JEE Chemistry Redox Reactions MCQs Set 16

Chemistry Objective Questions and Answers: Redox Reactions

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Multiple Choice Questions

Question 1: 3 g of an oxide of a metal is converted completely into the chloride and gives 5 g of chloride. The equivalent weight of the metal is
(a) 3.325
(b) 12
(c) 20
(d) 33.25
Show Answer & Explanation

Answer: (d) 33.25

Explanation:
1. Let the equivalent weight be E. Oxide: \( 3/(E+8) \) equivalents; chloride: \( 5/(E+35.5) \) equivalents.
2. Equal equivalents: \( 3(E+35.5)=5(E+8) \), so \( 2E=66.5 \) and E = 33.25.

Teacher's Note:
1. The same number of equivalents of metal is present in the oxide and in the chloride.
2. Replacing one equivalent of oxide (8 g) by one equivalent of chloride (35.5 g) increases the mass by 27.5 g.

Question 2: An ion exchange resin \( \ce{RH2} \) replaces \( \ce{Ca^{2+}} \): \( \ce{RH2 + Ca^{2+} -> RCa + 2H+} \). A 1.0 L hard water sample passed through the resin gives eluent of pH 2.0. The hardness as ppm of \( \ce{Ca^{2+}} \) is
(a) 125
(b) 50
(c) 200
(d) 100
Show Answer & Explanation

Answer: (c) 200

Explanation:
1. [H+] = 0.01 mol, so Ca2+ = 0.005 mol.
2. Mass = 0.005 x 40 = 0.2 g per litre = 200 mg/L = 200 ppm.

Teacher's Note:
1. Each calcium ion releases two protons.

Question 3: 10 mL of sulphuric acid solution (specific gravity 1.84) contains 98% acid by mass. The volume of 2 N NaOH needed to neutralise it is
(a) 18.4 mL
(b) 184 mL
(c) 92 mL
(d) 9.2 mL
Show Answer & Explanation

Answer: (b) 184 mL

Explanation:
1. Mass of acid = 10 x 1.84 x 0.98 = 18.03 g.
2. Equivalents = 18.03/49 = 0.368.
3. Volume = 368 meq/2 N = 184 mL.

Teacher's Note:
1. Convert percentage composition to equivalents before using \( N_1V_1=N_2V_2 \).

Question 4: The number of equivalents of Mg that must react to liberate \( 4N_A \) electrons (\( \ce{Mg -> Mg^{2+} + 2e^-} \)) is
(a) 4
(b) 8
(c) 1
(d) 2
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. One mole of electrons is one equivalent.
2. \( 4N_A \) electrons = 4 mol = 4 equivalents (2 mol of Mg).

Teacher's Note:
1. Equivalents equal moles of electrons exchanged.

Question 5: 125 mL of 63% (w/v) \( \ce{H2C2O4.2H2O} \) solution is mixed with 125 mL of 40% (w/v) NaOH. The resulting solution (ignoring hydrolysis) is
(a) strongly acidic
(b) alkaline
(c) acidic
(d) neutral
Show Answer & Explanation

Answer: (d) neutral

Explanation:
1. Oxalic acid: 125 x 0.63 = 78.75 g /126 = 0.625 mol = 1.25 eq.
2. NaOH: 125 x 0.40 = 50 g /40 = 1.25 eq.
3. Equivalents are equal, so the solution is neutral.

Teacher's Note:
1. Compare equivalents of acid and base to decide the nature of the final solution.

Question 6: In (I) \( \ce{5H2O2 + 2MnO4^- + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O} \) and (II) \( \ce{H2O2 + Ag2O -> 2Ag + H2O + O2} \), hydrogen peroxide acts as
(a) reducing in I as well as in II
(b) reducing in I and oxidising in II
(c) oxidising in I as well as in II
(d) oxidising in I and reducing in II
Show Answer & Explanation

Answer: (a) reducing in I as well as in II

Explanation:
1. In both, oxygen goes from -1 to 0 and is oxidised, so peroxide is the reductant.

Teacher's Note:
1. Evolution of oxygen from peroxide always means oxidation of peroxide.

