JEE Chemistry Redox Reactions MCQs Set 17

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Chapter-wise Objective Questions: Redox Reactions

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Multiple Choice Questions

Question 1: When 25 g of \( \ce{Na2SO4} \) is dissolved in \( 10^{3} \) kg of solution, the concentration is
(a) 2.5 ppm
(b) 25 ppm
(c) 250 ppm
(d) 100 ppm
Show Answer & Explanation

Answer: (b) 25 ppm

Explanation:
1. \( 10^{3} \) kg = \( 10^{6} \) g.
2. ppm = 25/\( 10^{6} \) x \( 10^{6} \) = 25.

Teacher's Note:
1. ppm is grams of solute per million grams of solution.

Question 2: In \( \ce{Na2S2O3 + 4Cl2 + 5H2O -> Na2SO4 + H2SO4 + 8HCl} \), the equivalent weight of \( \ce{Na2S2O3} \) (molar mass M) is
(a) \( M/8 \)
(b) \( M/2 \)
(c) \( M/4 \)
(d) \( M/1 \)
Show Answer & Explanation

Answer: (a) \( M/8 \)

Explanation:
1. Each S goes from +2 to +6: 4 electrons; two S atoms: n = 8.
2. Equivalent weight = M/8.

Teacher's Note:
1. Thiosulphate has n = 8 with chlorine but n = 1 with iodine.

Question 3: The moles of \( \ce{KMnO4} \) needed to react completely with one mole of ferrous oxalate in acid are
(a) 3/5
(b) 2/5
(c) 4/5
(d) 1
Show Answer & Explanation

Answer: (a) 3/5

Explanation:
1. Electrons: 1 (Fe) + 2 (oxalate) = 3.
2. Moles of permanganate = 3/5.

Teacher's Note:
1. Divide the electrons lost by 5.

Question 4: 22.7 mL of N/10 \( \ce{Na2CO3} \) neutralises 10.2 mL of dilute \( \ce{H2SO4} \). The volume of water to be added to 400 mL of this acid to make it exactly N/10 is
(a) 90.2 mL
(b) 490.2 mL
(c) 290.2 mL
(d) 890.2 mL
Show Answer & Explanation

Answer: (b) 490.2 mL

Explanation:
1. N of acid = 22.7 x 0.1/10.2 = 0.2225.
2. Final volume = 400 x 0.2225/0.1 = 890.2 mL.
3. Water added = 890.2 - 400 = 490.2 mL.

Teacher's Note:
1. Use \( N_1V_1=N_2V_2 \) for dilution and subtract the initial volume.

Question 5: x mmol of \( \ce{KMnO4} \) react completely with y mmol of \( \ce{MnSO4} \) in the presence of fluoride ions to give \( \ce{MnF4} \) quantitatively. Then
(a) 4x = y
(b) x = y
(c) x > y
(d) x < y
Show Answer & Explanation

Answer: (d) x < y

Explanation:
1. Mn(VII) to Mn(IV) gains 3 electrons per ion; Mn(II) to Mn(IV) loses 2.
2. Equal electron exchange: \( 3x=2y \), so \( y=1.5x \).
3. Hence y is larger than x, i.e. x < y.

Teacher's Note:
1. Equate the total electrons gained and lost, not the numbers of ions.

Question 6: The rate of \( \ce{MnO4^- + 8H+ + 5Fe^{2+} -> Mn^{2+} + 5Fe^{3+} + 4H2O} \) is best followed colorimetrically by monitoring
(a) \( \ce{MnO4^-} \)
(b) \( \ce{Mn^{2+}} \)
(c) \( \ce{Fe^{2+}} \)
(d) \( \ce{Fe^{3+}} \)
Show Answer & Explanation

Answer: (a) \( \ce{MnO4^-} \)

Explanation:
1. Permanganate is intensely purple while the other ions are nearly colourless in dilute solution.

Teacher's Note:
1. Choose the species with the strongest visible absorbance.

