{"id":17596,"date":"2022-05-24T08:00:43","date_gmt":"2022-05-24T02:30:43","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17596"},"modified":"2022-05-16T15:23:51","modified_gmt":"2022-05-16T09:53:51","slug":"mcq-questions-for-class-11-chemistry-chapter-8","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-chemistry-chapter-8\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers"},"content":{"rendered":"

Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers<\/a> are prepared as per the Latest Exam Pattern. Students can solve these Redox Reactions Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

Redox Reactions Class 11 MCQs Questions with Answers<\/h2>\n

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.<\/p>\n

Question 1.
\nKMnO4 reacts with oxalic acid according to the equation 2MnO4<\/sub>–<\/sup> + 5C2<\/sub>O4<\/sub>2-<\/sup> + 16H+ \u2192 2Mn2+<\/sup> + 10CO2<\/sub> + 8H2<\/sub>O Here 20 mL of 0.1 M KMnO4<\/sub> is equivalent to
\n(a) 50 mL of 0.5 M C2<\/sub>H2<\/sub>O4<\/sub>
\n(b) 20 mL of 0.1 M C2<\/sub>H2<\/sub>O4<\/sub>
\n(c) 20 mL of 0.5 M C2<\/sub>H2<\/sub>O4<\/sub>
\n(d) 50 mL of 0.1 M C2<\/sub>H2<\/sub>O4<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 50 mL of 0.1 MC2<\/sub>H2<\/sub>O4<\/sub>
\nExplanation:
\n2MnO4<\/sub>–<\/sup> + 5C2<\/sub>O4<\/sub>2-<\/sup> + 16H+ \u2192 2Mn2+<\/sup> + 10CO2<\/sub> + 8H2<\/sub>O
\nTherefore, 2 moles of MNO4<\/sub>–<\/sup> equivalent to 5 moles of C2<\/sub>O4<\/sub>2-<\/sup>
\n20 mL of 0.1 M KMnO4<\/sub> = 2 moles of KMnO4<\/sub>
\nAlso, 50 mL of 0.1 M C2<\/sub>H2<\/sub>O4<\/sub> equivalent to 5 mol of C2<\/sub>O4<\/sub>2-<\/sup>
\nTherefore, these are equivalent.<\/p>\n<\/details>\n

\n

Question 2.
\nWhich of the following is a redox reaction?
\n(a) NaCl + KNO3<\/sub> \u2192 NaNO3<\/sub> + KCl
\n(b) Mg(OH)2<\/sub> + 2NH4<\/sub>Cl \u2192 MgCl2<\/sub> + 2NH4<\/sub>OH
\n(c) CaC2<\/sub>O4<\/sub> + 2HCl \u2192 CaCl2<\/sub> + H2<\/sub>C2<\/sub>O4<\/sub>
\n(d) 2Zn + 2AgCN \u2192 2Ag + Zn(CN)2<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 2Zn + 2AgCN \u2192 2Ag + Zn(CN)2<\/sub>
\nExplanation:
\nIn all the three reaction there is no change in the oxidation states. These are simple ionic reactions.
\nBut in 2Zn + 2AgCN \u2192 2Ag + Zn(CN)2<\/sub> there is a change in oxidation state. Ag gains electrons and Zn lose electrons therefore it is a redox reaction.<\/p>\n<\/details>\n

\n

Question 3.
\nThe 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?
\n(a) O > N > M
\n(b) M > O > N
\n(c) M > N > O
\n(d) O > M > N<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) O > M > N
\nExplanation:
\nGiven Reduction Potential:
\nM \u2192 +2.46V
\nN \u2192 +1.13V
\nO \u2192 \u22123.13V
\nWe know that the electrode which has more reduction potential is a good oxidizing agent and has least reducing power.
\nWhile the electrode which has less reduction potential, it has more reducing power.
\nTherefore, Order of reducing power is: – O > M > N<\/p>\n<\/details>\n

\n

Question 4.
\nWhich of the following processes does not involve either oxidation or reduction?
\n(a) Formation of slaked lime from quick lime
\n(b) Heating Mercuric Oxide
\n(c) Formation of Manganese Chloride from Manganese oxide
\n(d) Formation of Zinc from Zinc blende<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Formation of slaked lime from quick lime
\nExplanation:
\nHere, in this reaction
\nCaO + H2<\/sub>\u200bO \u2192 Ca(OH)2<\/sub>\u200b
\nOxidation number doesnt change so its not a redox reaction.<\/p>\n<\/details>\n