Question 7: The volume of 3 M \( \ce{Ba(OH)2} \) needed to neutralise 120 mL of 1.5 M \( \ce{H3PO4} \) completely is
(a) 90 mL
(b) 30 mL
(c) 120 mL
(d) 60 mL
Show Answer & Explanation

Answer: (a) 90 mL

Explanation:
1. Acid: 120 x 1.5 x 3 = 540 meq.
2. Base: 3 M = 6 N; V = 540/6 = 90 mL.

Teacher's Note:
1. Convert both molarities to normalities before equating.

Question 8: A bottle of \( \ce{H3PO4} \) contains 70% acid, density 1.54 g cm-3. The volume needed to prepare 1 L of 1 N solution is
(a) 90 mL
(b) 45 mL
(c) 23 mL
(d) 30 mL
Show Answer & Explanation

Answer: (d) 30 mL

Explanation:
1. Equivalent weight = 98/3 = 32.67 g.
2. Mass of solution = 32.67/0.70 = 46.7 g; volume = 46.7/1.54 = 30.3 mL.

Teacher's Note:
1. First find the mass of pure acid for one equivalent, then correct for purity and density.

Question 9: 1.250 g of metal carbonate \( \ce{MCO3} \) is treated with 500 mL of 0.1 M HCl. The unreacted HCl needs 50.0 mL of 0.500 M NaOH. The metal M is
(a) Ca
(b) Sr
(c) Mg
(d) Ba
Show Answer & Explanation

Answer: (a) Ca

Explanation:
1. HCl taken = 50 mmol; excess = 25 mmol; used = 25 mmol.
2. Carbonate = 12.5 mmol, so molar mass = 1.250/0.0125 = 100.
3. M = 100 - 60 = 40, calcium.

Teacher's Note:
1. Back titration gives the acid used by the sample.

Question 10: The volume of water to be added to 1600 mL of 0.205 N solution to make it 0.2 N is
(a) 40 mL
(b) 100 mL
(c) 50 mL
(d) 20 mL
Show Answer & Explanation

Answer: (a) 40 mL

Explanation:
1. Final volume = 1600 x 0.205/0.2 = 1640 mL.
2. Water added = 40 mL.

Teacher's Note:
1. Dilution keeps equivalents constant.

Question 11: A river water sample gives no change with dilute HCl, a white precipitate with aqueous \( \ce{Na2CO3} \) and no change with \( \ce{Na2SO4} \). The sample likely contains
(a) \( \ce{Li+} \)
(b) \( \ce{Ba^{2+}} \)
(c) \( \ce{Cu^{2+}} \)
(d) \( \ce{Mg^{2+}} \)
Show Answer & Explanation

Answer: (d) \( \ce{Mg^{2+}} \)

Explanation:
1. Barium would precipitate with sulphate, copper is coloured, lithium carbonate is not precipitated from dilute solution.
2. Magnesium gives a white carbonate but a soluble sulphate.

Teacher's Note:
1. Use the solubility of sulphates and carbonates to identify the cation.

Question 12: One gram of an equimolar mixture of \( \ce{Na2CO3} \) and \( \ce{NaHCO3} \) is neutralised completely by 0.1 N HCl. The volume of acid needed is
(a) 198.4 mL
(b) 295.5 mL
(c) 15.78 mL
(d) 157.8 mL
Show Answer & Explanation

Answer: (d) 157.8 mL

Explanation:
1. Mass of one mole of each = 106 + 84 = 190 g, so 1 g has 1/190 mol of each.
2. Equivalents = (2 + 1)/190 = 0.01579.
3. Volume = 15.79 meq/0.1 = 157.9 mL.

Teacher's Note:
1. Carbonate consumes two equivalents of acid and bicarbonate one.

Question 13: The volume of 0.0154 M \( \ce{KMnO4} \) needed for 25.0 mL of 0.0208 M \( \ce{H2C2O4} \) (5 : 2 stoichiometry with permanganate) is
(a) 84.4 mL
(b) 13.5 mL
(c) 18.5 mL
(d) 33.8 mL
Show Answer & Explanation

Answer: (b) 13.5 mL

Explanation:
1. Oxalic acid = 0.52 mmol; permanganate = 0.52 x 2/5 = 0.208 mmol.
2. Volume = 0.208/0.0154 = 13.5 mL.