Question 7: In strongly basic medium \( \ce{MnO4^-} \) is reduced to \( \ce{MnO4^{2-}} \). The n-factor and equivalent weight of \( \ce{KMnO4} \) (M = 158) are
(a) 1 and 158
(b) 5 and 31.6
(c) 3 and 52.7
(d) 2 and 79
Show Answer & Explanation

Answer: (a) 1 and 158

Explanation:
1. Mn goes from +7 to +6: n = 1.
2. Equivalent weight = 158.

Teacher's Note:
1. Permanganate has n = 5 in acid, 3 in neutral and 1 in strongly basic medium.

Question 8: The oxidation number of chromium in \( \ce{CrO5} \) (which has one oxo oxygen and two peroxo groups) is
(a) +10
(b) +5
(c) +4
(d) +6
Show Answer & Explanation

Answer: (d) +6

Explanation:
1. Peroxo O: -1 each (4 atoms) = -4; oxo O = -2.
2. Cr = +6.

Teacher's Note:
1. Chromium(VI) peroxide has no change in Cr oxidation state from dichromate.

Question 9: In the half-reaction \( \ce{2BrO3^- + 12H+ + 10e^- -> Br2 + 6H2O} \), the equivalent weight of \( \ce{KBrO3} \) (molar mass M) is
(a) \( M/2 \)
(b) \( M/5 \)
(c) \( M/6 \)
(d) \( M/4 \)
Show Answer & Explanation

Answer: (b) \( M/5 \)

Explanation:
1. Two bromate ions take 10 electrons, so one bromate takes 5 (Br goes from +5 to 0).
2. n = 5, so equivalent weight = M/5.

Teacher's Note:
1. Always count electrons per formula unit of the species whose equivalent weight is asked, not per balanced equation.

Question 10: In the unbalanced reaction \( \ce{P4S3 + NO3^- -> H3PO4 + SO4^{2-} + NO} \) (acidic medium), the number of moles of water needed per mole of \( \ce{P4S3} \) is
(a) 8
(b) 8/3
(c) 18
(d) 28
Show Answer & Explanation

Answer: (b) 8/3

Explanation:
1. Balanced: \( \ce{3P4S3 + 38NO3^- + 20H+ + 8H2O -> 12H3PO4 + 9SO4^{2-} + 38NO} \)
2. Water per mole of \( \ce{P4S3} \) = 8/3.

Teacher's Note:
1. Balance electrons (38 per 3 \( \ce{P4S3} \) for 114 total), then charge with H+ and O with water.

Question 11: The volume of \( \ce{O2} \) at STP liberated when 5.6 mL of '20 V' \( \ce{H2O2} \) decomposes is
(a) 224 mL
(b) 112 mL
(c) 56 mL
(d) 168 mL
Show Answer & Explanation

Answer: (b) 112 mL

Explanation:
1. 1 mL of 20 V solution releases 20 mL of oxygen.
2. 5.6 x 20 = 112 mL.

Teacher's Note:
1. Volume strength is the volume of oxygen at STP released from one volume of solution.

Question 12: One litre of solution contains 18.9 g of \( \ce{HNO3} \) and one litre of another contains 3.2 g of NaOH. The volume ratio in which the acid and base solutions must be mixed to give a neutral solution is
(a) 15 : 4
(b) 3 : 8
(c) 8 : 3
(d) 4 : 15
Show Answer & Explanation

Answer: (c) 8 : 3

Explanation:
1. N of acid = 18.9/63 = 0.3; N of base = 3.2/40 = 0.08.
2. Neutral: \( 0.3V_1=0.08V_2 \), so \( V_1:V_2=8:3 \).

Teacher's Note:
1. Neutralisation needs equal equivalents of acid and base, so \( N_1V_1=N_2V_2 \) fixes the volume ratio.
2. The more dilute base must be taken in the larger volume.

Question 13: The mass of wet NaOH containing 15% water needed to prepare 70 L of 0.5 N solution is
(a) 140 kg
(b) 16.5 kg
(c) 1.65 kg
(d) 1.4 kg
Show Answer & Explanation

Answer: (c) 1.65 kg

Explanation:
1. Pure NaOH needed = 70 x 0.5 = 35 eq x 40 g = 1400 g.
2. Wet mass = 1400/0.85 = 1647 g = 1.65 kg.

