Redox Reactions Test

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Redox Reactions Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation state of molybdenum in its oxo complex species$$ [Mo_2O_4(C_2H_4)_2(H_2O_2)]^{2-}]$$.

A
+2

B
+3

C
+4

D
+5

Solution

Let the oxidation number of molybdenum $$(Mo)$$ be $$x$$.
Oxidation number of $$H_2O_2=C_2H_4=0$$, $$O=-2$$.
In $$[Mo_2O_4(C_2H_4)_2(H_2O_2)]^{2-}$$
$$2x + 4 \times (-2) + 2\times 0 + 0 = -2$$
$$2x - 8 = -2$$ or$$x = +3$$
The oxidation state of molybdenum in its oxo complex species $$[Mo_2O_4 (C_2H_4)_2(H_2O_2)]^{2−}$$ is +3
Question 2
1 / -0

Which of the following reaction shows reduction of water?

A
$$2H_2O+2Na\rightarrow 2NaOH+H_2$$

B
$$6CO_2+12H_2O\rightarrow C_6H_12O_6+6H_2O+6O_2$$

C
$$2F_2+2H_2O\rightarrow 4H^++4F^-+O_2$$

D
$$P_4O_10+6H_2O\rightarrow 4H_3PO_4$$

Solution

Hence the reaction shows reduction of water _.
Question 3
1 / -0

The oxidation number of $$Cr$$ in $$CrO_5$$ which has the following structure is:

A
+4

B
+5

C
+6

D
+3

Solution

In the structure of $$CrO_5$$, it has four $$O$$ atoms as peroxide which have oxidation number = -1 and one $$O$$ as oxide (double bonded $$O$$) atom with oxidation number = -2.
Let the oxidation number of $$Cr$$ is $$x$$.
Then, $$x+4 \times (-1) + 1 \times (-2) = 0$$
or $$x = +6$$
The oxidation number of $$Cr$$ in $$CrO_5$$ is +6.
Question 4
1 / -0

Which of the following arrangements represent increasing oxidation number of the central atom ?

A
$$CrO_{2}^{-}$$, $$ClO_{3}^{-}$$ ,$$CrO_{4}^{2-}$$, $$MnO_{4}^{-}$$,

B
$$ClO_{3}^{-}$$, $$CrO_{4}^{2-}$$, $$MnO_{4}^{-}$$, $$CrO_{2}^{-}$$,

C
$$CrO_{2}^{-}$$,$$ClO_{3}^{-}$$,$$MnO_{4}^{-}$$,$$CrO_{4}^{2-}$$,

D
$$CrO_{4}^{2-}$$,$$MnO_{4}^{-}$$,$$CrO_{2}^{-}$$,$$ClO_{3}^{-}$$,
Solution
Hint: In the oxides, the oxidation number of oxygen is $$-2$$
Correct Answer: Option A
Explanation:
- Let the oxidation number of the central atom be $$x$$
- $$CrO^{-}_{2}$$:
$$x\ +\ (2\times (-2))=-1$$
$$\therefore$$ $$x=+3$$
- $$CrO_{4}^{2-}$$:
$$x\ +\ (4\times (-2))=-2$$
$$\therefore$$ $$x=+6$$
- $$ClO^{-}_{3} $$:
$$x\ +\ (3\times (-2))=-1$$
$$\therefore$$ $$x = +5$$
- $$MnO_{4}^{-}$$:
$$ x + (4\times (-2))=-1$$
$$\therefore$$ $$x = +7$$

Hence, correct answer is $$CrO^{-}_{2}$$, $$ClO^{-}_{3} $$ , $$CrO_{4}^{2-}$$, $$MnO_{4}^{-}$$
Question 5
1 / -0

Which of the following is correct representation of a given molecular equation in ionic form?
$$K_2Cr_2O_7 + 7H_2SO_4 + 6FeSO_4 \rightarrow 3Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + K_2SO_4 + 7H_2O$$