Teacher's Note:
1. Use the 5 : 2 mole ratio.

Question 14: Some '20 V' \( \ce{H2O2} \) is mixed with excess acidified KI; the iodine needs 200 mL of 0.1 N hypo. The volume of the \( \ce{H2O2} \) solution taken is
(a) 5.6 mL
(b) 37.2 mL
(c) 22.4 mL
(d) 11.2 mL
Show Answer & Explanation

Answer: (a) 5.6 mL

Explanation:
1. 20 V means N = 20/5.6 = 3.571.
2. Equivalents = 200 x 0.1 = 20 meq.
3. Volume = 20/3.571 = 5.6 mL.

Teacher's Note:
1. Normality of H2O2 = volume strength / 5.6.

Question 15: In \( \ce{2MnO4^- + 5C2O4^{2-} + 16H+ -> 2Mn^{2+} + 10CO2 + 8H2O} \), 20 mL of 0.1 M \( \ce{KMnO4} \) reacts quantitatively with
(a) 50 mL of 0.1 M oxalate
(b) 20 mL of 0.1 M oxalate
(c) 50 mL of 0.25 M oxalate
(d) 40 mL of 0.1 M oxalate
Show Answer & Explanation

Answer: (a) 50 mL of 0.1 M oxalate

Explanation:
1. Permanganate = 2 mmol, oxalate needed = 5 mmol.
2. Only 50 mL of 0.1 M gives 5 mmol.

Teacher's Note:
1. Use the 2 : 5 mole ratio.

Question 16: An oxide of a metal contains 40% oxygen by mass. The equivalent weight of the metal is
(a) 8
(b) 12
(c) 16
(d) 24
Show Answer & Explanation

Answer: (b) 12

Explanation:
1. Per 40 g of oxygen there are 60 g of metal.
2. Per 8 g of oxygen: 60 x 8/40 = 12 g.

Teacher's Note:
1. Equivalent weight is the mass of metal combining with 8 g of oxygen.

Question 17: 25 mL of 0.1 M solution of a stable transition metal cation Z reacts exactly with 25 mL of 0.04 M acidified \( \ce{KMnO4} \). The most likely change in oxidation state of Z is
(a) \( \ce{Z+ -> Z^{2+}} \)
(b) \( \ce{Z^{2+} -> Z^{3+}} \)
(c) \( \ce{Z^{3+} -> Z^{4+}} \)
(d) \( \ce{Z^{2+} -> Z^{4+}} \)
Show Answer & Explanation

Answer: (b) \( \ce{Z^{2+} -> Z^{3+}} \)

Explanation:
1. Z: 2.5 mmol; permanganate: 1 mmol x 5 = 5 meq.
2. Electrons per Z = 5/2.5 = 2, so the oxidation number rises by 2.
3. Among the options, only \( \ce{Z^{2+} -> Z^{4+}} \) changes by 2.

Teacher's Note:
1. Compute electrons per ion from the titration data and match with the options.

Question 18: Hydrazine (\( \ce{N2H4} \)) acts as a reducing agent (to \( \ce{N2} \)). The mass needed to prepare 100 mL of 2 N hydrazine solution is
(a) 6.4 g
(b) 1.6 g
(c) 0.8 g
(d) 3.2 g
Show Answer & Explanation

Answer: (b) 1.6 g

Explanation:
1. n = 4 (N from -2 to 0, two atoms), so equivalent weight = 32/4 = 8.
2. Mass = 2 x 0.1 x 8 = 1.6 g.

Teacher's Note:
1. Count total electrons lost per molecule to get n.

Question 19: The quantity of electricity needed to reduce 0.05 mol of \( \ce{MnO4^-} \) to \( \ce{Mn^{2+}} \) in acid is
(a) 0.15 F
(b) 0.01 F
(c) 0.25 F
(d) 0.05 F
Show Answer & Explanation

Answer: (c) 0.25 F

Explanation:
1. Each ion takes 5 electrons.
2. 0.05 x 5 = 0.25 mol of electrons = 0.25 F.

