Redox Reactions Test

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Redox Reactions Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Determine the. oxidation state of chromium in the finfll product formed by the reaction between KI and acidified potassium dichromate solution.

A
+ 3

B
+2

C
+1

D
-3

Solution

K would not be expected to be in the final NET equation since K+ is a spectator ion; i.e., it does not react. When you put KI into water it dissolves and splits up into K+ and I- ions. Only the I- participates in the redox reaction. It wouldn't matter whether you used KI or NaI.
Oxidation: $$2I^- \rightarrow I^2 + 2e^- $$
Reduction: $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$

The final product formed is :
$$6I^- + Cr_2O_7^{2-} + 14H^+ ----> 3I_2 + 2Cr^{3+} + 7H_2O$$
Calculating the oxidation state of chromium we get + 3
Question 2
1 / -0

Which of the following processes does not involve either oxidation or reduction ?

A
Formation of slaked lime from quick lime

B
Heating mercuric oxide

C
Formation of manganese chloride from manganese oxide

D
Formation of zinc from zinc blende

Solution

Here, in this reaction
$$CaO+ H_2O \rightarrow Ca(OH)_2$$
oxidation number doesn't change so its not a redox reaction.
Question 3
1 / -0

The oxidation number of $$Mn$$ is $$+8$$ in the compound:

A
$${K}_{2}Mn{O}_{4}$$

B
$$Mn{O}_{4}$$

C
$$KMn{O}_{4}$$

D
$${Mn}_{2}{O}_{7}$$

Solution

In $$MnO_4$$ manganese oxidation number is $$+8$$.
Oxidation number of oxygen is $$-2$$. $$MnO_4$$ is neutral molecule so total oxidation number of molecule is zero.
Oxidation number of $$Mn+4(-2)=0$$
Oxidation number of $$Mn =+8$$.
So answer is B.
Question 4
1 / -0

Arrange the following in correct order of oxidation number of Mn
(i) $$Mn^{2+}$$ (ii) $$MnO_2$$
(III) $$KMnO_4$$ (iv) $$K_2MnO_4$$

A
(i) > (ii) > (iii) > (iv)

B
(i) < (ii) < (iv)< (iii)

C
(ii) < (iii < (i) < (iv)

D
(iii) < (i) < (iv) < (ii)

Solution

i) Oxidation number of Mn in $$Mn^{2+} = +2$$
(ii) Let oxidation number of Mn in $$MnO_2 = x$$
$$\therefore x + (-2 \times 2) = 0 , x = +4$$
(iii) Let oxidation number of Mn in KMno_4 = x
$$\therefore +1 +x +(-1 \times 4 ) = 0, \therefore x += + 7$$
(iv) Let oxidation number of Mn in $$K_2MnO_4$$ = x
$$\therefore (+1 \times 2 ) + x + (-2 \times \times 4 _ = 0$$
or x + 2 - 8 = 0
$$\therefore $$ x = +6
$$\therefore $$ the order of oxidation state is (i) < (ii) < (iv) < (iii)
Question 5
1 / -0

The oxidation number of oxygen in hydrogen peroxide is:

A
$$+1$$

B
$$-1$$

C
$$+2$$

D
$$-2$$
Question 6
1 / -0

Identify redox reaction among the following:

A
acid-base neutralization

B
precipitation reaction

C
metal displacement reaction

D
all the above

Solution

Metal displacement reactions are redox reactions in which both oxidation and reduction takes place.Example; $$Mg + ZnSO_{4} \rightarrow MgSO_{4} + Zn$$
Here Mg is undergoing oxidation with an increase in oxidation number from 0 to +2 and Zn is undergoing reduction with a decrease in oxidation number.
Question 7
1 / -0

The oxidation number of chlorine in $$NaCl{O}_{3}$$ is:

A
$$-5$$

B
$$+2$$

C
$$+5$$

D
$$-3$$

Solution

The overall charge of molecule is zero (because it's not an ion). Since sodium is a 1A family member, you can assume that the charge is $$+1$$. The charge of oxygen is almost always $$-2$$ so you can assume that as well.
$$(+1) +$$ oxidation of $$Cl + 3(-2) = 0 $$
So the oxidation of chlorine in that problem is $$+5$$.
So answer is C.
Question 8
1 / -0

Identify X in the equation given below. $$FeCl_3 + SO_2 + H_2O \rightarrow FeCl_2 + HCl + X$$

A
$$H_2 S$$

B
$$SO_3$$

C
$$H_2SO_4$$

D
$$H_2SO_3$$

Solution

The balanced equation can be written as:
$$2FeCl_{3} + SO_{2} + 2H_{2}O \rightarrow 2FeCl_{2} + 2HCl + H_{2}SO_{4}$$
Thus X = $$ H_{2}SO_{4}$$
Question 9
1 / -0

The oxidation state of nitrogen in $$Na{NO}_{3}$$ is :

A
$$+3$$

B
$$+5$$

C
$$-4$$

D
$$-5$$

Solution

The overall charge is zero (because it is not an ion). Since sodium is a 1A family member, the charge is +1. The charge of oxygen is almost always -2.
Hence, $$1+ OS \ of\ N + (3 \times -2)=0$$
So, the oxidation of nitrogen in that problem is +5.
Question 10
1 / -0

Oxidation number of S in $$SO_4^{2-}$$:

A
+3

B
+6

C
+2

D
-2

Solution

Let oxidation state of S in $$SO_4^{2-} = x$$
$$\therefore x+ (4 \times - 2) = - 2$$
or $$x - 8 = - 2$$ $$\therefore$$ $$x = +6$$

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Redox Reactions Test - 28

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Redox Reactions Test - 28

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SHARING IS CARING

Question 1

+ 3

+2

+1

-3

Solution

Question 2

Formation of slaked lime from quick lime

Heating mercuric oxide

Formation of manganese chloride from manganese oxide

Formation of zinc from zinc blende

Solution

Question 3

$${K}_{2}Mn{O}_{4}$$

$$Mn{O}_{4}$$

$$KMn{O}_{4}$$

$${Mn}_{2}{O}_{7}$$

Solution

Question 4

(i) > (ii) > (iii) > (iv)

(i) < (ii) < (iv)< (iii)

(ii) < (iii < (i) < (iv)

(iii) < (i) < (iv) < (ii)

Solution

Question 5

$$+1$$

$$-1$$

$$+2$$

$$-2$$

Question 6

acid-base neutralization

precipitation reaction

metal displacement reaction

all the above

Solution

Question 7

$$-5$$

$$+2$$

$$+5$$

$$-3$$

Solution

Question 8

$$H_2 S$$

$$SO_3$$

$$H_2SO_4$$

$$H_2SO_3$$

Solution

Question 9

$$+3$$

$$+5$$

$$-4$$

$$-5$$

Solution

Question 10

+3

+6

+2

-2

Solution

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