\n

Question 5.
\nThe number of moles of KMnO4<\/sub> reduced by one mole of KI in alkaline medium is
\n(a) One
\n(b) Two
\n(c) Five
\n(d) One fifth.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Two
\nExplanation:
\nIn alkaline medium the reduction of KMnO4<\/sub> with KI will takes place as
\n2 KMnO4<\/sub> + H2<\/sub>O \u2192 2 KOH + 2 MnO2<\/sub>
\nKI + 3[O] \u2192 KIO3<\/sub>
\nHence the overall reaction is
\nKI + 2KMnO4<\/sub> + H2<\/sub>O \u2192 KIO3<\/sub> + 2 KOH + 2 MnO2<\/sub>
\nSo, one mole of KI will reduced two moles of KMnO4<\/sub><\/p>\n<\/details>\n

\n

Question 6.
\nWhat is known as Autooxidation?
\n(a) Formation of H2<\/sub>O by the oxidation of H2<\/sub>O2<\/sub>.
\n(b) Formation of H2<\/sub>O2<\/sub> by the oxidation of H2<\/sub>O.
\n(c) Both (1) and (2) are true
\n(d) None of the above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Formation of H2<\/sub>O2<\/sub> by the oxidation of H2<\/sub>O.
\nExplanation:
\nAutoxidation 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.<\/p>\n<\/details>\n

\n

Question 7.
\nWhich of the following statements regarding sulphur is incorrect ?
\n(a) S2<\/sub> molecule is paramagnetic.
\n(b) The vapour at 200\u00b0 C consists mostly of S8<\/sub> rings.
\n(c) At 600\u00b0C the gas mainly consists of S2<\/sub> molecules.
\n(d) The oxidation state of sulphur is never less than +4 in its compounds.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) The oxidation state of sulphur is never less than +4 in its compounds.
\nExplanation:
\nOxidation state of oxygen family
\nOxygen shows -2, +2 and -1
\nOxidation states other elements show +2, +4 and +6 oxidation states
\nIn H2<\/sub>S, the oxidation state of S is -2. Oxidation state of S lie between -2 to +6.
\nOption 1)
\nS2<\/sub> molecule is paramagnetic.
\nThis option is incorrect.
\nOption 2)
\nThe vapour at 200\u00b0 C consists mostly of S8<\/sub> rings.
\nThis option is incorrect.
\nOption 3)
\nAt 600\u00b0 C the gas mainly consists of S2<\/sub> molecules.
\nThis option is incorrect.
\nOption 4)
\nThe oxidation state of sulphur is never less than +4 in its compounds.
\nThis option is correct.<\/p>\n<\/details>\n

\n

Question 8.
\nThe oxidation number of Xe in BaXeO6<\/sub> is
\n(a) 8
\n(b) 6
\n(c) 4
\n(d) 10<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 10
\nExplanation:
\nOxidation state of Ba in general = +2 and of O = \u22122
\nApplying formula, Sum of total oxidation state of all atoms = Overall charge on the compound.
\nLet oxidation state of Xe in BaXeO6<\/sub> be x.
\n2 + x + 6(\u22122) = 0,
\nx = 10
\nBut 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.<\/p>\n<\/details>\n

\n

Question 9.
\nCrO5<\/sub> has structure as shown, The oxidation number of chromium in the compound is?
\n\"MCQ
\n(a) +10
\n(b) +6
\n(c) +4
\n(d) +5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) +6
\nExplanation:
\nFrom the above structure we can observe that 4 oxygen atoms are linked by peroxide linkage. So there oxidation state is -1 as in peroxide.
\nOne oxygen atom is attached normally so its oxidation state is -2. So oxidation state of Cr is x + 4(\u22121) + (\u22122) = 0, x = +6<\/p>\n<\/details>\n

\n

Question 10.
\nPure water is bad conductor of electricity because
\n(a) It has high boiling point
\n(b) It is almost unionised
\n(c) Its molecules are associated with H- bonds
\n(d) Its pH is 7 at 25\u00b0C<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) It is almost unionised
\nExplanation:
\nDistilled 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.<\/p>\n<\/details>\n

\n

Question 11.
\nThe oxidation process involves
\n(a) Increase in oxidation number
\n(b) Decrease in oxidation number
\n(c) No change in oxidation number
\n(d) none of the above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Increase in oxidation number
\nExplanation:
\nOxidation process Involves:-
\nAddition of O2<\/sub> or electronegative element
\nRemoval of H\/ electropositive element
\nLoss of electrons
\nIncrease in oxidation number<\/p>\n<\/details>\n

\n

Question 12.
\nThe ionic mobility of alkali metal ions in aqueous solution is maximum for
\n(a) Li+<\/sup>
\n(b) Na+<\/sup>
\n(c) K+<\/sup>
\n(d) Rb+<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) Rb+<\/sup>
\nExplanation:
\nThe smaller is the ion, the more is hydration, the larger is size, lesser is the mobility.<\/p>\n<\/details>\n