A
$$Cr_2{ O }_{ 7 }^{ 2- } + 14H^+ + 6Fe^{2+} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$$

B
$$2K^+ + Cr_2{ O }_{ 7 }^{ 2- } + 7S{ O }_{ 4}^{ 2- } + 6Fe^{2+} \rightarrow 3Fe^{3+} + S{ O }_{ 4 }^{ 2- } + Cr^{3+} + K^+ + 7H_2O$$

C
$$Cr_2{ O }_{ 7 }^{ 2- } + 2K^+ + 7H^+ + 6Fe^{2+} \rightarrow 6Fe^{3+} + 6Cr^{3+} + K^+$$

D
$$Cr_2{ O }_{ 7 }^{ 2- } + 7H^+ +6Fe^{2+} \rightarrow 3Fe^{2+} + 2Cr^{3+} + 2K^+ + 7H_2O$$

Solution

Given reaction:
$$K_2Cr_2O_7 + 7H_2SO_4 + 6FeSO_4 \rightarrow 3Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + 7H_2O + K_2SO_4$$
In ionic form :
$$2K^+ + Cr_2O_7^{ 2- } + 14H^+ + 7SO_4^{ 2- } + 6Fe^{2+} + 6S{ O }_{ 4}^{ 2- } \rightarrow 6Fe^{3+} + 9SO_4^{ 2- } + 2Cr^{ 3+} + 3S{ O }_{ 4 }^{ 2- } + 2K^+ + SO_4^{ 2- } +7H_2O$$
OR
$$2K^+ + Cr_2O_7^{ 2- } + 14H^+ + 13SO_4^{ 2- } + 6Fe^{2+} \rightarrow 6Fe^{3+} + 13SO_4^{ 2- } + 2Cr^{ 3+} + 2K^+ +7H_2O$$
Eliminating the common species $$K^+$$ and $$SO_4^{ 2- }$$ from both sides of the reactions we get
$$Cr_2{ O }_{ 7 }^{ 2- } + 14H^+ + 6Fe^{2+} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$$
It is the ionic form of the given chemical reaction.
Question 6
1 / -0

In the reaction : $$I_2+2S_2{ O }_{ 3 }^{ 2- }\rightarrow 2I^- +S_4{ O }_{ 6 }^{ 2- }$$?

A
$$ I_2 $$ is reducing agent

B
$$ I_2 $$ is oxidising agent and $$ S_2{ O }_{ 3 }^{ 2- } $$ is reducing agent

C
$$ S_2{ O }_{ 3 }^{ 2- } $$ is oxidising agent

D
$$ I_2 $$ is reducing agent and $$ S_2{ O }_{ 3 }^{ 2- } $$ is oxidising agent

Solution

$$\overset { 0 }{ I }_2 \rightarrow 2\overset { 0 }{ I }^-$$
$$I_2$$ undergoes reduction hence acts as oxidising agents.
$${ S }_{ 2 }^{ +2 }{ O}_{ 6 }^{ 2- } \rightarrow { S }_{ 4}^{ +2.5 }{ O }_{ 6 }^{ 2- }$$
It undergoes oxidation hence acts as a reducing agent.
Question 7
1 / -0

What is the correct representation of reaction occurring when HCl is heated with $$MnO_2$$ ?

