Redox Reactions Test

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Redox Reactions Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

In the reaction, $$4Fe+{ 3O }_{ 2 }\rightarrow { 4Fe }^{ 3+ }+{ 6O }^{ 2- } $$ which of the following statement is incorrect?

A
It is redox reaction

B
Metallic iron is a reducing agent

C
$${ Fe }^{ 3+ }$$ is an oxidising agent

D
Metallic iron is reduced to $${ Fe }^{ 3+ }$$

Solution

$$ \begin{array}{l} \text { The given reaction is redox reaction } \\ \text { as Fe oxidises to Fe }^{3+} \text { and } \mathrm{O}_{2} \\ \text { reduces to } O^{2-} \text { . } \end{array} $$
$$\begin{array}{l}\text{In this reaction metallic iron }\\ \text { is acting like a reducing agent. }\end{array} $$
$$ \begin{array}{l} \text { If, we reverse the given reaction, } \\\text{Fe}^{3+}\text{ will behave as an oxidizing }\\ \text { agent. } \end{array} $$
Question 2
1 / -0

Which of the following statements is wrong?

A
Oxidation number of oxygen is $$+1$$ in peroxides

B
Oxidation number of oxygen is $$+2$$ in oxygen difluoride

C
Oxidation number of oxygen is $$-1/2$$ super oxides

D
Oxidation number of oxygen is $$-2$$ in most of its compounds

Solution
Question 3
1 / -0

Given the standard half-cell potentials $$(E^o)$$ of the following as

$$Zn=Zn^{2+}+2e^-$$; $$E^o=+0.76$$V
$$Fe=Fe^{2+}+2e^{-}$$; $$E^o=+0.41$$V

Then the standard e.m.f. of the cell with the reaction $$Fe^{2+}+Zn\rightarrow Zn^{2+}+Fe$$ is?

A
$$-0.35$$V

B
$$+0.35$$V

C
$$+1.17$$V

D
$$-1.17$$V

Solution

$$Zn^{+2}+2e^-\rightarrow Zn$$ ; $$E^o=-0.76\ V$$ (SRP) (Anode)

$$Fe^{+2}+2e^-\rightarrow Fe$$ ; $$E^o=-0.41\ V$$ (SRP) (Cathode)

$$E^o_{cell}=E^+_{Fe^{+2}/Fe}-E^o_{Zn^{+2}/Zn}$$

$$=-0.41+0.76=0.35V$$.

Hence, the correct option is $$\text{B}$$
Question 4
1 / -0

In the chemical reaction
$${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+x{ H }_{ 2 }{ SO }_{ 4 }+y{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }+z{ H }_{ 2 }O$$
Values of x, y and z are:

A
1, 3, 1

B
4, 1, 4

C
3, 2, 3

D
2, 1, 2

Solution
Question 5
1 / -0

The values of coefficients to balance the following reaction are:
$$Cr(OH)_3 + ClO^- + OH^- \rightarrow Cr{ O }_{ 4 }^{ 2- } + Cl^- +H_2O$$?

