Redox Reactions Test

Result

Redox Reactions Test - 40

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Oxidation number of P in $$Ba\, (H_2PO_2)_2$$ is:

A
+3

B
+2

C
+1

D
-1

Solution

$$Ba(H_{2}PO_{2})_{2}$$ is
$$Ba(H_{2}PO_{2})_{2}\Rightarrow (H_{2}PO_{2})_{2}^{2-}$$
$$(+1)\times 2+(x)-2\times 2=-1$$
$$+2+x-4=-1$$
$$x=-1+4-2$$
$$\boxed{x=+1}$$
So oxidation number of $$P=+1$$
Question 2
1 / -0

What is the oxidation state of $$Th$$ in organometallic compound $$[ThHOR(C_5Me_5)_2]$$?

A
$$+1$$

B
$$+2$$

C
$$+4$$

D
$$+3$$

Solution

Given compound $$[ThHOR(C_5Me_5)_2]$$ $$[ROH^+ ; (+1) - (H); (C_5Me_5)-(-1)]$$
$$x +1 -2 =0$$
$$\Rightarrow x - 1 =0$$
$$\Rightarrow x = +1$$
Hence, oxidation state of $$Th$$ is $$+1$$
Question 3
1 / -0

Predict the highest and lowest oxidation state of $$(a)\ Ti$$ and $$(b)\ Tl$$ in combined state.

A
$$a[0,+3]\ b[0,+2]$$

B
$$a[+3,0]\ b[+4,0]$$

C
$$a[+4,0]\ b[+4,0]$$

D
$$a[+4,+2]\ b[+3,+1]$$

Solution

Outer shell configuration of $$Ti$$ is :
$$[Ar]3d^24s^2$$
It minimum state it will remove outer electrons
and at maximum it will remove both d and
b structures.
For $$Tl : [Xe]4f^{14}5d^{10}6s^26p^1$$
For minimum states it will remove p electrons
and at maximum all 3 bth shall
electrons will be removed.
Question 4
1 / -0

In Redox reaction:
$$Fe + HNO_{3}\rightarrow Fe(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O$$ the coefficient of $$HNO_{3}, Fe(NO_{3})_{2}, NH_{4}NO_{3}$$ is:

A
$$1 : 10 : 4$$

B
$$10 : 4 : 1$$

C
$$4:10:4$$

D
$$4:1:4$$

Solution

After balancing the equation, we get
$$ 4Fe + 10HNO_3 \rightarrow 4Fe(NO_3)_2 + NH_4NO_3 + 3H_2O$$
Therefore, Ratio of $$HNO_3,\ Fe(NO_3)_2\ and\ NH_4NO_3 $$is :
10:4:1.
Hence, Option B is correct.
Question 5
1 / -0

The oxidation number of iron in potassium ferricyanide $$[K_{3}Fe(CN)_{6}]$$ is :

A
Two

B
Six

C
Three

D
None

Solution

To calculate: Oxidation state of iron in potassium ferro cyanide.
Solution :
Oxidation state of Potassium , K = +1
Oxidation state of Cyanide ion =-1
Let oxidation state of Iron be x
We know that sum of all the oxidation states in the compound is equal to overall charge of the compound.
Therefore, (+1)3 +[ x +6(-1)] = 0
$$\Rightarrow 3+ [ x-6] =0$$
$$\Rightarrow x= +6-3 =+3$$
Hence, the correct option is C.
Question 6
1 / -0

The ratio of number of moles of $$KMnO_4$$ and $$K_2Cr_2O_7$$ required to oxidise $$0.1$$mol $$Sn^{+4}$$ in acidic medium :

A
$$6 : 5$$

B
$$5 : 6$$

C
$$1 : 2$$

D
$$2 : 1$$

Solution

$$\overset { +7 }{ KMnO_4 } \quad \underrightarrow { acidic \quad medium }\quad \overset { +2 }{ Mn^{+2} } $$
$$\overset { +6 }{ K_2Cr_2O_7 } \quad \underrightarrow { acidic \quad medium }\quad \overset { +3 }{ Cr^{+3} } $$
$$\overset { +4 }{Sn^{+4} } \quad \underrightarrow { reducing \quad agent}\quad \overset { +2 }{ Sn^{+2} } $$
Number of equivalents $$= moles \times nf$$
$$n(Sn^{+4})=0.1$$
$$nf(Sn^{+4})=2$$
Equating number of equivalents of $$KMn{O}_{4}$$ and $${Sn^{+4}}$$
$${ n }_{ 1 }{ nf }_{ 1 }={ n }_{ 2 }{ nf }_{ 2 }$$
$$0.1\times 2=5n_{ 2 }\\ n_{ 21 }=0.04$$
Equating number of equivalents $$K_2$$ $$Cr_2$$ $$O_7$$ and $${Sn^{+4}}$$
$$n_1f_1=n_22f_22$$
$$0.1 \times2=6n_{22}$$
$$n_{22} =\cfrac{0.1} {3}$$
Ratio of moles of $$KMnO_3$$ and $$K_2 Cr_2 O_7=\cfrac{0.04} {0.1}=\cfrac{6} {5}$$
Question 7
1 / -0