\n

Question 13.
\nPure water is bad conductor of electricity because
\n(a) It has high boiling point
\n(b) It is almost unionised
\n(c) Its molecules are associated with H- bonds
\n(d) Its pH is 7 at 25\u00b0C<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) It is almost unionised
\nExplanation:
\nDistilled 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.
\nImpurities in water get ionised to conduct electricity. Hence pure water cannot conduct electricity.<\/p>\n<\/details>\n

\n

Question 14.
\nThe oxidation number of Fe in K4<\/sub> [Fe (CN)6<\/sub>] is
\n(a) 3
\n(b) 4
\n(c) 2
\n(d) Zero<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 2
\nExplanation:
\nThe oxidation number of Fe in K4<\/sub> Fe (CN)6<\/sub> can be calculated as follows,
\nOxidation state of K = 1, CN = -1.
\nLet Oxidation state of Fe be x. so
\n4(+1) + x + 6(-1) = 0
\nHence x = +2<\/p>\n<\/details>\n

\n

Question 15.
\nA standard hydrogen electrode has zero electrode potential because
\n(a) Hydrogen is easiest to oxidise
\n(b) This electrode potential is assumed to be zero
\n(c) Hydrogen atom has only one electron
\n(d) Hydrogen is the lightest element<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) This electrode potential is assumed to be zero
\nExplanation:
\nThe electrode potential of a standard hydrogen electrode is arbitrarily assumed to be zero.<\/p>\n<\/details>\n

\n

Question 16.
\nBurning of lime to give calcium oxide and carbon dioxide is
\n(a) An Oxidation Process
\n(b) A Reduction Process
\n(c) Disproportionation
\n(d) Decomposition.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) Decomposition.
\nExplanation:
\nLime water formula is Calcium hydroxide (Ca(OH)2<\/sub>)
\nCa(OH)2<\/sub> in the presence of excess heat gives calcium oxide(CaO) , Carbon dioxide (CO2<\/sub>), and water(H2<\/sub>O).
\nIn this process excess of heat is given and lime water breaks down in different compounds, therefore it undergoes Thermal decomposition reaction.<\/p>\n<\/details>\n

\n

Question 17.
\nThe colourless solution of silver nitrate slowly turns blue on adding copper chips to it because of
\n(a) Dissolution of Copper
\n(b) Oxidation of Ag+<\/sup> \u2192 Ag
\n(c) Reduction of Cu2+<\/sup> ions
\n(d) Oxidation of Cu atoms.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) Oxidation of Cu atoms.
\nExplanation:
\nWhen 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.<\/p>\n<\/details>\n

\n

Question 18.
\nThe oxidation number of carbon in CH2<\/sub> Cl2<\/sub> is
\n(a) 0
\n(b) +2
\n(c) +3
\n(d) +5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0
\nExplanation:
\nThe oxidation state of carbon in dichloromethane as x.
\nAlso the charges on H and Cl are +1 and \u22121 respectively.
\nTherefore, CH2<\/sub>\u200bCl2<\/sub> \u200b\u2192 x + 2(+1) +2(\u22121) = 0
\n\u21d2 x = 0<\/p>\n<\/details>\n

\n

Question 19.
\nThe oxidation state of I in IPO4<\/sub> is
\n(a) +1
\n(b) +3
\n(c) +5
\n(d) +7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) +3
\nExplanation:
\nLet oxidation state of iodine be x.
\nx \u2212 3 = 0, x = +3,
\nBecause PO4<\/sub>3-<\/sup> has combined oxidation number \u22123.
\nTherefore, x \u2212 3 = 0
\n\u2234 x = +3
\nThus oxidation state of iodine is +3.<\/p>\n<\/details>\n

\n

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

\nAnswer<\/span><\/summary>\n

Answer: (c) Nernst Equation
\nExplanation:
\nThe relationship between electrode potentials and concentrations of the substances involved in half cell reaction is given by Nernst Equation.
\nE = E\u00b0\u2212(2.303RT)\/(nF) log[Mn+<\/sup>]\/[M]
\nWhere
\nE = cell potential (V) under specific conditions
\nE\u00b0= cell potential at standard-state conditions
\nR = ideal gas constant = 8.314 J\/mol-k<\/sup>
\nT = temperature (kelvin), which is generally 25C (298 K)
\nn = number of moles of electrons transferred in the balanced equation
\nF = Faradays constant, the charge on a mole of electrons = 95,484.56 C\/mol
\n[M] and [Mn+<\/sup>] are molar concentrations of element and its cation resp.<\/p>\n<\/details>\n

\n

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Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers are prepared as per the Latest Exam Pattern. Students can solve these Redox Reactions Class 11 MCQs Questions with Answers and assess their …<\/p>\n

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