A
$$Cr(OH)_3 - ( 2), ClO^- - (3), Cr{ O }_{ 4 }^{ 2- } - (3), Cl^- - (3)$$

B
$$Cr(OH)_3 - ( 2), ClO^- - (4), Cr{ O }_{ 4 }^{ 2- } - (3), Cl^- - (2)$$

C
$$Cr(OH)_3 - ( 2), ClO^- - (4), Cr{ O }_{ 4 }^{ 2- } - (4), Cl^- - (2)$$

D
$$Cr(OH)_3 - ( 2), ClO^- - (3), Cr{ O }_{ 4 }^{ 2- } - (2), Cl^- - (3)$$

Solution

$$Cr(OH)_3+ClO^-+OH^- \rightarrow CrO_4^{2-}+Cl^-+H_2O$$
in a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
$$\overset {+3}{Cr}(OH)_3+ \overset {+1}{Cl}O^-+OH^- \rightarrow \overset {+6}{Cr}O_4^{2-} + \overset {-1}{Cl^-}+H_2O$$
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: $$\overset {+1}{Cl}O^- \rightarrow \overset {-1}{Cl^-}$$
Oxidation: $$\overset {+3}{Cr}(OH)_3 \rightarrow \overset {+6}{Cr}O_4^{2-}$$
Step 3: oxidation-number change is:
Reduction: $$\overset {+1}{Cl}O^- \rightarrow \overset {-1}{Cl^-}$$: Gain of 2 electron
Oxidation: $$\overset {+3}{Cr}(OH)_3 \rightarrow \overset {+6}{Cr}O_4^{2-}$$: Loss of total 3 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: $$\overset {+1}{Cl}O^- \rightarrow \overset {-1}{Cl^-} \times 3$$: Gain of 6 electron
Oxidation: $$\overset {+3}{Cr}(OH)_3 \rightarrow \overset {+6}{Cr}O_4^{2-} \times 2$$: Loss of total 6 electrons
OR
Reduction: $$3\overset {+1}{Cl}O^- \rightarrow 3\overset {-1}{Cl^-}$$
Oxidation: $$2\overset {+3}{Cr}(OH)_3 \rightarrow 2\overset {+6}{Cr}O_4^{2-}$$
Step 5: Balance O atoms in reduction reaction by adding $$H_2O$$ and then balance H by $$H^+$$ as:
Reduction: $$3\overset {+1}{Cl}O^- +6H^+ \rightarrow 3\overset {-1}{Cl^-}+3H_2O$$
Oxidation: $$2\overset {+3}{Cr}(OH)_3+2H_2O\rightarrow 2\overset {+6}{Cr}O_4^{2-}+4H^+$$
Step 6: For base catalysed reaction add $$OH^-$$ to both side to neutralize $$H^+$$ as:
Reduction: $$3\overset {+1}{Cl}O^- +6H^++6OH^- \rightarrow 3\overset {-1}{Cl^-}+3H_2O+6OH^-$$
Oxidation: $$2\overset {+3}{Cr}(OH)_3+2H_2O +4OH^-\rightarrow 2\overset {+6}{Cr}O_4^{2-}+4H^+ +4OH^-$$
OR
Reduction: $$3\overset {+1}{Cl}O^- +3H_2O \rightarrow 3\overset {-1}{Cl^-}+6OH^-$$
Oxidation: $$2\overset {+3}{Cr}(OH)_3 +4OH^-\rightarrow 2\overset {+6}{Cr}O_4^{2-}+2H_2O$$
Thus overall reaction is:
$$3\overset {+1}{Cl}O^- +3H_2O + 2\overset {+3}{Cr}(OH)_3 +4OH^- \rightarrow 3\overset {-1}{Cl^-}+6OH^- +2\overset {+6}{Cr}O_4^{2-}+2H_2O$$
OR
$$3ClO^- +H_2O + 2Cr(OH)_3 \rightarrow 3Cl^-+2OH^- +2CrO_4^{2-}$$
thus the values of coefficients of balanced reaction are:
$$Cr(OH)_3−(2),ClO^−−(3),CrO_4^{2−}-(2),Cl^−−(3)$$

Question 6

1 / -0

Match the column I with column II with the type of reaction and mark the appropriate choice.

	Column I		Column II
$$(A)$$	$$3Mg_(s) + N_{2(g)}\xrightarrow{\Delta } Mg_3N_{2(s)}$$	$$(i)$$	Displacement
$$(B)$$	$$NaH_(s) + H_2O_(l)\rightarrow NaOH_{(aq)} + H_{2(g)}$$	$$(ii)$$	Decomposition
$$(C)$$	$$3Cl{ O }_{( aq) }^{ - }\rightarrow 2C{ l}_{( aq) }^{ - } + Cl{ O_3 }_{ (aq) }^{ - }$$	$$(iii)$$	Combination
$$(D)$$	$$2KClO_{3(s)}\rightarrow 2KCl_(s) + 3O_{2(g)}$$	$$(iv)$$	Disproportionation

$$(A)\rightarrow (i) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (iv)$$

$$(A)\rightarrow (iv) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (i)$$

$$(A)\rightarrow (ii) , (B)\rightarrow (i) , (C)\rightarrow (iii) ,D\rightarrow (iv)$$

$$(A)\rightarrow (iii) , (B)\rightarrow (i) , (C)\rightarrow (iv) ,D\rightarrow (ii)$$