Teacher's Note:
1. Correct for the percentage of impurity or water by dividing by the purity fraction.

Question 14: The number of moles of oxalate ions oxidised by one mole of \( \ce{MnO4^-} \) in acidic medium is
(a) 3/5
(b) 2/5
(c) 5/3
(d) 5/2
Show Answer & Explanation

Answer: (d) 5/2

Explanation:
1. One \( \ce{MnO4^-} \) accepts 5 electrons; each oxalate gives 2.
2. Moles of oxalate = 5/2.

Teacher's Note:
1. The balanced equation 2MnO4- + 5C2O4(2-) shows the same 2 : 5 ratio.

Question 15: The oxidation state of chromium in the final product of the reaction between KI and acidified potassium dichromate is
(a) +4
(b) +3
(c) +2
(d) +6
Show Answer & Explanation

Answer: (b) +3

Explanation:
1. Dichromate is reduced to \( \ce{Cr^{3+}} \) in acid.

Teacher's Note:
1. Acidified dichromate always ends in Cr(III), the green ion.

Question 16: In \( \ce{xMnO4^- + yC2O4^{2-} + zH+ -> xMn^{2+} + 2yCO2 + (z/2)H2O} \), the values of x, y and z are respectively
(a) 5, 2, 16
(b) 2, 5, 8
(c) 2, 5, 16
(d) 5, 2, 8
Show Answer & Explanation

Answer: (c) 2, 5, 16

Explanation:
1. Electrons: Mn gains 5 and oxalate loses 2, so x : y = 2 : 5.
2. Charge balance: -x - 2y + z = 2x gives z = 3x + 2y = 6 + 10 = 16.

Teacher's Note:
1. Balance electrons first, then charge using H+.

Question 17: Pond water contains 1.62 mg of calcium bicarbonate per 100 mL. The lime, \( \ce{Ca(OH)2} \), needed to remove the hardness of 60 L is
(a) 0.222 g
(b) 0.444 g
(c) 4.44 g
(d) 2.22 g
Show Answer & Explanation

Answer: (b) 0.444 g

Explanation:
1. Calcium bicarbonate in 60 L = \( 1.62\times10^{-3}\times10\times60=0.972 \) g = 6 mmol (M = 162).
2. \( \ce{Ca(HCO3)2 + Ca(OH)2 -> 2CaCO3 + 2H2O} \): 1 : 1, so 6 mmol lime = 6 x 74 mg = 0.444 g.

Teacher's Note:
1. Clark's method removes temporary hardness by adding exactly the lime needed to convert bicarbonate to carbonate.

Question 18: If one mole of \( \ce{H2SO4} \) reacts with one mole of NaOH, the equivalent weight of \( \ce{H2SO4} \) in that reaction is
(a) 49
(b) 96
(c) 48
(d) 98
Show Answer & Explanation

Answer: (d) 98

Explanation:
1. \( \ce{H2SO4 + NaOH -> NaHSO4 + H2O} \): only one proton is replaced.
2. n = 1, so equivalent weight = 98/1 = 98.

Teacher's Note:
1. Equivalent weight depends on the reaction.
2. Here the acid acts as monobasic, so its equivalent weight equals its molar mass.

Question 19: The volume of 0.05 M \( \ce{Ca(OH)2} \) needed to convert 10 mL of 0.1 M \( \ce{H3PO4} \) completely into \( \ce{Ca(H2PO4)2} \) is
(a) 5 mL
(b) 10 mL
(c) 30 mL
(d) 20 mL
Show Answer & Explanation

Answer: (b) 10 mL

Explanation:
1. \( \ce{2H3PO4 + Ca(OH)2 -> Ca(H2PO4)2 + 2H2O} \)
2. 1 mmol acid needs 0.5 mmol base = 10 mL of 0.05 M.

Teacher's Note:
1. Only one proton per acid molecule is neutralised in this conversion.