Teacher's Note:
1. One faraday carries one mole of electrons.

Question 20: An ion is reduced to the element when it absorbs \( 6\times10^{20} \) electrons. The number of equivalents of the ion is about
(a) 0.01
(b) 0.1
(c) 0.0001
(d) 0.001
Show Answer & Explanation

Answer: (d) 0.001

Explanation:
1. One equivalent corresponds to one mole of electrons, \( 6.022\times10^{23} \).
2. \( 6\times10^{20}/6.022\times10^{23}\approx 10^{-3} \) equivalents.

Teacher's Note:
1. Equivalents of any species equal the moles of electrons it exchanges.
2. Dividing the number of electrons by Avogadro's number converts to moles.

Question 21: Oxalic acid is titrated with KMnO4 in the presence of \( \ce{H2SO4} \). The titration gives unsatisfactory results with HCl because HCl
(a) oxidises oxalic acid to carbon dioxide and water
(b) gets oxidised by oxalic acid to chlorine
(c) furnishes H+ ions in addition to those from oxalic acid
(d) reduces permanganate to Mn2+
Show Answer & Explanation

Answer: (d) reduces permanganate to Mn2+

Explanation:
1. Chloride is oxidised to chlorine by permanganate (HCl acts as a reductant).
2. Extra permanganate is consumed, so the volume at the end point is too high.

Teacher's Note:
1. Use only sulphuric acid for permanganate titrations, since sulphate cannot be oxidised.

Question 22: Battery acid (\( \ce{H2SO4} \), density 1.285 g cm-3), 10.0 cm3, is diluted to 1 L. 25.0 cm3 of the dilute solution needs 25.0 cm3 of 0.1 N NaOH. The percentage of acid by mass in the battery acid is
(a) 19
(b) 49
(c) 98
(d) 38
Show Answer & Explanation

Answer: (d) 38

Explanation:
1. The dilute solution is 0.1 N, so 1 L holds 0.1 eq, all from the 10 cm3 of battery acid.
2. Acid mass = 0.1 x 49 = 4.9 g.
3. Mass of 10 cm3 = 12.85 g; % = 4.9/12.85 = 38.

Teacher's Note:
1. Dilution preserves equivalents.

Question 23: 1 mol of hydroxide ions is obtained from 85 g of the hydroxide of a metal. The equivalent weight of the metal is
(a) 85
(b) 68
(c) 102
(d) 17
Show Answer & Explanation

Answer: (b) 68

Explanation:
1. 85 g hydroxide supplies 1 eq of \( \ce{OH^-} \) (17 g).
2. Metal part = 85 - 17 = 68 g per equivalent.

Teacher's Note:
1. Equivalent weight of the compound is the sum of the equivalent weights of its parts.

Question 24: The moles of CaO required to remove the hardness from 1000 L of water containing 324 ppm calcium bicarbonate and 74.5 ppm KCl is
(a) 3
(b) 1
(c) 4
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. 324 g \( \ce{Ca(HCO3)2} \) in 1000 L = 2 mol (M = 162).
2. \( \ce{Ca(HCO3)2 + CaO -> 2CaCO3 + H2O} \): 1 : 1, so 2 mol CaO.
3. KCl does not contribute to hardness.

Teacher's Note:
1. Only dissolved calcium and magnesium bicarbonates are removed by lime.

Question 25: 1.4 g of an organic compound is digested by Kjeldahl's method and the ammonia absorbed in 60 mL of M/10 \( \ce{H2SO4} \). The unreacted acid needs 20 mL of M/10 NaOH. The percentage of nitrogen is
(a) 5%
(b) 10%
(c) 6%
(d) 3%
Show Answer & Explanation

Answer: (b) 10%

Explanation:
1. Acid taken = 60 x 0.1 x 2 = 12 meq; NaOH = 2 meq.
2. Acid used for ammonia = 10 meq, so N = 10 mmol = 140 mg.
3. % N = 0.14/1.4 x 100 = 10%.

Teacher's Note:
1. Remember that sulphuric acid has two equivalents per mole.

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