MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers

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Redox Reactions Class 11 MCQs Questions with Answers

Solving the Redox Reactions Multiple Choice Questions of Class 11 Chemistry Chapter 8 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on Redox Reactions Class 11 with answers pave for a quick revision of the Chapter thereby helping you to enhance subject knowledge. Have a glance at the MCQ of Chapter 8 Chemistry Class 11 and cross-check your answers during preparation.

Question 1.
KMnO4 reacts with oxalic acid according to the equation 2MnO4 + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O 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 MC2H2O4
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 2.
Which of the following is a redox reaction?
(a) NaCl + KNO3 → NaNO3 + KCl
(b) Mg(OH)2 + 2NH4Cl → MgCl2 + 2NH4OH
(c) CaC2O4 + 2HCl → CaCl2 + H2C2O4
(d) 2Zn + 2AgCN → 2Ag + Zn(CN)2

Answer

Answer: (d) 2Zn + 2AgCN → 2Ag + Zn(CN)2
Explanation:
In all the three reaction there is no change in the oxidation states. These are simple ionic reactions.
But in 2Zn + 2AgCN → 2Ag + Zn(CN)2 there is a change in oxidation state. Ag gains electrons and Zn lose electrons therefore it is a redox reaction.


Question 3.
The reduction potential values of M, N and O are +2.46 V, -1.13 V, -3.13 V respectively. Which of the following orders is correct regarding their reducing property?
(a) O > N > M
(b) M > O > N
(c) M > N > O
(d) O > M > N

Answer

Answer: (d) O > M > N
Explanation:
Given Reduction Potential:
M → +2.46V
N → +1.13V
O → −3.13V
We know that the electrode which has more reduction potential is a good oxidizing agent and has least reducing power.
While the electrode which has less reduction potential, it has more reducing power.
Therefore, Order of reducing power is: – O > M > N


Question 4.
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 doesnt change so its not a redox reaction.


Question 5.
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 6.
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 7.
Which of the following statements regarding sulphur is incorrect ?
(a) S2 molecule is paramagnetic.
(b) The vapour at 200° C consists mostly of S8 rings.
(c) At 600°C the gas mainly consists of S2 molecules.
(d) The oxidation state of sulphur is never less than +4 in its compounds.

Answer

Answer: (d) The oxidation state of sulphur is never less than +4 in its compounds.
Explanation:
Oxidation state of oxygen family
Oxygen shows -2, +2 and -1
Oxidation states other elements show +2, +4 and +6 oxidation states
In H2S, the oxidation state of S is -2. Oxidation state of S lie between -2 to +6.
Option 1)
S2 molecule is paramagnetic.
This option is incorrect.
Option 2)
The vapour at 200° C consists mostly of S8 rings.
This option is incorrect.
Option 3)
At 600° C the gas mainly consists of S2 molecules.
This option is incorrect.
Option 4)
The oxidation state of sulphur is never less than +4 in its compounds.
This option is correct.


Question 8.
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 9.
CrO5 has structure as shown, The oxidation number of chromium in the compound is?
MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers 1
(a) +10
(b) +6
(c) +4
(d) +5

Answer

Answer: (b) +6
Explanation:
From the above structure we can observe that 4 oxygen atoms are linked by peroxide linkage. So there oxidation state is -1 as in peroxide.
One oxygen atom is attached normally so its oxidation state is -2. So oxidation state of Cr is x + 4(−1) + (−2) = 0, x = +6


Question 10.
Pure water is bad conductor of electricity because
(a) It has high boiling point
(b) It is almost unionised
(c) Its molecules are associated with H- bonds
(d) Its pH is 7 at 25°C

Answer

Answer: (b) It is almost unionised
Explanation:
Distilled water is a poor conductor of electricity because it does not contain any dissolved salts in it which can provide it ions to conduct electricity. Impurities in water get ionised to conduct electricity. Hence pure water cannot conduct electricity.


Question 11.
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 12.
The ionic mobility of alkali metal ions in aqueous solution is maximum for
(a) Li+
(b) Na+
(c) K+
(d) Rb+

Answer

Answer: (d) Rb+
Explanation:
The smaller is the ion, the more is hydration, the larger is size, lesser is the mobility.


Question 13.
Pure water is bad conductor of electricity because
(a) It has high boiling point
(b) It is almost unionised
(c) Its molecules are associated with H- bonds
(d) Its pH is 7 at 25°C

Answer

Answer: (b) It is almost unionised
Explanation:
Distilled water is a poor conductor of electricity because it does not contain any dissolved salts in it which can provide it ions to conduct electricity.
Impurities in water get ionised to conduct electricity. Hence pure water cannot conduct electricity.


Question 14.
The oxidation number of Fe in K4 [Fe (CN)6] is
(a) 3
(b) 4
(c) 2
(d) Zero

Answer

Answer: (c) 2
Explanation:
The oxidation number of Fe in K4 Fe (CN)6 can be calculated as follows,
Oxidation state of K = 1, CN = -1.
Let Oxidation state of Fe be x. so
4(+1) + x + 6(-1) = 0
Hence x = +2


Question 15.
A standard hydrogen electrode has zero electrode potential because
(a) Hydrogen is easiest to oxidise
(b) This electrode potential is assumed to be zero
(c) Hydrogen atom has only one electron
(d) Hydrogen is the lightest element

Answer

Answer: (b) This electrode potential is assumed to be zero
Explanation:
The electrode potential of a standard hydrogen electrode is arbitrarily assumed to be zero.


Question 16.
Burning of lime to give calcium oxide and carbon dioxide is
(a) An Oxidation Process
(b) A Reduction Process
(c) Disproportionation
(d) Decomposition.

Answer

Answer: (d) Decomposition.
Explanation:
Lime water formula is Calcium hydroxide (Ca(OH)2)
Ca(OH)2 in the presence of excess heat gives calcium oxide(CaO) , Carbon dioxide (CO2), and water(H2O).
In this process excess of heat is given and lime water breaks down in different compounds, therefore it undergoes Thermal decomposition reaction.


Question 17.
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 18.
The oxidation number of carbon in CH2 Cl2 is
(a) 0
(b) +2
(c) +3
(d) +5

Answer

Answer: (a) 0
Explanation:
The oxidation state of carbon in dichloromethane as x.
Also the charges on H and Cl are +1 and −1 respectively.
Therefore, CH2​Cl2 ​→ x + 2(+1) +2(−1) = 0
⇒ x = 0


Question 19.
The oxidation state of I in IPO4 is
(a) +1
(b) +3
(c) +5
(d) +7

Answer

Answer: (b) +3
Explanation:
Let oxidation state of iodine be x.
x − 3 = 0, x = +3,
Because PO43- has combined oxidation number −3.
Therefore, x − 3 = 0
∴ x = +3
Thus oxidation state of iodine is +3.


Question 20.
The relationship between electrode potentials and concentrations of the substances involved in half cell reaction is given by
(a) Habers process
(b) Hess Law
(c) Nernst Equation
(d) None of the Above

Answer

Answer: (c) Nernst Equation
Explanation:
The relationship between electrode potentials and concentrations of the substances involved in half cell reaction is given by Nernst Equation.
E = E°−(2.303RT)/(nF) log[Mn+]/[M]
Where
E = cell potential (V) under specific conditions
E°= cell potential at standard-state conditions
R = ideal gas constant = 8.314 J/mol-k
T = temperature (kelvin), which is generally 25C (298 K)
n = number of moles of electrons transferred in the balanced equation
F = Faradays constant, the charge on a mole of electrons = 95,484.56 C/mol
[M] and [Mn+] are molar concentrations of element and its cation resp.


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