JEE Chemistry Redox Reactions MCQs Set 15

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Multiple Choice Questions

Question 1: The number of moles of \( \ce{KMnO4} \) needed to oxidise a mixture of 1 mol each of \( \ce{FeSO4} \) and \( \ce{FeC2O4} \) in acid is
(a) 4/5
(b) 3/4
(c) 5/4
(d) 5/3
Show Answer & Explanation

Answer: (a) 4/5

Explanation:
1. Electrons: \( \ce{FeSO4} \) = 1, \( \ce{FeC2O4} \) = 3; total 4.
2. Moles of \( \ce{KMnO4} \) = 4/5.

Teacher's Note:
1. Add the electrons lost by each component and divide by 5.

Question 2: A metal weighing \( x \) g forms \( y \) g of its oxide. The equivalent weight of the metal is
(a) \( \frac{xy}{x}\times 8 \)
(b) \( \frac{x}{y-x}\times 8 \)
(c) \( \frac{y-x}{x}\times 8 \)
(d) \( \frac{x}{y}\times 8 \)
Show Answer & Explanation

Answer: (b) \( \frac{x}{y-x}\times 8 \)

Explanation:
1. Mass of oxygen combined = \( y-x \) g.
2. Equivalent weight is the mass of metal that combines with 8 g of oxygen.
3. \( E=\frac{x}{y-x}\times 8 \).

Teacher's Note:
1. Equivalent weight is always defined per 8 g of oxygen, 1.008 g of hydrogen or 35.5 g of chlorine.
2. Write the mass ratio first and then scale it to 8 g of oxygen.

Question 3: A 5.0 cm3 solution of \( \ce{H2O2} \) liberates 0.508 g of iodine from acidified KI. The volume strength of the solution at STP is
(a) 1.12 V
(b) 8.96 V
(c) 2.24 V
(d) 4.48 V
Show Answer & Explanation

Answer: (d) 4.48 V

Explanation:
1. 0.508 g \( \ce{I2} \) = 2 mmol, so \( \ce{H2O2} \) = 2 mmol in 5 mL = 0.4 M.
2. Volume strength = 0.4 x 11.2 = 4.48 V.

Teacher's Note:
1. A 1 M hydrogen peroxide solution has volume strength 11.2 V.

Question 4: 25 mL of \( \ce{H2O2} \) solution, treated with excess acidified KI, liberates iodine needing 20 mL of 0.3 N hypo. Which are correct? (i) mass of \( \ce{H2O2} \) in 25 mL is 0.102 g; (ii) molarity is 0.12 M; (iii) mass in 1 L is 0.816 g; (iv) volume strength is 1.344
(a) (i), (ii) and (iv) only
(b) (i) and (iii) only
(c) (ii), (iii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Show Answer & Explanation

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Hypo = 20 x 0.3 = 6 meq, so \( \ce{H2O2} \) = 6 meq = 3 mmol = 0.102 g: (i) true.
2. 3 mmol/25 mL = 0.12 M: (ii) true.
3. Mass in 1 L = 0.102 x 40 = 4.08 g: (iii) false.
4. N = 0.24; volume strength = 0.24 x 5.6 = 1.344: (iv) true.

Teacher's Note:
1. Volume strength = 5.6 x normality.

Question 5: In alkaline medium, \( \ce{2KMnO4 + 2KOH -> 2K2MnO4 + H2O + [O]} \). The equivalent weight of \( \ce{KMnO4} \) is
(a) 63.2
(b) 126.4
(c) 31.6
(d) 158
Show Answer & Explanation

Answer: (d) 158

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

Teacher's Note:
1. One electron is gained per permanganate in strongly alkaline medium.

Question 6: To find the normality of an alkali solution you need to know
(a) the mass of solution, mass of alkali and its equivalent weight
(b) the mass of solution, volume of alkali and its equivalent weight
(c) the volume of solution, volume of the alkali and its formula weight
(d) the volume of solution, mass of alkali present and its equivalent weight
Show Answer & Explanation

Answer: (d) the volume of solution, mass of alkali present and its equivalent weight

Explanation:
1. \( N=\dfrac{\text{mass}}{E\times V(\text{L})} \).
2. So the solute mass, its equivalent weight and the solution volume are needed.

Teacher's Note:
1. Normality is defined per litre of solution, so volume of solution (not mass) is required.

