Redox Reactions Test

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Redox Reactions Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Oxidation state of carbon in $$HCOOH$$ will be:

A
$$+1$$

B
$$+2$$

C
$$-4$$

D
$$0$$

Solution

Let the oxidation state of carbon be x.

The oxidation state of oxygen and hydrogen in $$HCOOH$$ is -2 and +1 respectively.

Therefore, $$2+x-4=0$$

$$x=2$$.
Question 2
1 / -0

Oxidation states of chlorine in $$HClO_4$$ and $$HClO_3$$ are:

A
$$+4, +3$$

B
$$+7, +5$$

C
$$+3, +4$$

D
$$+5, +7$$

Solution

Let the oxidation state of chlorine be x an y in $$HClO_4$$ and $$HClO_3$$.

The oxidation state of $$H$$ and $$O$$ in $$HClO_4$$ and $$HClO_3$$ is +1 and -2 respectively.

$$HClO_4$$ and $$HClO_3$$.

$$HClO_4: +1+x-8=0, x=+7$$

$$HClO_3: +1+x-6=0, y=+5$$.
Question 3
1 / -0

Which of the following has least oxidation state of $$Fe$$?

A
$$K_3[Fe(OH)_6]$$

B
$$K_2[FeO_4]$$

C
$$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$

D
$$[Fe(CN)_6]^{3-}$$

Solution

In mohr salt $$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$, oxidation state of iron is $$+2$$ which is least.
$$K_3[Fe(OH)_6]$$ $$+3+x-6=0$$ $$x=+3$$
$$K_2[FeO_4]$$ $$+2+x-8=0$$ $$x=+6$$
$$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$ $$+2$$ state in $$FeSO_4$$ $$(x-2=0)$$
$$Fe(CN)^{3-}_6$$ $$x-6=-3$$ $$x=+3$$
Question 4
1 / -0

Select the compound in which chlorine shows oxidation state $$+7$$.

A
$$HClO_4$$

B
$$HClO_3$$

C
$$HClO_2$$

D
$$HClO$$

Solution

Let oxidation state of chlorine is $$x$$.
In $$HCl{ O }_{ 4 }$$, $$1+x+4(-2)=0$$
$$x=+7$$
In $$HCl{ O }_{ 3 }$$, $$1+x+3(-2)=0$$
$$x=+5$$
In $$HCl{ O }_{ 2 }$$, $$1+x+2(-2)=0$$
$$x=+3$$
In $$HCl{ O }_{ }$$, $$1+x+1(-2)=0$$
$$x=+1$$
So $$HCl{ O }_{ 4 }$$ is correct answer.
Question 5
1 / -0

The oxidation state of phoshorus varies from

A
$$-1$$ to $$+1$$

B
$$-3$$ to $$+3$$

C
$$-3$$ to $$+5$$

D
$$-5$$ to $$+1$$

Solution

Electronic shell configuration of phosphorus:$${ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 3 }$$
Electrons present in outermost shell is 5 so they need three more electrons to complete their octet so they can show oxidation state of -3.They can also show -1, and -2 oxidation states.They can also lose electrons and maximum number of electrons they can lose is 5 so they can show maximum oxidation state of +5.Phosphorus can also show lower oxidation states such as +1,+2 etc.
So the oxidation state of phosphorus varies from -3 to +5.
Question 6
1 / -0

Phosphorus has the oxidation state of $$+3$$ in:

A
Ortho phosphoric acid

B
Phosphorus acid

C
Meta phosphoric acid

D
Pyrophosphoric acid
Solution
Hint:

The number of electrons an atom gains or losses to form a bond with other elements to make a molecule is called as oxidation state.

Explanation:
- Ortho phosphoric acid has a chemical formula $$H_3PO_4$$
Let the oxidation state of $$P$$ be $$X$$

$$+3+X+ (4 \times {( -2 )})=0$$

$$\Rightarrow X=+5$$
- The chemical formula of Phosphorus acid is $$H_3PO_3$$
Similarly, Let the oxidation state of $$P$$ be $$X$$

$$+3+X+ (3 \times {( -2)})=0$$

$$\Rightarrow X=+3$$
- The chemical formula of Pyro phosphoric acid is $$H_4P_2O_7$$
Similarly, Let the oxidation state of $$P$$ be $$X$$

$$+4+2X+ {(7 \times {( -2 )})}=0$$

$$\Rightarrow X=+5$$
- The chemical formula of meta phosphoric acid is $$HPO_3$$
Similarly, Let the oxidation state of $$P$$ be $$X$$

$$+1+X+ (3 \times {( -2)})=0$$

$$\Rightarrow X=+5$$

Hence, phosphorous acid has the oxidation state of $$+3$$.

Therefore, the correct answer is option B.
Question 7
1 / -0

The oxidation number of Fe in $$K_4Fe(CN)_6$$ is?

A
$$+6$$

B
$$+4$$

C
$$+3$$

D
$$+2$$

Solution

Let oxidation state of Fe is x,
$$4\times (1)+x+6\times (-1)=0\\ x=+2$$
Question 8
1 / -0

Oxidation number of $$P$$ in $$PO^{3-}_4$$, of S in $$SO^{2-}_4$$ and that of $$Cr$$ in $$Cr_2O^{2-}_7$$ are respectively.

