JEE Chemistry Redox Reactions MCQs Set 19

Answer: (c) X + Y = Z

Explanation:
1. In a redox titration the total equivalents (milliequivalents) of the reducing agent equal those of the oxidising agent.
2. Both \( \ce{Fe^{2+}} \) and \( \ce{C2O4^{2-}} \) are oxidised, so their milliequivalents add up.
3. Hence \( X+Y=Z \).

Teacher's Note:
1. Equivalents of oxidant = sum of equivalents of all reductants.
2. Equivalents are additive, moles are not.

Answer: (a) \( \ce{Ba(MnO4)2 ->[H+] Mn^{2+}} \) ; M/5

Explanation:
1. FeS2: Fe +2 to +3 (1 e) and 2 S from -1 to +6 (14 e) gives 15 e, so M/15 is correct.
2. P2H4: balanced as \( 5P_2H_4 \to P_4H_2 + 6PH_3 \), 6 electrons lost per 5 molecules, so 5M/6 is correct.
3. Two Mn(VII): one goes to +2 (5 e) and one to +6 (1 e), total 6 e, so M/3 is correct.
4. Ba(MnO4)2 contains two Mn(VII), each going to Mn(II): 10 e per formula unit, so the equivalent mass is M/10, not M/5. This pair is incorrect.

Teacher's Note:
1. Count the electrons per formula unit, not per atom.
2. Ba(MnO4)2 has two permanganate ions, doubling the n-factor to 10.

Answer: (d) 0.24 g

Explanation:
1. Milliequivalents of oxalic acid = \( 35\times0.16=5.6 \).
2. MnO2 goes from +4 to +2 (2 e), so its equivalent mass = \( 87/2=43.5 \).
3. Mass = \( 5.6\times43.5\ \text{mg}=243.6 \) mg \( \approx0.24 \) g.

Teacher's Note:
1. Milliequivalents of oxidant = milliequivalents of reductant.
2. Remember mg versus g: meq x equivalent mass gives mg.

Answer: (b) 6.32 g

Explanation:
1. Equivalents = \( 4\times0.05=0.2 \).
2. In acidic medium, \( \ce{MnO4- -> Mn^{2+}} \) is a 5-electron change, so equivalent mass = \( 158/5=31.6 \).
3. Mass = \( 0.2\times31.6=6.32 \) g.

Teacher's Note:
1. The equivalent mass of KMnO4 depends on the medium: 158/5 in acid, 158/3 in neutral or weakly basic, 158/1 in strongly basic solution.

Answer: (b) \( \ce{X3(YZ4)3} \)

Explanation:
1. In \( \ce{X3(YZ4)3} \): X contributes \( 3\times3=+9 \); each \( \ce{YZ4} \) group has \( +5+4(-2)=-3 \), and three groups give -9.
2. The total is zero, so the formula is neutral and possible.
3. For XYZ: \( 3+5-2\neq0 \); for \( \ce{Y2(XZ3)2} \): \( 10+2(3-6)=4\neq0 \); for \( \ce{X3(Y4Z)2} \): \( 9+2(20-2)\neq0 \).

Teacher's Note:
1. Test each formula by adding the oxidation numbers; the sum must be zero for a neutral compound.

Answer: (d) \( \ce{KMnO4} \)

Explanation:
1. Mn in KMnO4 is +7, the maximum for Mn, so it can only be reduced.
2. Mn in K2MnO4 is +6 (can be oxidised or reduced); O in H2O2 is -1 (intermediate); S in SO2 is +4 (intermediate).

Teacher's Note:
1. An element in its highest oxidation state can only gain electrons.
2. Intermediate oxidation states can act as both oxidant and reductant.

Answer: (a) 5 ; 5/2 ; 5/3

Explanation:
1. One mole of KMnO4 accepts 5 mol of electrons in acidic medium.
2. FeSO4 loses 1 e per mole: 5 mol. H2C2O4 loses 2 e: 5/2 mol. FeC2O4 loses 3 e (1 from Fe, 2 from oxalate): 5/3 mol.

