CBSE Class 11 Chemistry – MCQ and Online Tests – Unit 8 – Redox Reactions
Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.
These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.
CBSE Class 11 Chemistry – MCQ and Online Tests – Unit 8 – Redox Reactions
Question 1.
What is n-factor?
(a) Equal to product of Number of moles of electrons when Lost or gained by one mole of reductant or oxidant
(b) When Number of moles of electrons Lost or gained by one mole of reductant or oxidant is not same.
(c) Equal to Number of moles of electrons Lost or gained by one mole of reductant or oxidant
(d) None of the above
Answer
Answer: (c) Equal to Number of moles of electrons Lost or gained by one mole of reductant or oxidant.
Explanation:
For redox reaction it is considered as change in their oxidation number or change in their reduction number in both side of a chemical reaction.
Question 2.
The oxidation number of Mn is maximum in
(a) MnO2
(b) K2MnO4
(c) Mn3O4
(d) KMnO4.
Answer
Answer: (d) KMnO4.
Explanation:
The electronic configuration of Mn is:
Mn(25) = [Ar]3d5 4s2, 4p0
In excited state, it can lose its all 7 electrons.
Hence, maximum oxidation sate exhibited by Mn is +7 which is in KMnO4.
Question 3.
The tendency of an electrode to lose electrons is known as
(a) Electrode Potential
(b) Reduction Potential
(c) Oxidation Potential
(d) E.M.F.
Answer
Answer: (c) Oxidation Potential
Explanation:
The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons. i.e., it is a measure of the relative tendency to undergo oxidation (loss of electrons) or reduction (gain of electrons).
M ? Mn+ + ne– (oxidation potential)
Mn+ + ne– ? M (reduction potential)
Question 4.
If equal volumes of 1M KMnO4 and 1M K2Cr2O7 solutions are allowed to oxidize Fe2+ in acidic medium. The amount of iron oxidized will be:
(a) More with KMnO2
(b) More with K2Cr2O7
(c) Equal with both oxidising agents
(d) Cannot be determined
Answer
Answer: (b) More with K2Cr2O7
Explanation:
The reason due to which the amount of Fe oxidised will be more with ?K2Cr2O7 is:
the change in the oxidation state (or number) or n factor is greater with KMnO4
Also, ?K2Cr2O7 is a very strong oxidising agent and holds the ability to take the electrons but ?KMnO4 is more stronger than ?K2Cr2O7.
Question 5.
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
Answer
Answer: (a) Formation of slaked lime from quick lime
Explanation:
Here, in this reaction
CaO + H2?O ?Ca(OH)2
Oxidation number doesn’t change so its not a redox reaction.
Question 6.
One mole of N2H4 loses ten moles of electrons to form a new compound A. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in A? (There is no change in the oxidation state of hydrogen.)
(a) -1
(b) -3
(c) +3
(d) +5
Answer
Answer: (c) +3
Explanation:
First to find oxidation number of Nitrogen in N2H4
Oxidation number of H = +1
Let oxidation number of nitrogen be x
2x + 4(1) = 0
2x = -4
x = -2
Each nitrogen atom has -2 oxidation number. So taken both nitrogen atoms in account gives oxidation number -4.
Change in oxidation number of nitrogen on losing 10 mol of electrons (considering no change in oxidation number of hydrogen atoms)
-4 – (-10) = +6
Therefore, oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, oxidation number of each nitrogen atom will be +3 in new compound Y.
Question 7.
How many millilitres of 0.5 M H2SO4 are needed to dissolve 0.5 g of copper(II)carbonate?
(a) 6.01
(b) 4.5
(c) 8.1
(d) 11.1
Answer
Answer: (c) 8.1
Explanation:
The volume can be calculated :
N1V1 = N2V2
N1 = Normality of H2SO4 = 0.5 × 2 = 1 N
V1 = Volume of H2SO4
Molar mass of copper(II) carbonate = 123.5 g
N2 = Normality of copper (II) carbonate = (0.5×2)/(123.5) N
V2 = Volume of copper (II) carbonate = 1000 mL
So, after applying the formula,
1 × V1 = (0.5×2)/(123.5)×1000
Hence, V1 = 8.09 mL
= approx. 8.1 mL
Question 8.
