the oxidation numbers of the carbon in $${C}_{6}{H}_{5}CHO.$$

Now,
$$7\times x+6\times(+1)+(-2)=0;\ \Rightarrow x=-\dfrac{4}{7}$$

So, the sum of the oxidation numbers of all the carbons in $${C}_{6}{H}_{5}CHO$$ is $$-4.$$
Question 6
1 / -0

In which of the following compounds the oxidation number of carbon is maximum?

A
$$HCHO$$

B
$$CH{Cl}_{3}$$

C
$${CH}_{3}OH$$

D
$${C}_{12}{H}_{22}{O}_{11}$$
Solution
The oxidation state of $$C$$ in the following compounds is as follows:
$$HCHO\rightarrow{0}$$
$$CHCl_3\rightarrow{+2}$$
$$CH_3OH\rightarrow{-2}$$
$$C_{12}H_{22}O_{11}\rightarrow{0}$$

Hence option $$(B)$$ is correct.
Question 7
1 / -0

Oxidation number of $$As$$ atoms in $${H}_{3}As{O}_{4}$$ is

A
$$-3$$

B
$$+4$$

C
$$+6$$

D
$$+5$$

Solution

$${H}_{3}As{O}_{4}$$
$$+3+x-2\times 4=0;x=+5$$
Question 8
1 / -0

If $$H{NO}_{3}$$ changes into $${N}_{2}O$$, the oxidation number is changed by

A
$$+2$$

B
$$-1$$

C
$$0$$

D
$$+4$$

Solution

$$H{NO}_{3}\rightleftharpoons \overset { \ast }{ {N}_{2} } O$$
$$1+x-6=0;2x'-2=0$$
$$x=+5$$; $$x'=+1$$

Hence the oxidation no. is changed by +4.
Question 9
1 / -0

Which of the following reaction is a redox reaction?

A
$${P}_{2}{O}_{5}+2{H}_{2}O\rightarrow {H}_{4}{P}_{2}{O}_{7}$$

B
$$2Ag{NO}_{3}+Ba{Cl}_{2}\rightarrow 2AgCl+Ba{({NO}_{3})}_{2}$$

C
$$Ba{Cl}_{2}+{H}_{2}{SO}_{4}\rightarrow Ba{SO}_{4}+2HCl$$

D
$$Cu+2Ag{NO}_{3}\rightarrow 2A+Cu{({NO}_{3})}_{2}$$

Solution

Above reaction is a redox reaction.
Question 10
1 / -0

Oxidation number of osmiium $$(Os)$$ in $$Os{O}_{4}$$ is

A
$$+4$$

B
$$+6$$

C
$$+7$$

D
$$+8$$

Solution

$$\overset { \ast }{ O } s{O}_{4}$$
$$x+4(-2)=0;x=+8$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 57

Result

Redox Reactions Test - 57

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SHARING IS CARING

Question 1

$$-2$$

$$+2$$

$$-4$$

$$+4$$

Solution

Question 2

$$4$$, $$-2$$

$$2$$, $$-1$$

$$3$$, $$-1.5$$

$$3$$, $$-1$$

Solution

Question 3

$$+6$$

$$+2$$

$$1$$

$$+4$$

Solution

Question 4

$$+2$$

zero

$$+4$$

$$-2$$

Solution

Question 5

$$+2$$

$$0$$

$$+4$$

$$-4$$

Solution

Question 6

$$HCHO$$

$$CH{Cl}_{3}$$

$${CH}_{3}OH$$

$${C}_{12}{H}_{22}{O}_{11}$$

Solution

Question 7

$$-3$$

$$+4$$

$$+6$$

$$+5$$

Solution

Question 8

$$+2$$

$$-1$$

$$0$$

$$+4$$

Solution

Question 9

$${P}_{2}{O}_{5}+2{H}_{2}O\rightarrow {H}_{4}{P}_{2}{O}_{7}$$

$$2Ag{NO}_{3}+Ba{Cl}_{2}\rightarrow 2AgCl+Ba{({NO}_{3})}_{2}$$

$$Ba{Cl}_{2}+{H}_{2}{SO}_{4}\rightarrow Ba{SO}_{4}+2HCl$$

$$Cu+2Ag{NO}_{3}\rightarrow 2A+Cu{({NO}_{3})}_{2}$$

Solution

Question 10

$$+4$$

$$+6$$

$$+7$$

$$+8$$

Solution

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