Teacher's Note:
1. Moles oxidised = 5 / (electrons lost per mole of reductant).
2. FeC2O4 has two oxidisable parts, so its n-factor is 3.

Answer: (c) \( \ce{H2S} \)

Explanation:
1. F2, KMnO4 and HNO3 are strong oxidising agents; with SO2 they oxidise it to sulfate or SO3 and are themselves reduced.
2. H2S is oxidised to sulfur by SO2: \( \ce{SO2 + 2H2S -> 3S + 2H2O} \), so H2S acts as the reducing agent.

Teacher's Note:
1. Here SO2 is the oxidant, which is its less common role.
2. Sulfur in SO2 (+4) is reduced to 0 in this reaction.

Answer: (c) \( \ce{BaC2 + N2 -> Ba(CN)2} \)

Explanation:
1. Disproportionation means the same element is simultaneously oxidised and reduced.
2. P4 goes to +1 and -3; mercury(I) goes to +2 and 0; chlorine goes to +1 and -1: all are disproportionations.
3. In \( \ce{BaC2 + N2 -> Ba(CN)2} \), carbon goes from -1 to +2 (oxidised) and nitrogen from 0 to -3 (reduced): two different elements, so it is not disproportionation.

Teacher's Note:
1. Look for one element appearing in two different oxidation states among the products.
2. Here the oxidised and reduced elements are different (C and N).

Answer: (a) a-S, b-R, c-Q, d-P

Explanation:
1. AlF3: Al = +3 (S).
2. P4O10: each P = +5 (R).
3. HClO4: \( 1+x-8=0 \), Cl = +7 (Q).
4. S4O6\(^{2-}\): \( 4x-12=-2 \), so \( x=+2.5 \) (P).

Teacher's Note:
1. Fractional oxidation numbers are averages, as in the tetrathionate ion.

Answer: (d) 3M/5

Explanation:
1. \( \ce{BrO3- + 5Br- + 6H+ -> 3Br2 + 3H2O} \).
2. 3 mol of Br2 are formed using 5 electrons, so the n-factor of Br2 is \( 5/3 \).
3. Equivalent weight = \( \frac{M}{5/3}=\frac{3M}{5} \).

Teacher's Note:
1. For a product formed by comproportionation, divide the electrons exchanged by the moles of that product formed.

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

Explanation:
1. Electrons accepted = \( 1.68\times10^{-3}\times6=10.08\times10^{-3} \) mol.
2. Electrons lost per mole of ABD = \( 10.08/3.36=3 \).
3. A changes from -n to \( -n+3=3-n \).

Teacher's Note:
1. Each \( \ce{Cr2O7^{2-}} \) takes 6 electrons.
2. The oxidation number rises by the number of electrons lost.

Answer: (b) \( \frac{2x_1-x_2}{6} \)

Explanation:
1. Each N goes from 0 to -3: for NH3, n-factor = 3, so \( y_1=x_1/3 \).
2. For N2 (two N atoms) n-factor = 6, so \( y_2=x_2/6 \).
3. \( y_1-y_2=\frac{x_1}{3}-\frac{x_2}{6}=\frac{2x_1-x_2}{6} \).

Teacher's Note:
1. N2 has two nitrogen atoms, so its n-factor is double that of one N.

Answer: (d) M/11

Explanation:
1. Cu: +1 to +2, 3 atoms lose 3 e.
2. P: -3 to +5, loses 8 e.
3. Total n-factor = 3 + 8 = 11, so equivalent weight = M/11.

Teacher's Note:
1. Add the electron losses of every atom that is oxidised.

Answer: (a) \( \ce{PbO2 + H2O -> PbO + H2O2} \)

Explanation:
1. Cl2 gives Cl- and ClO-; H2O2 gives H2O (O at -2) and O2 (O at 0); P4 gives phosphorus in two different states: all are disproportionations.
2. In \( \ce{PbO2 + H2O -> PbO + H2O2} \), Pb is reduced (+4 to +2) and O is oxidised (-2 to -1), but they are different elements, so it is not a disproportionation.

