Redox Reactions Test

Result

Redox Reactions Test - 38

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

The sum of oxidation number of nitrogen in $$NH_4NO_3$$ is:

A
+1

B
+2

C
+4

D
none of these

Solution

Given $$NH_4NO_3$$ in the ionic state is present in the form of $$NH_4^++{NO_3}^-$$.

$$\Rightarrow x+4=+1$$ $$\Rightarrow x-6=-1$$
$$\Rightarrow x= -3$$ $$\Rightarrow x=+5$$

Sum of two oxidation states of $$N$$ gives $$+2$$ .
Question 2
1 / -0

The oxidation number of cobalt in $$K[Co(CO)_4]$$ is:

A
$$-1$$

B
$$-3$$

C
$$+1$$

D
$$+3$$

Solution

The oxidation number of colbalt in $$K[Co(CO)_4]$$ is:

$$+1+x+0\times 4=0$$

$$\therefore x=-1$$.

Hence, option $$A$$ is correct.
Question 3
1 / -0

Oxalic acid dihydrate, $$H_{2}C_{2}O_{4}\cdot 2H_{2}O(s)$$ is often used as a primary reagent to standardise sodium hydroxide solution. Which of these facts are reasons to choose this substance as a primary standard?
I. It is diprotic.
II. It is a stable compound that can be weighed directly in air.
III. It is available in pure form.

A
III only

B
I and II only

C
II and III only

D
I, II and III

Solution

$$\begin{array}{l} { H_{ 2 } }{ C_{ 2 } }{ O_{ 4 } }\cdot 2{ H_{ 2 } }O\left( s \right) \to oxadic\, acid\, dihydrate\, \\ Since,\, it\, is\, very\, stable\, and\, also\, available\, in\, pressure \\ from\, in\, the\, laboratory\, i.e.\, it's\, always\, used\, as\, primary\, s\tan dard\, in\, the\, lab. \\ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$$
Question 4
1 / -0

The oxidation number of sulphur in $$H_2S_2O_8$$ is:

A
$$+2$$

B
$$+4$$

C
$$+6$$

D
$$+7$$

Solution

$$H-O-\overset{\overset{\displaystyle O}{\displaystyle ||}}{\underset{\underset{\displaystyle O}{\displaystyle ||}}{\displaystyle S}}-O-O-\overset{\overset{\displaystyle O}{\displaystyle ||}}{\underset{\underset{\displaystyle O}{\displaystyle ||}}{\displaystyle S}}-O-H$$

$$\because$$ $$2\ O$$-atoms are in peroxide linkage
$$\therefore$$ Their oxidation state$$=-1$$
and, oxidation state of remaining $$6\ O$$-atoms $$=-2$$
Oxidation state of $$H-$$ atoms $$=+1$$

Let oxidation state of $$S$$ atom $$=x$$
$$S_2$$ $$H_2$$ $$O_6$$ $$O_2$$

$$\therefore 2x+2(+1)+6(-2)+2(-1)=0$$

$$2x+2-12-2=0$$

$$\therefore x=+6$$.
Question 5
1 / -0

Oxidation state of phosphorus in cyclotrimetaphosphoric acid is:

A
+5

B
+4

C
+3

D
none of these

Solution

Formula of Trimetaphosphoric acid $$(HPO_3)_3$$
Oxidation state of $$P$$ is: (Refer to Image)
$$3(+1+x+3(-2))=0$$
$$\Rightarrow x-6+1=0$$
$$\Rightarrow x= +5$$
$$+5$$ is the oxidation state of phosphorus in Trimetaphosphoric acid.
Question 6
1 / -0

A compound of Xe and F is found to have $$53.5\%$$ Xe. What is the oxidation number of Xe in this compound?

A
$$-4$$

B
$$0$$

C
$$+4$$

D
$$+6$$

Solution

Hint: The total number of electrons that an atom acquires or loses in order to form a chemical connection with another atom is known as the oxidation number.

Explanation:

Let's say the compound's total weight is $$100\ grams$$.

As a result, $$53,5\ grams$$ of $$Xe$$ and $$46.7\ grams$$ of $$F$$ are present.

$$\therefore$$ The number of moles of $$Xe$$ $$=\frac{53.5}{133}=0.4$$
$$\therefore$$ The number of moles of $$F$$ $$=\frac{46.7}{19}=2.45$$.

Achieving a simple whole-number ratio by dividing both by $$0.4$$.