A
$$-3, +6$$ and $$+6$$

B
$$+5, +6$$ and $$+6$$

C
$$+3, +6$$ and $$+5$$

D
$$+5, +3$$ and $$+6$$

Solution

$$\bullet$$ Oxidation no. of $$P$$ in $$PO^{3-}_4$$ be $$x$$.
Oxidation state of $$O$$ in general $$=-2$$
Applying formula, $$\text{Sum of total oxidation state of all atoms } = \text{ Overall charge on the compound}.\\ x+4(-2)=-3,\\x=+5$$
$$\therefore$$ Oxidation state of $$P=+5.$$
Oxidation no. of $$S$$ in $$SO^{2-}_4$$ be $$x$$.
Applying above formula,
$$x+4\times (-2)=-2\\x=+6$$
$$\therefore$$ Oxidation state of $$S=+6.$$
Oxidation no. of $$Cr$$ in $$Cr^{2-}_7$$ be $$x$$.
Applying above formula,
$$2x+7\times (-2)=-2\\x=+6$$
$$\therefore$$ Oxidation state of $$Cr=+6.$$
Question 9
1 / -0

In which of the following reactions, the underlined element has decreased its oxidation number during the reaction?

A
$$\underline{Fe}+CuSO_4\rightarrow Cu+FeSO_4$$

B
$$\underline{H_2}+Cl_2\rightarrow 2HCl$$

C
$$\underline{C}+H_2O\rightarrow CO+H_2$$

D
$$\underline{MnO_2}+4HCl\rightarrow MnCl_2+Cl_2+2H_2O$$

Solution

In A) $$Fe$$ oxidation state changes from 0 to +2.
In B) $$H$$ oxidation state changes from 0 to +1.
In C) $$C$$ oxidation state changes from 0 to +2.
In D) $$Mn$$ oxidation state changes from +4 to +2 so in this reaction underlined element has decreased its oxidation number.
Question 10
1 / -0

The highest state of Mn is shown in.

A
$$K_2MnO_4$$

B
$$KMnO_4$$

C
$$MnO_2$$

D
$$Mn_2O_3$$

Solution

Oxygen always show oxidation state of -2 except in peroxides and superoxides. Alkali metal ($$K$$) has an oxidation state of +1 int these given compounds.Let oxidation state of $$Mn$$ is x,so we found that:
In $${ K }_{ 2 }Mn{ O }_{ 4 }\\ $$, $$2(1)+x+4(-2)=0,x=+6$$
In $${ K }_{ }Mn{ O }_{ 4 }\\ $$, $$1+x+4(-2)=0,x=+7$$
In $${ }_{ }Mn{ O }_{ 2 }\\ $$, $$x+2(-2)=0,x=+4$$
In $${Mn}_{2}{ O }_{ 3 }\\ $$, $$2(x)+3(-2)=0,x=+3$$
In $${ K }_{ }Mn{ O }_{ 4 }$$, $$Mn$$ shows its highest oxidation state of +7.

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Re-Attempt Weekly Quiz Competition

$$K_3[Fe(OH)_6]$$	$$+3+x-6=0$$	$$x=+3$$
$$K_2[FeO_4]$$	$$+2+x-8=0$$	$$x=+6$$
$$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$	$$+2$$ state in $$FeSO_4$$	$$(x-2=0)$$
$$Fe(CN)^{3-}_6$$	$$x-6=-3$$	$$x=+3$$

Redox Reactions Test - 33

Result

Redox Reactions Test - 33

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Question 1

$$+1$$

$$+2$$

$$-4$$

$$0$$

Solution

Question 2

$$+4, +3$$

$$+7, +5$$

$$+3, +4$$

$$+5, +7$$

Solution

Question 3

$$K_3[Fe(OH)_6]$$

$$K_2[FeO_4]$$

$$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$

$$[Fe(CN)_6]^{3-}$$

Solution

Question 4

$$HClO_4$$

$$HClO_3$$

$$HClO_2$$

$$HClO$$

Solution

Question 5

$$-1$$ to $$+1$$

$$-3$$ to $$+3$$

$$-3$$ to $$+5$$

$$-5$$ to $$+1$$

Solution

Question 6

Ortho phosphoric acid

Phosphorus acid

Meta phosphoric acid

Pyrophosphoric acid

Solution

Question 7

$$+6$$

$$+4$$

$$+3$$

$$+2$$

Solution

Question 8

$$-3, +6$$ and $$+6$$

$$+5, +6$$ and $$+6$$

$$+3, +6$$ and $$+5$$

$$+5, +3$$ and $$+6$$

Solution

Question 9

$$\underline{Fe}+CuSO_4\rightarrow Cu+FeSO_4$$

$$\underline{H_2}+Cl_2\rightarrow 2HCl$$

$$\underline{C}+H_2O\rightarrow CO+H_2$$

$$\underline{MnO_2}+4HCl\rightarrow MnCl_2+Cl_2+2H_2O$$

Solution

Question 10

$$K_2MnO_4$$

$$KMnO_4$$

$$MnO_2$$

$$Mn_2O_3$$

Solution

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