Teacher's Note:
1. Disproportionation needs ONE element to be both oxidised and reduced.

Answer: (c) \( \ce{As2S3 + HNO3 -> H3AsO4 + H2SO4 + NO} \)

Explanation:
1. In an intermolecular redox reaction, one substance is oxidised and a different substance is reduced.
2. In \( \ce{As2S3 + HNO3} \): As2S3 is oxidised (As +3 to +5, S -2 to +6) and HNO3 is reduced (N +5 to +2): two different substances.
3. The glyoxal and benzaldehyde reactions are disproportionations of one substance, and the decomposition of CrO5 involves a single compound.

Teacher's Note:
1. Intermolecular: oxidant and reductant are different species. Disproportionation and intramolecular redox involve only one species.

Answer: (c) \( \ce{A3(BC4)2} \)

Explanation:
1. For \( \ce{A3(BC4)2} \): \( 3(+2)+2[+5+4(-2)]=6+2(-3)=0 \).
2. The other options do not add up to zero.

Teacher's Note:
1. Sum of oxidation numbers in a neutral compound is zero.

Answer: (a) \( \ce{CrO2Cl2} \) and \( \ce{MnO4-} \)

Explanation:
1. \( \ce{CrO2Cl2} \): \( x-4-2=0 \), so Cr = +6, the maximum for Cr. \( \ce{MnO4-} \): Mn = +7, the maximum for Mn.
2. MnO2 (+4), MnO (+2), Fe(III) and Co(III) are all below their group maxima.

Teacher's Note:
1. The maximum oxidation state of a d-block element equals its number of valence (n-1)d and ns electrons, up to Mn(+7).

Answer: (d) \( 1.13\times10^{-3} \)

Explanation:
1. \( \ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O} \).
2. Moles of H2O2 = \( 1.2\times10^{-3}/34=3.53\times10^{-5} \); permanganate (\( 1.5\times10^{-3} \) mol) is in large excess, so H2O2 is limiting.
3. 1 mol H2O2 gives 1 mol O2: mass = \( 3.53\times10^{-5}\times32=1.13\times10^{-3} \) g.

Teacher's Note:
1. Always check which reagent is limiting.
2. Here every H2O2 molecule gives one O2 molecule.

Answer: (b) 0

Explanation:
1. Cr(II) to Cr(III) loses 1 e: \( 25\times0.1=2.5 \) mmol of electrons.
2. SeO2 present = \( 12.5\times0.05=0.625 \) mmol, each gaining \( (4-x) \) electrons: \( 0.625(4-x)=2.5 \).
3. \( 4-x=4 \), so \( x=0 \): elemental selenium.

Teacher's Note:
1. Equate millimoles of electrons lost to millimoles of electrons gained.

Answer: (b) 40%

Explanation:
1. Moles of KMnO4 = \( 0.6\times0.1=0.06 \), which accepts \( 0.06\times5=0.30 \) mol of electrons.
2. Only \( \ce{Fe^{2+}} \) in FeSO4 gives up electrons (1 e each), so FeSO4 = 0.30 mol.
3. Fe2(SO4)3 = \( 0.5-0.3=0.2 \) mol, i.e. \( \frac{0.2}{0.5}\times100=40\% \).

Teacher's Note:
1. Iron(III) cannot be oxidised further by permanganate, so only FeSO4 consumes it.

Answer: (d) M/28

Explanation:
1. Arsenic: +3 to +5, 2 atoms lose 4 e.
2. Sulfur: -2 to +6, 3 atoms lose \( 3\times8=24 \) e.
3. Total n-factor = 28, so the equivalent mass is M/28.

Teacher's Note:
1. Count the electrons lost by every atom of each element that changes oxidation state.