Practice MCQs for JEE Chemistry Redox Reactions
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Multiple Choice Questions
(a) (a), (c)
(b) (b), (d)
(c) (a), (b)
(d) (c), (d)
Show Answer & Explanation
Answer: (b) (b), (d)
Explanation:
1. Reducing action means loss of electrons giving O2: (b) and (d).
2. (a) and (c) are electron gain, so H2O2 is oxidising there.
Teacher's Note:
1. Look for O2 as the product.
(a) \( \ce{FeCl3} \) is the oxidant
(b) \( \ce{FeCl3} \) and \( \ce{H2S} \) are both oxidised
(c) \( \ce{FeCl3} \) is oxidised and \( \ce{H2S} \) is reduced
(d) \( \ce{H2S} \) is the oxidant
Show Answer & Explanation
Answer: (a) \( \ce{FeCl3} \) is the oxidant
Explanation:
1. Fe: +3 to +2 (reduced, FeCl3 is oxidant).
2. S: -2 to 0 (oxidised, H2S is reductant).
Teacher's Note:
1. The substance reduced is the oxidant.
(a) \( \ce{Mn^2+ + 5Fe^3+ + 4H2O -> MnO4^- + 5Fe^2+ + 8H^+} \)
(b) \( \ce{MnO4^- + 5Fe^2+ + 8H^+ -> Mn^2+ + 5Fe^3+ + 4H2O} \)
(c) \( \ce{MnO4^- + 5Fe^3+ + 8H^+ -> Mn^2+ + 5Fe^2+ + 4H2O} \)
(d) \( \ce{Mn^2+ + 5Fe^2+ + 4H2O -> MnO4^- + 5Fe^3+ + 8H^+} \)
Show Answer & Explanation
Answer: (b) \( \ce{MnO4^- + 5Fe^2+ + 8H^+ -> Mn^2+ + 5Fe^3+ + 4H2O} \)
Explanation:
1. Permanganate (E = 1.51 V) is a stronger oxidant than Fe^3+ (0.77 V), so it oxidises Fe^2+.
2. Reaction (b) is the spontaneous direction.
Teacher's Note:
1. The stronger oxidant reacts with the stronger reductant.
(a) 5 N
(b) 10 N
(c) 30 N
(d) 60 N
Show Answer & Explanation
Answer: (c) 30 N
Explanation:
1. N = M x n-factor = 5 x 6.
2. N = 30 N.
Teacher's Note:
1. The n-factor of dichromate is 6 (two Cr, 3 electrons each).
(a) 1.6
(b) 1.0
(c) 0.8
(d) 2.5
Show Answer & Explanation
Answer: (a) 1.6
Explanation:
1. Cu2S: 2Cu (+1 to +2) = 2 e, S (-2 to +4) = 6 e; n-factor 8.
2. 1 x 8 = n x 5, so n = 1.6.
Teacher's Note:
1. Add up all electrons lost by every changing atom in the formula.
(a) 100 mL
(b) 150 mL
(c) 197.33 mL
(d) 250 mL
Show Answer & Explanation
Answer: (c) 197.33 mL
Explanation:
1. n(NaOH) = 1.184/40 = 0.0296 mol.
2. V = 0.0296/0.15 = 0.19733 L = 197.33 mL.
Teacher's Note:
1. Volume = moles / molarity.
(a) 0.5679 g
(b) 0.6681 g
(c) 1.06 g
(d) 0.4827 g
Show Answer & Explanation
Answer: (b) 0.6681 g
Explanation:
1. Meq of Na2CO3 = 45.6 x 0.235 = 10.716; E = 53.
2. Pure mass = 10.716 x 53/1000 = 0.5679 g.
3. Sample mass = 0.5679 x 100/85 = 0.6681 g.
Teacher's Note:
1. Correct for purity at the end.
(a) \( \frac{6}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{4}{5} \)
(d) 1
Show Answer & Explanation
Answer: (a) \( \frac{6}{5} \)
Explanation:
1. Each mole of ferrous oxalate gives 3 mol of electrons: 6 in total.
2. n(KMnO4) = 6/5.
Teacher's Note:
1. Moles of KMnO4 = moles of electrons / 5.
Assertion–Reason Questions
Assertion (A): \( \ce{NH4NO3} \): the oxidation numbers of the two nitrogen atoms are not equal.
Reason (R): One N atom is present in the ammonium ion while the other is in the nitrate ion.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Ammonium N is -3 and nitrate N is +5.
2. The different ions explain the different states.
Teacher's Note:
1. Always split ionic compounds into their ions.
Assertion (A): The oxidation state of hydrogen is +1 in \( \ce{H2O} \) while -1 in \( \ce{CaH2} \).
Reason (R): \( \ce{CaH2} \) is a metal hydride, and in metal hydrides hydrogen is assigned the oxidation number -1.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Both statements are true; the rule about metal hydrides explains the assertion.
Teacher's Note:
1. Hydrogen takes -1 only when bonded to a less electronegative metal.
Assertion (A): The oxidation number of Ni in \( \ce{Ni(CO)4} \) is zero.