Question 20: For the reaction \( \ce{2Fe + 3CdCl2 -> 2FeCl3 + 3Cd} \) as written, which statement is correct?
(a) Cd is the reducing agent
(b) Fe undergoes oxidation
(c) Fe is the oxidising agent
(d) Cd undergoes oxidation
Show Answer & Explanation

Answer: (b) Fe undergoes oxidation

Explanation:
1. Fe goes from 0 to +3: it is oxidised and is the reducing agent.
2. Cd(II) goes to Cd(0): it is reduced.

Teacher's Note:
1. The species that loses electrons is oxidised and acts as the reducing agent.

Question 21: An element A in compound ABD has oxidation number -n. It is oxidised by \( \ce{Cr2O7^{2-}} \) in acid. For \( 3.36\times10^{-3} \) mol of ABD, \( 1.68\times10^{-3} \) mol of \( \ce{K2Cr2O7} \) is used. The new oxidation number of A is
(a) \( n-3 \)
(b) \( 3 \)
(c) \( +n \)
(d) \( 3-n \)
Show Answer & Explanation

Answer: (d) \( 3-n \)

Explanation:
1. Electrons gained = \( 1.68\times10^{-3}\times 6=10.08\times10^{-3} \) mol.
2. Per mole of ABD = \( 10.08/3.36=3 \) electrons lost.
3. New oxidation number = \( -n+3=3-n \).

Teacher's Note:
1. Divide total electrons transferred by the moles of the substance oxidised to get electrons per formula unit, then add to the initial oxidation number.

Question 22: In \( \ce{H2{}^{18}O2 + O3 -> H2O + O2} \) (the oxygen of ozone is ordinary oxygen), radioactivity (heavy isotope) will appear in
(a) water
(b) oxygen
(c) both water and oxygen
(d) neither
Show Answer & Explanation

Answer: (b) oxygen

Explanation:
1. Hydrogen peroxide acts as a reductant here and its O-O bond is not broken.
2. The marked oxygen atoms therefore end up in the oxygen gas evolved.

Teacher's Note:
1. In peroxide oxidations the oxygen gas comes from the peroxide, not from water or the oxidant.

Question 23: After treating 1 L of hard water with washing soda, 10 g of insoluble \( \ce{CaCO3} \) is precipitated. The concentration of \( \ce{Ca^{2+}} \) in the sample was
(a) 0.2 M
(b) 0.4 M
(c) 0.1 M
(d) 0.3 M
Show Answer & Explanation

Answer: (c) 0.1 M

Explanation:
1. 10 g of \( \ce{CaCO3} \) = 0.1 mol.
2. All calcium precipitated, so \( \ce{[Ca^{2+}]} \) = 0.1 mol per litre = 0.1 M.

Teacher's Note:
1. Every mole of precipitate comes from one mole of calcium ions.

Question 24: A 500 g toothpaste sample contains 0.4 g fluoride. The concentration in ppm is
(a) 500
(b) 800
(c) 200
(d) 400
Show Answer & Explanation

Answer: (b) 800

Explanation:
1. ppm = 0.4/500 x \( 10^{6} \) = 800.

Teacher's Note:
1. ppm is parts per million by mass.

Question 25: A 0.2 g sample containing copper(II) was analysed iodometrically: \( \ce{2Cu^{2+} + 4I^- -> 2CuI + I2} \). If 20 mL of 0.1 M \( \ce{Na2S2O3} \) is required for the liberated iodine, the percentage of copper is
(a) 31.75%
(b) 63.5%
(c) 53%
(d) 37%
Show Answer & Explanation

Answer: (b) 63.5%

Explanation:
1. Hypo used = 2 mmol, equal to the iodine equivalents, which equal the \( \ce{Cu^{2+}} \) equivalents (n = 1).
2. Cu = 2 mmol = 0.127 g; % = 0.127/0.2 x 100 = 63.5.

Teacher's Note:
1. In iodometry, one mole of thiosulphate corresponds to one mole of electrons, so millimoles of hypo equal millimoles of the oxidising species (per electron).

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