Solution

	Column I	Column II	Reason
$$A$$	$$3Mg(s)+N_2(g) \rightarrow Mg_3N_2(s)$$	(iii) Combination	A combination reaction is defined as a chemical reaction in which two or more substances combine to form a single substance under suitable conditions.
$$B$$	$$NaH(s)+H_2O(l)\rightarrow NaOH(aq)+H_2(g)$$	(i) Displacement	Displacement reaction is a chemical reaction in which a more reactive element (NaH) displaces a less reactive element ($$H_2O \rightarrow H_2$$) from its compound. Both metals and non-metals take part in displacement reactions.
$$C$$	$$3ClO^−(aq) \rightarrow 2Cl^−(aq)+ClO_3^−(aq)$$	(iv) Disproportionation	Disproportionation reaction is a redox reaction in which a compound of intermediate oxidation state converts to two different compounds, one of higher and one of lower oxidation states as: $$3\overset {+1}{Cl}O^−(aq) \rightarrow 2\overset {-1}{Cl^−}(aq)+\overset {+5}{Cl}O_3^−(aq)$$
$$D$$	$$2KClO_3(s) \rightarrow 2KCl(s)+3O_2(g)$$	(ii) Decomposition	A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds.

$$(A)-(iii),\ (B)-(i),\ (C)-(iv),\ D-(ii)$$

Question 7
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Which of the following species has an atom with $$+6$$ oxidation state?

A
$${ MnO }_{ 4 }^{ - }$$

B
$$Cr{ (CN) }_{ 6 }^{ 3- }$$

C
$${ NiF }_{ 6 }^{ 2- }$$

D
$$CrO_2Cl_2$$

Solution

The oxidation state of chromium is $$+6$$.
$$CrO_2Cl_2 \rightarrow x + (-4) + (-2) = 0 \Rightarrow x = +6$$
The oxidation states of central atoms in other three compounds are:
$${ MnO }_{ 4 }^{ - }$$ $$=+7$$

$$Cr{ (CN) }_{ 6 }^{ 3- }$$ $$=+3$$

$${ NiF }_{ 6 }^{ 2- }$$ $$=+4$$
Question 8
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Arrange the oxides of nitrogen in increasing order of oxidation state of $$N$$ from +1 to +5:

A
$$ N_2O < N_2O_3 < NO_2 < N_2O_5 < NO $$

B
$$ N_2O < NO < N_2O_3 < NO_2 < N_2O_5 $$

C
$$ N_2O_5 < NO_2 < N_2O_3 < NO < N_2O $$

D
$$ NO < N_2O < NO_2 < N_2O_3 < N_2O_5 $$

Solution

We know that the charge on oxygen is $$-2$$.
Different oxides of nitrogen has different number of oxygen atoms. Hence the oxidation states of nitrogen in its Oxides are as follows
$$N_2O \rightarrow +1, NO \rightarrow +2$$
$$N_2O_3 \rightarrow +3 , NO_2 \rightarrow +4 , N_2O_5 \rightarrow +5$$
Therefore, option B is the correct answer of this question.

Question 9

1 / -0

Fill up the table from the given choice.

Element	Oxidation number
Oxygen	-2 in most compounds ____(i) in $$H_2O_2$$ and _____(ii) in $$ OF_2$$
Halogen	-1 for_____(iii) in all its compounds
Hydrogen	_____(iv) in most of its compounds ______(v)in binary metallic hydrides
Sulphur	______(vi) in all sulphides

$$(i) +1 (ii) +1 (iii) Cl (iv) +1 (v) -1 (vi) +2 $$

$$(i) -1 (ii) +2 (iii) F (iv) +1 (v) -1 (vi) -2 $$

$$(i) -1 (ii) +1 (iii) F (iv) +1 (v) +2 (vi) +2$$

$$(i) +1 (ii) +2 (iii) Cl (iv) +1 (v) +1 (vi) +6$$

Solution

Oxygen: -2 in most compounds ($$2 \times 1+2x=0 \Rightarrow x=-1)\ \underline { -1 }$$ (i) in $$H_2O_2$$ and ($$x-2 \Rightarrow x=+2)\ \underline { +2 }$$ (ii) in $$OF_2$$

Halogen: -1 for $$\underline { Fluorine(F) }$$ (iii) in all its compounds

Hydrogen: $$\underline { +1 }$$ (iv) in most of its compounds $$\underline { -1 }$$ (v) in binary metallic hydrides

Sulphur: $$\underline { -2 }$$ (vi) in all sulphides.