The oxidation state of Cr in Cr (CO)6 is
(a) 0
(b) 2
(c) 2
(d) 6
Answer
Answer: (a) 0
Explanation:
CO (carbonyl) is a neutral ligand, hence oxidation state of Cr in Cr (CO)6 is zero
Question 9.
Which of the following processes does not involve oxidation of iron?
(a) Formation of Fe(CO)5 from Fe.
(b) Liberation of H2 from steam by iron at high temperature.
(c) Rusting of iron sheets.
(d) Decolourisation of blue CuSO4 solution by iron.
Answer
Answer: (a) Formation of Fe(CO)5 from Fe.
Explanation:
Oxidation number of Fe in Fe(CO)5 is zero.
In both Fe and Fe(CO)5, the oxidation state of iron is zero.
3Fe + 4H2O ? Fe3O4 + 4H2
team
rusting
Fe ? Fe2O3.xH2O
(+3)
CuSO4(aq) + Fe (s) ? FeSO4(aq) + Cu(s)
(0) (+2)
Question 10.
The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is
(a) One
(b) Two
(c) Five
(d) One fifth.
Answer
Answer: (b) Two
Explanation:
In alkaline medium the reduction of KMnO4 with KI will takes place as
2 KMnO4 + H2O ? 2 KOH + 2 MnO2
KI + 3[O] ? KIO3
Hence the overall reaction is
KI + 2KMnO4 + H2O ? KIO3 + 2 KOH + 2 MnO2
So, one mole of KI will reduced two moles of KMnO4
Question 11.
Which of the following reactions does not involve either oxidation or reduction?
(a) VO2+ ? V2O3
(b) Na ? Na+
(c) CrO2-4 ? Cr2O2-7
(d) Zn2+ ?Zn
Answer
Answer: (c) CrO2-4 ? Cr2O2-7
Explanation:
In VO2+ ?V2?O3, V is reduced from +4 to +3 oxidation state.
In Na ? Na2+, Na is oxidised from to +1 oxidation state.
In CrO4-2? ? Cr2O7-2, Cr remains in same oxidation state +6.
In Zn+2 ? Zn, Zn is reduced from +2 to 0 oxidation state.
Question 12.
The oxidation number of Cl in Cl2O7 is
(a) + 7
(b) + 5
(c) + 3
(d) – 7
Answer
Answer: (a) + 7
Explanation:
Cl show different oxidation state as -1 to +7 due to vacant d orbital. As oxygen is more electronegative than Cl. Oxygen size is small hence its more electronegative and show -2 oxidation states.
Here Cl2O7 then equation is: 2x + 7 × (-2) = 0
x = +7 hence oxidation state of Cl is +7. I think you get your answer how to find oxidation state.
Question 13.
What is known as Autooxidation?
(a) Formation of H2O by the oxidation of H2O2.
(b) Formation of H2O2 by the oxidation of H2O.
(c) Both (1) and (2) are true
(d) None of the above
Answer
Answer: (b) Formation of H2O2 by the oxidation of H2O.
Explanation:
Autoxidation is any oxidation that occurs in presence of oxygen. The term is usually used to describe the degradation of organic compounds in air (as a source of oxygen). Autoxidation produces hydroperoxides and cyclic organic peroxides. These species can react further to form many products. The process is relevant to many phenomena including aging, paint, and spoilage of foods, degradation of petrochemicals, and the industrial production of chemicals. Autoxidation is important because it is a useful reaction for converting compounds to oxygenated derivatives, and also because it occurs in situations where it is not desired (as in the destructive cracking of the rubber in automobile tires or in rancidification).
Water automatically gets oxidised to hydrogen peroxide.
Question 14.
The colourless solution of silver nitrate slowly turns blue on adding copper chips to it because of
(a) Dissolution of Copper
(b) Oxidation of Ag+ ? Ag
(c) Reduction of Cu2+ ions
(d) Oxidation of Cu atoms.
Answer
Answer: (d) Oxidation of Cu atoms.