A
$$ Mn{ O }_{ 4}^{ - } + 5Cl^- + 8 H^+ \rightarrow Mn^{2+} + 5Cl^- + 5H_2O$$

B
$$ MnO_2 + 2Cl^- + 4H^+ \rightarrow Mn^{2+} + Cl_2 + 2H_2O$$

C
$$ 2MnO_2 + 4Cl^- + 8H^+ \rightarrow 2Mn^{2+} + 2Cl_2 + 4H_2O$$

D
$$ MnO_2 + 4HCl \rightarrow MnCl_4 + Cl_2 + H_2O $$

Solution

Reaction involved is:
$$MnO_2 + HCl \rightarrow MnCl_2 + Cl_2 + H_2O $$
Assign the oxidation states we get:
$$\overset {+4}{Mn}O_2 + H\overset {-1}{Cl} \rightarrow \overset {+2}{Mn}Cl_2 + \overset {0}{Cl_2}$$
thus electron change in oxidation of Cl and reduction of Mn are same-2 electrons each
balance Cl
$$\overset {+4}{Mn}O_2 + 4H\overset {-1}{Cl} \rightarrow \overset {+2}{Mn}Cl_2 + \overset {0}{Cl_2}$$
Balance O by adding $$H_2O$$
$$\overset {+4}{Mn}O_2 + 4H\overset {-1}{Cl} \rightarrow \overset {+2}{Mn}Cl_2 + \overset {0}{Cl_2}+2H_2O$$
$$\therefore$$ balanced chemical reaction is:
$$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O $$
The reaction can be represented in the ionic form as:
$$MnO_2 + 4Cl^- + 4H^+ \rightarrow Mn^{2+} +2Cl^-+ Cl_2 + 2H_2O$$
OR
$$MnO_2 + 2Cl^- + 4H^+ \rightarrow Mn^{2+} + Cl_2 + 2H_2O$$ is the correct representation of reaction occurring when HCl is heated with $$MnO_2$$
Question 8
1 / -0

What is the oxidation number of carbon in $$C_3O_2$$ (carbon suboxide )?

A
$$+4/3$$

B
$$+10/4$$

C
$$+2$$

D
$$+2/3$$

Solution

The structure of $$C_3O_2$$ is:
$$O=C=C=C=O$$
assigning the oxidation number each $$C$$ in the molecule by considering oxidation number of $$O=-2$$ we get oxidation state as:
$$O = \overset{+2}{C}= \overset{0}{C} = \overset{+2}{C} = O$$
In $$C_3O_2$$ , two atoms linked with oxygen atoms are present in +2 oxidation state and central carbon has zero oxidation state. So, the average oxidation state of $$C =$$ $$\dfrac{2+0+2}{3}= \dfrac{ +4}{3}$$
Question 9
1 / -0

The standard reduction potential for the half-cell reaction, $$Cl_2+2e^-\rightarrow 2Cl^-$$ will be:
$$Pt^{2+}+2Cl^-\rightarrow Pt+Cl_2, E^o_{cell}=-0.15$$V;
$$Pt^{2+}+2e^-\rightarrow Pt, E^o=1.20$$V

A
$$-1.35$$V

B
$$+1.35$$V

C
$$-1.05$$V

D
$$+1.05$$V

Solution

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

The given reaction is, $$Pt^{2+}+2Cl^{-}\rightarrow Pt+Cl_2,E^{0}_{cell}=-0.15V$$

On reversing the above reaction, $$Pt+Cl_2\rightarrow Pt^{2+}+2Cl^{-},E^{0}_{cell}=0.15V$$

Also given,$$Pt^{2+}+2e^{-}\rightarrow Pt,E^{0}_{cell}=1.20V$$

As we know the relation,

$$E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{anode}$$

So we get,

$$E^{0}_{Cl_2/2Cl^{-}}=0.15-(-1.20)=1.35V$$

Hence, the correct option is $$\text{B}$$
Question 10
1 / -0

Which of the following is the correct cell representation for the given cell reaction?
$$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$$

A
$$Zn|Zn^{2+}||H^+|H_2$$

B
$$Zn|Zn^{2+}||H^+, H_2|Pt$$

C
$$Zn|ZnSO_4||H_2SO_4|Zn$$

D
$$Zn|H_2SO_4||ZnSO_4|H_2$$

Solution

In the given reaction the oxidation state of $$Zn$$ is changing from 0 to +2 so it is undergoing oxidation. Whereas the oxidation state of $$H^{+}$$ is changing from +1 to 0 which implies it is undergoing reduction. And in cell representation oxidation is represented in left part and the reduction in right. It has to be remembered that $$H_2$$ electrode has Pt too in it.
So, the correct representation of the cell is,
$$Zn|Zn^{2+}||H^{+},H_2|Pt$$
Hence option (B) is correct