Question 10
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What will be the balanced equation in acidic medium for the given reaction ?
$$Cr_2{ O }_{ 7(aq) }^{ 2- } + SO_2{(g)} \rightarrow C{ r }_{ (aq) }^{ 3+ } + S{ O }_{ 4(aq) }^{ 2- }$$

A
$$Cr_2{ O }_{ 7(aq) }^{ 2- } + 3SO_2{(g}) + 2{ H }_{ (aq) }^{ + }\rightarrow 2C{ r}_{ (aq) }^{ 3+ } + 3S{ O }_{ 4(aq) }^{ 2- } + H_2O{(l)}$$

B
$$2Cr_2{ O }_{ 7(aq) }^{ 2- } + 3SO_2{(g}) + 4{ H }_{ (aq) }^{ + }\rightarrow 4C{ r}_{ (aq) }^{ 3+ } + 3S{ O }_{ 4(aq) }^{ 2- } + 2H_2O{(l)}$$

C
$$Cr_2{ O }_{ 7(aq) }^{ 2- } + 3SO_2{(g}) + 14{ H }_{ (aq) }^{ + }\rightarrow 2C{ r}_{ (aq) }^{ 3+ } + 3S{ O }_{ 4(aq) }^{ 2- } + 7H_2O{(l)}$$

D
$$Cr_2{ O }_{ 7(aq) }^{ 2- } + 6SO_2{(g}) + 7{ H }_{ (aq) }^{ + }\rightarrow 2C{ r}_{ (aq) }^{ 3+ } + 6S{ O }_{ 4(aq) }^{ 2- } + 7H_2O{(l)}$$

Solution

In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
$$\overset {+6}{Cr_2}{ O_7 }^{ 2- }(aq) + \overset {+4}{S}O_2 \rightarrow \overset{+3}{Cr} + \overset {+6}{S}O_4^{2-}$$
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: $$\overset {+6}{Cr_2}{ O_7 }^{ 2- }(aq) \rightarrow 2\overset{+3}{Cr}$$
Oxidation: $$\overset {+4}{S}O_2 \rightarrow \overset {+6}{S}O_4^{2-}$$
Step 3: oxidation-number change is:
Reduction: $$\overset {+6}{Cr_2}{ O_7 }^{ 2- } \rightarrow 2\overset{+3}{Cr}$$: Gain of 6 electrons total
Oxidation: $$\overset {+4}{S}O_2 \rightarrow \overset {+6}{S}O_4^{2-}$$ Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: $$\overset {+6}{Cr_2}{ O_7 }^{ 2- } \rightarrow 2\overset{+3}{Cr} \times 1$$: Gain of 6 electron total
Oxidation: $$\overset {+4}{S}O_2 \rightarrow \overset {+6}{S}O_4^{2-} \times 3$$ : Loss of 6 electron
$$\therefore$$ Reduction: $$\overset {+6}{Cr_2}{ O_7 }^{ 2- } \rightarrow 2\overset{+3}{Cr}$$
Oxidation: $$3\overset {+4}{S}O_2 \rightarrow 3\overset {+6}{S}O_4^{2-}$$
Step 5: Balance O atoms in reduction reaction by adding $$H_2O$$ and then balance H by $$H^+$$ as:
Reduction: $$\overset {+6}{Cr_2}{ O_7 }^{ 2- }+14H^+ \rightarrow 2\overset{+3}{Cr}+7H_2O$$
Oxidation: $$3\overset {+4}{S}O_2+6H_2O \rightarrow 3\overset {+6}{S}O_4^{2-}+12H^+$$
Thus overall Balanced reaction is:
$$\overset {+6}{Cr_2}{ O_7 }^{ 2- }+14H^++3\overset {+4}{S}O_2+6H_2O \rightarrow 2\overset{+3}{Cr}+7H_2O+3\overset {+6}{S}O_4^{2-}+12H^+$$
or $$\overset {+6}{Cr_2}{ O_7 }^{ 2- }+2H^++3\overset {+4}{S}O_2 \rightarrow 2\overset{+3}{Cr}+H_2O+3\overset {+6}{S}O_4^{2-}$$