Explanation:
When copper turnings are added to silver nitrate solution, the solution becomes brown in color after sometime because copper is more reactive than silver so it displaces silver from silver nitrate solution and form copper nitrate solution.
Question 15.
A standard reduction electrode potentials of four metals are A = -0.250 V, B = -0.140 V, C = -0.126 V, D = -0.402 V The metal that displaces A from its aqueous solution is:
(a) A
(b) B
(c) C
(d) D
Answer
Answer: (d) D
Explanation:
Reduction potential of D is minimum i.e. -0.402 V. Thus oxidation potential of D is maximum i.e. to +0.402 V. D can oxidise itself and reduce other.
The aqueous solution A will be present in its ionic form and can be reduced by D as its reduction potential is higher than D.
Thus D can replace A from its Aqueous solution.
Question 16:
The oxidation process involves
(a) Increase in oxidation number
(b) Decrease in oxidation number
(c) No change in oxidation number
(d) none of the above
Answer
Answer: (a) Increase in oxidation number
Explanation:
Oxidation process Involves:-
Addition of O2 or electronegative element
Removal of H/electropositive element
Loss of electrons
Increase in oxidation number
Question 17.
Metals generally react with dilute acids to produce hydrogen gas. Which one of the following metals does not react with dilute hydrochloric acid?
(a) Copper
(b) Magnesium
(c) Iron
(d) Silver
Answer
Answer: (b) Magnesium
Explanation:
Most of the metals such as Al, Cu, Fe etc. reacts with dilute acids to produce hydrogen gas but magnesium is an exception. Magnesium being an active metal liberates dihydrogen gas as it is allowed to react with dilute HCl. Thus all the given metals react with dilute acids.
Mg + 2HCl ? MgCl2 + H2
Question 18.
The oxidation number of Xe in BaXeO6 is
(a) 8
(b) 6
(c) 4
(d) 10
Answer
Answer: (d) 10
Explanation:
Oxidation state of Ba in general = +2 and of O = -2
Applying formula, Sum of total oxidation state of all atoms = Overall charge on the compound.
Let oxidation state of Xe in BaXeO6 be x.
2 + x + 6(-2) = 0,
x = 10
But oxidation state 10 is not possible for Xe. In this case the oxidation state of Xe is equal to maximum possible oxidation state for Xe = +8.
Question 19.
KMnO4? reacts with oxalic acid according to the equation 2MnO4–? + 5C2?O4?2-? + 16H+ ? 2Mn2+? +10CO2? + 8H2?O Here 20 mL of 0.1 M KMnO4? is equivalent to
(a) 50 mL of 0.5 M C2H2O4
(b) 20 mL of 0.1 M C2H2O4
(c) 20 mL of 0.5 M C2H2O4
(d) 50 mL of 0.1 M C2H2O4
Answer
Answer: (d) 50 mL of 0.1 M C2H2O4
Explanation:
2MnO4– + 5C2O42- + 16H+ ? 2Mn2+ + 10CO2 + 8H2O
Therefore, 2 moles of MNO4– equivalent to 5 moles of C2O42-
20 mL of 0.1 M KMnO4 = 2 moles of KMnO4
Also, 50 mL of 0.1 M C2H2O4 equivalent to 5 mol of C2O42-
Therefore, these are equivalent.
Question 20.
One mole of N2H4 loses ten moles of electrons to form a new compound A. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in A? (There is no change in the oxidation state of hydrogen.)
(a) -1
(b) -3
(c) +3
(d) +5
Answer
Answer: (c) +3
Explanation:
First to find oxidation number of Nitrogen in N2H4
Oxidation number of H = +1
Let oxidation number of nitrogen be x
2x + 4(1) = 0
2x = -4
x = -2
Each nitrogen atom has -2 oxidation number. So taken both nitrogen atoms in account gives oxidation number -4.
Change in oxidation number of nitrogen on losing 10 mol of electrons (considering no change in oxidation number of hydrogen atoms)
-4 – (-10) = +6
Therefore, oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, oxidation number of each nitrogen atom will be +3 in new compound Y.
0 comments:
Post a Comment