Question 7: The volume of 1 M \( \ce{KMnO4} \) in acid that reacts with 5.08 g of potassium acid oxalate \( \ce{K2C2O4.3H2C2O4.4H2O} \) (molar mass 508) is
(a) 16 mL
(b) 64 mL
(c) 32 mL
(d) 8 mL
Show Answer & Explanation

Answer: (a) 16 mL

Explanation:
1. 5.08 g = 0.01 mol, which contains 4 oxalate units (1 + 3).
2. Each oxalate gives 2 electrons: 0.01 x 8 = 0.08 eq = 80 meq.
3. KMnO4 has n = 5, so 80/5 = 16 mmol = 16 mL of 1 M.

Teacher's Note:
1. Count all oxalate units in the double salt, including the one present as the potassium salt.

Question 8: 40 mL of 0.05 M sodium sesquicarbonate (\( \ce{Na2CO3.NaHCO3.2H2O} \)) is titrated with 0.05 M HCl. x mL is used with phenolphthalein and y mL with methyl orange. The value of (y - x) is
(a) 120 mL
(b) 30 mL
(c) 180 mL
(d) 80 mL
Show Answer & Explanation

Answer: (d) 80 mL

Explanation:
1. Salt = 2 mmol: contains 2 mmol carbonate and 2 mmol bicarbonate.
2. Phenolphthalein: carbonate to bicarbonate: 2 mmol = 40 mL (x).
3. Methyl orange: 2 + 2 + 2 = 6 mmol = 120 mL (y).
4. y - x = 80 mL.

Teacher's Note:
1. Phenolphthalein sees only half the neutralisation of carbonate; methyl orange sees all.

Question 9: In \( \ce{28NO3^- + 3As2S3 + 4H2O -> 6AsO4^{3-} + 28NO + 9SO4^{2-} + 8H+} \), the equivalent mass of \( \ce{As2S3} \) (molar mass M) is
(a) \( M/2 \)
(b) \( M/4 \)
(c) \( M/24 \)
(d) \( M/28 \)
Show Answer & Explanation

Answer: (d) \( M/28 \)

Explanation:
1. Electrons from \( \ce{NO3^-} \) (+5 to +2): 28 x 3 = 84 for 3 \( \ce{As2S3} \).
2. Per \( \ce{As2S3} \): 84/3 = 28, so equivalent mass = M/28.

Teacher's Note:
1. Total electrons transferred divided by the number of formula units of the substance gives its n-factor.

Question 10: The normality of the mixture of 100 mL of 0.2 M \( \ce{H2SO4} \) and 200 mL of 0.2 M HCl is
(a) 0.0267
(b) 1.0267
(c) 1.1670
(d) 0.2670
Show Answer & Explanation

Answer: (d) 0.2670

Explanation:
1. Acid 1: 100 x 0.2 x 2 = 40 meq; acid 2: 200 x 0.2 x 1 = 40 meq.
2. N = 80/300 = 0.267.

Teacher's Note:
1. Add milliequivalents and divide by total volume.

Question 11: 50 mL of 0.1 M solution of a salt (metal oxidation number +3) reacts with 25 mL of 0.1 M sodium sulphite (\( \ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e^-} \)). The new oxidation number of the metal is
(a) +1
(b) +3
(c) +2
(d) +4
Show Answer & Explanation

Answer: (c) +2

Explanation:
1. Sulphite: 25 x 0.1 x 2 = 5 meq.
2. Metal: 5 mmol gains 5 meq, 1 electron each.
3. New oxidation number = +3 - 1 = +2.

Teacher's Note:
1. The electrons lost by sulphite equal the electrons gained by the metal ions.

Question 12: The volume of water to be added to 0.4 dm3 of 0.25 N oxalic acid to make it exactly decinormal is
(a) 0.8 dm3
(b) 0.6 dm3
(c) 0.2 dm3
(d) 0.4 dm3
Show Answer & Explanation

Answer: (b) 0.6 dm3

Explanation:
1. Final volume = 0.4 x 0.25/0.1 = 1.0 dm3.
2. Water added = 1.0 - 0.4 = 0.6 dm3.

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

Question 13: Equivalent mass of \( \ce{Ba(MnO4)2} \) in acidic medium (M = molar mass) is
(a) \( M/2 \)
(b) \( M/10 \)
(c) \( M/5 \)
(d) \( M/6 \)
Show Answer & Explanation

Answer: (b) \( M/10 \)

Explanation:
1. Two permanganate ions each gain 5 electrons: n = 10.
2. Equivalent mass = M/10.

Teacher's Note:
1. Count electrons per formula unit of the salt.