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Redox Reactions Test - 44

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Redox Reactions Test - 44

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Question 1

+2

+3

+4

+5

Solution

Question 2

$$2H_2O+2Na\rightarrow 2NaOH+H_2$$

$$6CO_2+12H_2O\rightarrow C_6H_12O_6+6H_2O+6O_2$$

$$2F_2+2H_2O\rightarrow 4H^++4F^-+O_2$$

$$P_4O_10+6H_2O\rightarrow 4H_3PO_4$$

Solution

Question 3

+4

+5

+6

+3

Solution

Question 4

$$CrO_{2}^{-}$$, $$ClO_{3}^{-}$$ ,$$CrO_{4}^{2-}$$, $$MnO_{4}^{-}$$,

$$ClO_{3}^{-}$$, $$CrO_{4}^{2-}$$, $$MnO_{4}^{-}$$, $$CrO_{2}^{-}$$,

$$CrO_{2}^{-}$$,$$ClO_{3}^{-}$$,$$MnO_{4}^{-}$$,$$CrO_{4}^{2-}$$,

$$CrO_{4}^{2-}$$,$$MnO_{4}^{-}$$,$$CrO_{2}^{-}$$,$$ClO_{3}^{-}$$,

Solution

Question 5

$$Cr_2{ O }_{ 7 }^{ 2- } + 14H^+ + 6Fe^{2+} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$$

$$2K^+ + Cr_2{ O }_{ 7 }^{ 2- } + 7S{ O }_{ 4}^{ 2- } + 6Fe^{2+} \rightarrow 3Fe^{3+} + S{ O }_{ 4 }^{ 2- } + Cr^{3+} + K^+ + 7H_2O$$

$$Cr_2{ O }_{ 7 }^{ 2- } + 2K^+ + 7H^+ + 6Fe^{2+} \rightarrow 6Fe^{3+} + 6Cr^{3+} + K^+$$

$$Cr_2{ O }_{ 7 }^{ 2- } + 7H^+ +6Fe^{2+} \rightarrow 3Fe^{2+} + 2Cr^{3+} + 2K^+ + 7H_2O$$

Solution

Question 6

$$ I_2 $$ is reducing agent

$$ I_2 $$ is oxidising agent and $$ S_2{ O }_{ 3 }^{ 2- } $$ is reducing agent

$$ S_2{ O }_{ 3 }^{ 2- } $$ is oxidising agent

$$ I_2 $$ is reducing agent and $$ S_2{ O }_{ 3 }^{ 2- } $$ is oxidising agent

Solution

Question 7

$$ Mn{ O }_{ 4}^{ - } + 5Cl^- + 8 H^+ \rightarrow Mn^{2+} + 5Cl^- + 5H_2O$$

$$ MnO_2 + 2Cl^- + 4H^+ \rightarrow Mn^{2+} + Cl_2 + 2H_2O$$

$$ 2MnO_2 + 4Cl^- + 8H^+ \rightarrow 2Mn^{2+} + 2Cl_2 + 4H_2O$$

$$ MnO_2 + 4HCl \rightarrow MnCl_4 + Cl_2 + H_2O $$

Solution

Question 8

$$+4/3$$

$$+10/4$$

$$+2$$

$$+2/3$$

Solution

Question 9

$$-1.35$$V

$$+1.35$$V

$$-1.05$$V

$$+1.05$$V

Solution

Question 10

$$Zn|Zn^{2+}||H^+|H_2$$

$$Zn|Zn^{2+}||H^+, H_2|Pt$$

$$Zn|ZnSO_4||H_2SO_4|Zn$$

$$Zn|H_2SO_4||ZnSO_4|H_2$$

Solution

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