Reason (R): Nickel is bonded to the neutral ligand carbonyl.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. CO is neutral, so Ni = 0, and the neutral ligand is the explanation.
Teacher's Note:
1. Metal carbonyls have zero-valent metals.
Assertion (A): The oxidation number of S in \( \ce{HSO3^-} \) is +4.
Reason (R): Sulphur is in different oxidation states in different compounds.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. 1 + x - 6 = -1, x = +4, so A is true. R is true but does not explain the +4.
Teacher's Note:
1. The charge on the ion goes on the right-hand side.
Assertion (A): The oxidation number of Cr in \( \ce{CrO5} \) is +6.
Reason (R): In \( \ce{CrO5} \), four oxygen atoms are involved in peroxide linkage.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. CrO5 has one O^2- and two peroxide groups: x + 4(-1) + (-2) = 0, x = +6.
2. Peroxide linkage explains why Cr is +6 and not +10.
Teacher's Note:
1. The maximum state of Cr is +6.
Assertion (A): In the reaction \( \ce{H2S + Cl2 -> 2HCl + S} \), Cl has been oxidised to \( \ce{Cl^-} \) while \( \ce{S^2-} \) has been reduced to S.
Reason (R): In a reaction the element whose oxidation number decreases is reduced and the element whose oxidation number increases is oxidised.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Cl is reduced (0 to -1) and S is oxidised (-2 to 0), so A is false.
2. R is a true statement.
Teacher's Note:
1. Do not confuse the direction of the change.
Assertion (A): \( \ce{HNO2} \) may act as an oxidising agent as well as a reducing agent.
Reason (R): The oxidation number of nitrogen remains the same in all the compounds.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. N is +3, an intermediate state, so A is true.
2. R is false: N shows states from -3 to +5.
Teacher's Note:
1. Intermediate oxidation states give dual behaviour.
Assertion (A): In a redox reaction the oxidation number of the oxidant decreases while that of the reductant increases.
Reason (R): The oxidant gains electron(s) while the reductant loses electron(s).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Gain of electrons is a decrease in oxidation number.
2. R explains A.
Teacher's Note:
1. OIL RIG: oxidation is loss, reduction is gain.
Assertion (A): \( \ce{H2SO4} \) cannot act as a reducing agent.
Reason (R): Sulphur cannot increase its oxidation number beyond +6.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. S in H2SO4 is already +6, its maximum.
2. Hence H2SO4 cannot lose electrons.
Teacher's Note:
1. The highest state of an element equals its valence-shell electron count.
Assertion (A): When \( \ce{Cl2} \) reacts with concentrated NaOH it forms NaCl and \( \ce{NaClO3} \).
Reason (R): \( \ce{Cl2} \) is an oxidising agent.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Hot concentrated NaOH gives chloride and chlorate (disproportionation).
2. Cl2 is an oxidising agent but this does not explain the formation of both products.
Teacher's Note:
1. Cl2 acts as both oxidant and reductant here.
Assertion (A): \( \ce{MnO4^-} \) is always reduced to \( \ce{Mn^2+} \).
Reason (R): Decrease in oxidation number or gaining of electrons means reduction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. In neutral medium Mn goes to MnO2, in strong alkali to MnO4^2-, so A is false.
2. R is a correct definition.
Teacher's Note:
1. The product depends on the pH of the medium.
Assertion (A): \( \ce{KClO3 -> KClO4 + KCl} \) is a disproportionation reaction.
Reason (R): The reaction in which a substance is oxidised or reduced is known as disproportionation.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Cl +5 goes to +7 and -1, so A is true.
2. R is false: disproportionation needs the same substance to be oxidised AND reduced.
Teacher's Note:
1. The word 'and' makes the difference.
Assertion (A): Bromide ion is serving as a reducing agent in \( \ce{2MnO4^- + Br^- + H2O -> 2MnO2 + BrO3^- + 2OH^-} \).
Reason (R): The oxidation number of Br increases from -1 to +5.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Br: -1 to +5 loses 6 electrons, so it is the reductant.
2. The increase in oxidation number is the reason.
Teacher's Note:
1. The species whose oxidation number increases is the reductant.
Assertion (A): The equivalent weight of \( \ce{NH3} \) in \( \ce{N2 -> NH3} \) is 17/3 while that of \( \ce{N2} \) is 28/6.
Reason (R): Equivalent weight = molecular weight / number of electrons lost or gained per mole.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Both values follow from the definition given in R.
2. NH3: 3 e per mole, N2: 6 e per mole.
Teacher's Note:
1. Use the n-factor per mole of the species.
Assertion (A): In acidic medium, the equivalent weight of \( \ce{K2Cr2O7} \) is 294/6.
Reason (R): In acidic medium, \( \ce{Cr2O7^2-} \) is reduced to \( \ce{Cr^3+} \).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Reduction to Cr^3+ involves 6 electrons per dichromate ion.
2. This gives E = 294/6 = 49.
Teacher's Note:
1. Dichromate acts as an oxidant only in acidic medium.
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