Question 14: Which statements are correct? (i) 1 mol of \( \ce{MnO4^-} \) oxidises 5 mol of \( \ce{Fe^{2+}} \); (ii) 1 mol of \( \ce{Cr2O7^{2-}} \) oxidises 6 mol of \( \ce{Fe^{2+}} \); (iii) 1 mol of \( \ce{Cu2S} \) (to \( \ce{Cu^{2+}} \) and \( \ce{SO2} \)) is oxidised by 1.6 mol of \( \ce{MnO4^-} \); (iv) 1 mol of \( \ce{Cu2S} \) is oxidised by 1.33 mol of \( \ce{Cr2O7^{2-}} \)
(a) (i) and (ii) only
(b) (i), (ii) and (iii) only
(c) (i), (ii), (iii) and (iv)
(d) (iii) and (iv) only
Show Answer & Explanation

Answer: (c) (i), (ii), (iii) and (iv)

Explanation:
1. (i) and (ii) follow from n = 5, 6 against 1.
2. \( \ce{Cu2S} \) loses 2 (Cu) + 6 (S: -2 to +4) = 8 electrons.
3. (iii) 8/5 = 1.6 and (iv) 8/6 = 1.33, so all four are correct.

Teacher's Note:
1. Find the n-factor of the reductant first, then divide by the oxidant's n-factor.

Question 15: Which compound can act as both an oxidising agent and a reducing agent?
(a) \( \ce{HCN} \)
(b) \( \ce{HCOOH} \)
(c) \( \ce{HNO2} \)
(d) \( \ce{HI} \)
Show Answer & Explanation

Answer: (c) \( \ce{HNO2} \)

Explanation:
1. N in \( \ce{HNO2} \) is +3, between -3 and +5, so it can be oxidised or reduced.
2. HI (I at -1, lowest) is only a reductant.

Teacher's Note:
1. A species with the element in an intermediate oxidation state is amphoteric in redox behaviour.

Question 16: In \( \ce{2CuSO4 + 4KI -> Cu2I2 + I2 + 2K2SO4} \), the equivalent weight of \( \ce{CuSO4} \) is
(a) 79.75
(b) 159.5
(c) 329
(d) none of these
Show Answer & Explanation

Answer: (b) 159.5

Explanation:
1. Cu changes from +2 to +1, n = 1.
2. Equivalent weight = 159.5.

Teacher's Note:
1. Copper(II) is reduced by one electron to copper(I) in iodometry.

Question 17: 10 g of \( \ce{V2O5} \) is dissolved in acid and reduced to \( \ce{V^{2+}} \) by zinc. The number of moles of \( \ce{I2} \) (reduced to iodide) that the resulting solution can reduce when it is oxidised to \( \ce{VO^{2+}} \) is (V = 51, O = 16)
(a) 0.11
(b) 0.055
(c) 0.44
(d) 0.22
Show Answer & Explanation

Answer: (a) 0.11

Explanation:
1. 10 g \( \ce{V2O5} \) = 10/182 mol, giving 0.1099 mol V.
2. \( \ce{V^{2+} -> VO^{2+}} \) (+2 to +4): 2 electrons per V = 0.2198 mol of electrons.
3. Each \( \ce{I2} \) takes 2 electrons, so 0.11 mol.

Teacher's Note:
1. Track electrons: the amount of vanadium is fixed, and each vanadium gives two electrons.

Question 18: An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH needed to neutralise 10 mL of this solution is
(a) 10 mL
(b) 20 mL
(c) 4 mL
(d) 40 mL
Show Answer & Explanation

Answer: (d) 40 mL

Explanation:
1. 6.3 g = 0.1 eq in 0.25 L, N = 0.4.
2. 10 mL x 0.4 = 4 meq; volume of 0.1 N NaOH = 40 mL.

Teacher's Note:
1. Use \( N_1V_1=N_2V_2 \): 0.4 x 10 = 0.1 x V.

Question 19: A substance that takes part readily in both acid-base and oxidation-reduction reactions is
(a) \( \ce{Na2CO3} \)
(b) \( \ce{KOH} \)
(c) \( \ce{H2C2O4} \)
(d) \( \ce{KMnO4} \)
Show Answer & Explanation

Answer: (c) \( \ce{H2C2O4} \)

Explanation:
1. Oxalic acid is a dibasic acid (acid-base).
2. Its carbon (+3) can be oxidised to +4, so it is also a reducing agent.

Teacher's Note:
1. Sodium carbonate and KOH only show acid-base behaviour; permanganate is only an oxidant in titrations.