Moles of $$Xe$$ $$=\frac{0.4}{0.4}=1$$

Moles of $$F$$ $$=\frac{2.45}{0.4}=6$$

As a result, the formula is$$Xe{{F}_{6}}$$.

$$\therefore$$ $$Xe$$ has an oxidation state of $$+6$$, as $$F$$ has an oxidation state of $$-1$$.

Final answer:

The correct answer is option $$(D).$$
Question 7
1 / -0

The value of n in the molecular formula $$Be_nAl_2Si_6O_{18}$$ is:

A
1

B
2

C
3

D
4
Question 8
1 / -0

Of the following reactions, only one is a redox reaction. Identify this reaction.

A
$$Ca{ \left( OH \right) }_{ 2 }+2HCl\longrightarrow Ca{ Cl }_{ 2 }+2{ H }_{ 2 }O$$

B
$$2{ S }_{ 2 }{ O }_{ 7 }^{ 2- }+2{ H }_{ 2 }O\longrightarrow 2S{ O }_{ 4 }^{ 2- }+4{ H }^{ + }$$

C
$$Ba{ Cl }_{ 2 }+MgS{ O }_{ 4 }\longrightarrow BaS{ O }_{ 4 }+Mg{ Cl }_{ 2 }$$

D
$${ Cu }_{ 2 }S+2FeO\longrightarrow 2Cu+2Fe+S{ O }_{ 2 }$$

Solution

Redox reaction means oxidation and reduction reaction must occur simultaneously.

$$Cu_2S+2FeO\longrightarrow 2Cu+2Fe+SO_2$$
Question 9
1 / -0

When $$6 \times 10^{22}$$ electrons are used in the electrolysis of a metalic salt, 1.9 gm of the metal is deposited at the cathode The atomic weight of that metal Is 57. So oxidation state of the metal in the salt is:

A
+2

B
+3

C
+1

D
+4

Solution

mol. of $${ e }^{ - }=\dfrac { 6\times { 10 }^{ 22 } }{ 6\times { 10 }^{ 23 } } =0.1$$ mol
moles of metal $$=\dfrac { 1.9 }{ 57 } =0.033$$ mol
$${ m }^{ n+ }+n{ e }^{ - }\longrightarrow m$$
$$\therefore$$ if $$0.1$$ of $${ e }^{ - }$$ is passed and $$0.033$$ mol of $$M$$ is deposited.
charge $$=\dfrac { 0.1 }{ 0.033 } =+3$$
Question 10
1 / -0

Which of the following is not a redox reaction?

A
$$CaCO_{3}\rightarrow CaO + CO_{2}$$

B
$$O_{2} + 2H_{2}\rightarrow 2H_{2}O$$

C
$$Na + H_{2}O\rightarrow NaOH + \dfrac {1}{2}H_{2}$$

D
$$MnCl_{3}\rightarrow MnCl_{2} + \dfrac {1}{2}Cl_{2}$$

Solution

$$CaCO_{3} \rightarrow CaO + CO_{2}$$

Since the reaction does not involve oxidation and reduction, it is not 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 - 38

Result

Redox Reactions Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

+1

+2

+4

none of these

Solution

Question 2

$$-1$$

$$-3$$

$$+1$$

$$+3$$

Solution

Question 3

III only

I and II only

II and III only

I, II and III

Solution

Question 4

$$+2$$

$$+4$$

$$+6$$

$$+7$$

Solution

Question 5

+5

+4

+3

none of these

Solution

Question 6

$$-4$$

$$0$$

$$+4$$

$$+6$$

Solution

Question 7

1

2

3

4

Question 8

$$Ca{ \left( OH \right) }_{ 2 }+2HCl\longrightarrow Ca{ Cl }_{ 2 }+2{ H }_{ 2 }O$$

$$2{ S }_{ 2 }{ O }_{ 7 }^{ 2- }+2{ H }_{ 2 }O\longrightarrow 2S{ O }_{ 4 }^{ 2- }+4{ H }^{ + }$$

$$Ba{ Cl }_{ 2 }+MgS{ O }_{ 4 }\longrightarrow BaS{ O }_{ 4 }+Mg{ Cl }_{ 2 }$$

$${ Cu }_{ 2 }S+2FeO\longrightarrow 2Cu+2Fe+S{ O }_{ 2 }$$

Solution

Question 9

+2

+3

+1

+4

Solution

Question 10

$$CaCO_{3}\rightarrow CaO + CO_{2}$$

$$O_{2} + 2H_{2}\rightarrow 2H_{2}O$$

$$Na + H_{2}O\rightarrow NaOH + \dfrac {1}{2}H_{2}$$

$$MnCl_{3}\rightarrow MnCl_{2} + \dfrac {1}{2}Cl_{2}$$

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.