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Redox Reactions Test - 67

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Redox Reactions Test - 67

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SHARING IS CARING

Question 1

It is redox reaction

Metallic iron is a reducing agent

$${ Fe }^{ 3+ }$$ is an oxidising agent

Metallic iron is reduced to $${ Fe }^{ 3+ }$$

Solution

Question 2

Oxidation number of oxygen is $$+1$$ in peroxides

Oxidation number of oxygen is $$+2$$ in oxygen difluoride

Oxidation number of oxygen is $$-1/2$$ super oxides

Oxidation number of oxygen is $$-2$$ in most of its compounds

Solution

Question 3

$$-0.35$$V

$$+0.35$$V

$$+1.17$$V

$$-1.17$$V

Solution

Question 4

1, 3, 1

4, 1, 4

3, 2, 3

2, 1, 2

Solution

Question 5

$$Cr(OH)_3 - ( 2), ClO^- - (3), Cr{ O }_{ 4 }^{ 2- } - (3), Cl^- - (3)$$

$$Cr(OH)_3 - ( 2), ClO^- - (4), Cr{ O }_{ 4 }^{ 2- } - (3), Cl^- - (2)$$

$$Cr(OH)_3 - ( 2), ClO^- - (4), Cr{ O }_{ 4 }^{ 2- } - (4), Cl^- - (2)$$

$$Cr(OH)_3 - ( 2), ClO^- - (3), Cr{ O }_{ 4 }^{ 2- } - (2), Cl^- - (3)$$

Solution

Question 6

$$(A)\rightarrow (i) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (iv)$$

$$(A)\rightarrow (iv) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (i)$$

$$(A)\rightarrow (ii) , (B)\rightarrow (i) , (C)\rightarrow (iii) ,D\rightarrow (iv)$$

$$(A)\rightarrow (iii) , (B)\rightarrow (i) , (C)\rightarrow (iv) ,D\rightarrow (ii)$$

Solution

Question 7

$${ MnO }_{ 4 }^{ - }$$

$$Cr{ (CN) }_{ 6 }^{ 3- }$$

$${ NiF }_{ 6 }^{ 2- }$$

$$CrO_2Cl_2$$

Solution

Question 8

$$ N_2O < N_2O_3 < NO_2 < N_2O_5 < NO $$

$$ N_2O < NO < N_2O_3 < NO_2 < N_2O_5 $$

$$ N_2O_5 < NO_2 < N_2O_3 < NO < N_2O $$

$$ NO < N_2O < NO_2 < N_2O_3 < N_2O_5 $$

Solution

Question 9

$$(i) +1 (ii) +1 (iii) Cl (iv) +1 (v) -1 (vi) +2 $$

$$(i) -1 (ii) +2 (iii) F (iv) +1 (v) -1 (vi) -2 $$

$$(i) -1 (ii) +1 (iii) F (iv) +1 (v) +2 (vi) +2$$

$$(i) +1 (ii) +2 (iii) Cl (iv) +1 (v) +1 (vi) +6$$

Solution

Question 10

$$Cr_2{ O }_{ 7(aq) }^{ 2- } + 3SO_2{(g}) + 2{ H }_{ (aq) }^{ + }\rightarrow 2C{ r}_{ (aq) }^{ 3+ } + 3S{ O }_{ 4(aq) }^{ 2- } + H_2O{(l)}$$

$$2Cr_2{ O }_{ 7(aq) }^{ 2- } + 3SO_2{(g}) + 4{ H }_{ (aq) }^{ + }\rightarrow 4C{ r}_{ (aq) }^{ 3+ } + 3S{ O }_{ 4(aq) }^{ 2- } + 2H_2O{(l)}$$

$$Cr_2{ O }_{ 7(aq) }^{ 2- } + 3SO_2{(g}) + 14{ H }_{ (aq) }^{ + }\rightarrow 2C{ r}_{ (aq) }^{ 3+ } + 3S{ O }_{ 4(aq) }^{ 2- } + 7H_2O{(l)}$$

$$Cr_2{ O }_{ 7(aq) }^{ 2- } + 6SO_2{(g}) + 7{ H }_{ (aq) }^{ + }\rightarrow 2C{ r}_{ (aq) }^{ 3+ } + 6S{ O }_{ 4(aq) }^{ 2- } + 7H_2O{(l)}$$

Solution

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