Question 20: Oxalic acid reacts with permanganate as \( \ce{5H2C2O4 + 2MnO4^- + 6H+ -> 2Mn^{2+} + 10CO2 + 8H2O} \). The volume of 0.0162 M \( \ce{KMnO4} \) needed for 25.0 mL of 0.022 M \( \ce{H2C2O4} \) is (mL)
(a) 84.4
(b) 13.6
(c) 33.8
(d) 18.5
Show Answer & Explanation

Answer: (b) 13.6

Explanation:
1. Oxalic acid = 25 x 0.022 = 0.55 mmol.
2. Permanganate = 0.55 x 2/5 = 0.22 mmol.
3. Volume = 0.22/0.0162 = 13.6 mL.

Teacher's Note:
1. Use the mole ratio from the balanced equation directly.

Question 21: In the titration of \( \ce{K2Cr2O7} \) (V1 mL, M1) against \( \ce{FeSO4} \) (V2 mL, M2) in acid, which relations are true? (i) \( 6M_1V_1=M_2V_2 \); (ii) \( M_1V_1=6M_2V_2 \); (iii) \( N_1V_1=N_2V_2 \)
(a) (i) and (iii) only
(b) (ii) only
(c) (ii) and (iii) only
(d) (i) only
Show Answer & Explanation

Answer: (a) (i) and (iii) only

Explanation:
1. Dichromate n = 6 and ferrous n = 1, so \( 6M_1V_1=M_2V_2 \).
2. The relation \( N_1V_1=N_2V_2 \) always holds at the end point.

Teacher's Note:
1. The normality relation is universal; for molarity multiply by n-factors.

Question 22: 1 mL of a \( \ce{KMnO4} \) solution reacts with 0.140 g of \( \ce{Fe^{2+}} \); 1 mL of \( \ce{KHC2O4.H2C2O4} \) solution reacts with 0.1 mL of the same \( \ce{KMnO4} \). How many mL of 0.20 M NaOH react with 1 mL of the oxalate solution (all three ionisable H+ counted)?
(a) 15/16 mL
(b) 13/16 mL
(c) 11/14 mL
(d) none of these
Show Answer & Explanation

Answer: (a) 15/16 mL

Explanation:
1. 0.140 g \( \ce{Fe^{2+}} \) = 2.5 meq, so 1 mL of KMnO4 = 2.5 meq; 0.1 mL = 0.25 meq.
2. The oxalate salt has two oxalate units (2 electrons each), n = 4: 0.25/4 = 0.0625 mmol.
3. It has 3 ionisable H+: 0.1875 meq base, volume = 0.1875/0.2 = 15/16 mL.

Teacher's Note:
1. Use the redox data to find the moles of the salt, then use its basicity for the acid-base step.

Question 23: The normality of '20 volume' \( \ce{H2O2} \) solution is about
(a) 2.0
(b) 2.5
(c) 3.0
(d) 3.5
Show Answer & Explanation

Answer: (d) 3.5

Explanation:
1. N = volume strength/5.6 = 20/5.6 = 3.57, about 3.5.

Teacher's Note:
1. Normality of hydrogen peroxide equals volume strength divided by 5.6.

Question 24: The equivalent weight of oxalic acid in \( \ce{H2C2O4 + Ca(OH)2 -> CaC2O4 + H2O} \) is
(a) 90
(b) 64
(c) 45
(d) 128
Show Answer & Explanation

Answer: (c) 45

Explanation:
1. Both protons are replaced: n = 2.
2. Equivalent weight = 90/2 = 45.

Teacher's Note:
1. A diprotic acid fully neutralised has n = 2.

Question 25: Which of the following is a balanced equation?
(a) \( \ce{6BiO3^- + 12H+ + 3Mn^{2+} -> 6Bi^{3+} + 6H2O + 3MnO4^-} \)
(b) \( \ce{2BiO3^- + 4H+ + Mn^{2+} -> 2Bi^{3+} + 2H2O + MnO4^-} \)
(c) \( \ce{5BiO3^- + 14H+ + 2Mn^{2+} -> 5Bi^{3+} + 7H2O + 2MnO4^-} \)
(d) \( \ce{5BiO3^- + 22H+ + Mn^{2+} -> 5Bi^{3+} + 7H2O + MnO4^-} \)
Show Answer & Explanation

Answer: (c) \( \ce{5BiO3^- + 14H+ + 2Mn^{2+} -> 5Bi^{3+} + 7H2O + 2MnO4^-} \)

Explanation:
1. Electrons: Bi(V) to Bi(III) gains 2; Mn(II) to Mn(VII) loses 5. So ratio Bi : Mn = 5 : 2.
2. Charge on the left: -5 + 14 + 4 = +13; on the right: +15 - 2 = +13. Atoms O and H also balance (15 + 0 = 7 + 8; 14 = 14).

Teacher's Note:
1. Balance the electrons first, then charge with H+, then H and O with water.

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