When benzaldehyde is oxidised to give benzoic acid then the oxidation state of carbon of aldehydic group is changed from:

A
$$+2$$ to $$+3$$

B
$$+1$$ to $$+3$$

C
Zero to $$+2$$

D
No change

Solution

Bonds between same atoms donot affect the oxidation state. Therefore only functional group affects the oxidation state of $$C$$
In benzaldehyde:
$$=O$$ has$$-2$$ and $$H$$ has $$+1$$, therefore in $$CHO$$ O.S. of $$C$$ is: $$x-2+1=0$$
$$C$$ is $$+1$$
In Benzoic acid $$=O$$ has$$-2$$ and $$OH$$ has $$-1$$, therefore in $$COOH$$ O.S. of $$C$$ in $$CO-OH$$ is: $$x-2-1=0$$
$$C$$ is $$+3$$
Thus O.S. of $$C$$ changes from $$+1$$ to $$+3$$. Option B is correct
Question 8
1 / -0

The reaction which proceeds in the forward direction is:

A
$$Fe_3O_4 + 6HCl \rightarrow2 FeCl_3 + 3H_2O$$

B
$$NH_3 + H_2O + NaCl\rightarrow NH_4Cl + NaOH$$

C
$$SnCl_4 +Hg_2Cl_2\rightarrow SnCl_2 + 2HgCl_2$$

D
$$2CuI + I_2 + 4 K^+ \rightarrow 2 Cu^{2+} + 4 KI$$

Solution

$$FeCl_3$$ and $$H_2O$$ do not react to give back $$Fe_3O_4$$ and $$HCl$$.
Question 9
1 / -0

Which of the following elements never show positive oxidation number?

A
$$O$$

B
$$Fe$$

C
$$Ga$$

D
$$F$$

Solution

F is electronegative element . Electronegetivity of F is 4. Therefore F never show positive oxidation number.
Question 10
1 / -0

Choose the redox reaction from the following.

A
$$Cu+2H_2SO_4 \rightarrow CuSO_4+SO_2+2H_2O$$

B
$$BaCl_2+H_2SO_4 \rightarrow BaSO_4+2HCl$$

C
$$2NaOH+H_2SO_4 \rightarrow Na_2SO_4+2H_2O$$

D
$$KNO_2+H_2SO_4 \rightarrow 2HNO_2+K_2SO_4$$

Solution

A redox reaction is the one in which oxidation and reduction take place simultaneously.
In option A,
$$\cfrac { Cu }{ O } +\cfrac { 2{ H }_{ 2 }{ SO }_{ 4 } }{ +6 } \longrightarrow \cfrac { CuS{ O }_{ 4 } }{ +2 } +\cfrac { { SO }_{ 2 } }{ +4 } +2H_{ 2 }O$$
Oxidation of $$Cu$$ from $$"0"$$ to $$"+2"$$ and reduction of S from $$"+6"$$ to $$"+4"$$ occurs in the same reaction and hence it is a redox reaction.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 40

Result

Redox Reactions Test - 40

Score

-

Rank

-

SHARING IS CARING

Question 1

+3

+2

+1

-1

Solution

Question 2

$$+1$$

$$+2$$

$$+4$$

$$+3$$

Solution

Question 3

$$a[0,+3]\ b[0,+2]$$

$$a[+3,0]\ b[+4,0]$$

$$a[+4,0]\ b[+4,0]$$

$$a[+4,+2]\ b[+3,+1]$$

Solution

Question 4

$$1 : 10 : 4$$

$$10 : 4 : 1$$

$$4:10:4$$

$$4:1:4$$

Solution

Question 5

Two

Six

Three

None

Solution

Question 6

$$6 : 5$$

$$5 : 6$$

$$1 : 2$$

$$2 : 1$$

Solution

Question 7

$$+2$$ to $$+3$$

$$+1$$ to $$+3$$

Zero to $$+2$$

No change

Solution

Question 8

$$Fe_3O_4 + 6HCl \rightarrow2 FeCl_3 + 3H_2O$$

$$NH_3 + H_2O + NaCl\rightarrow NH_4Cl + NaOH$$

$$SnCl_4 +Hg_2Cl_2\rightarrow SnCl_2 + 2HgCl_2$$

$$2CuI + I_2 + 4 K^+ \rightarrow 2 Cu^{2+} + 4 KI$$

Solution

Question 9

$$O$$

$$Fe$$

$$Ga$$

$$F$$

Solution

Question 10

$$Cu+2H_2SO_4 \rightarrow CuSO_4+SO_2+2H_2O$$

$$BaCl_2+H_2SO_4 \rightarrow BaSO_4+2HCl$$

$$2NaOH+H_2SO_4 \rightarrow Na_2SO_4+2H_2O$$

$$KNO_2+H_2SO_4 \rightarrow 2HNO_2+K_2SO_4$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.