JEE Chemistry Redox Reactions MCQs Set 12

Answer: (d) 2, 4

Explanation:
1. A reducing agent loses electrons and is oxidised.
2. In (1) and (3) \(\ce{H2O2}\) gains electrons, so it is an oxidant.
3. In (2) and (4) it loses electrons, forming \(\ce{O2}\) (oxygen goes from -1 to 0), so it is a reductant.

Teacher's Note:
1. Electrons on the left side of a half reaction mean reduction; electrons on the right mean oxidation.
2. Hydrogen peroxide can act both ways because oxygen is in the intermediate state -1.

Answer: (a) \(\ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}\)

Explanation:
1. Sulphuric acid is an oxidant when sulphur is reduced.
2. In (a) sulphur changes from +6 in \(\ce{H2SO4}\) to +4 in \(\ce{SO2}\), while iodide is oxidised to iodine.
3. The other reactions have no change in oxidation number.

Teacher's Note:
1. Look for a fall in oxidation number of sulphur.
2. Neutralisation and acid-displacement reactions are not redox reactions.

Answer: (c) 5, 1, 3, 4

Explanation:
1. Mn in \(\ce{MnO4^-}\) is +7.
2. Products: \(\ce{Mn^{2+}}\) is +2 (change 5); \(\ce{MnO4^{2-}}\) is +6 (change 1); \(\ce{MnO2}\) is +4 (change 3); \(\ce{Mn2O3}\) is +3 (change 4).
3. The sequence is 5, 1, 3, 4.

Teacher's Note:
1. Find the oxidation number of Mn in each product separately.
2. In acidic medium permanganate gives the largest change, 5.

Answer: (a) Bromine is a stronger oxidant than iodine

Explanation:
1. With iodine sulphur changes only from +2 to +2.5 (tetrathionate).
2. With bromine sulphur goes all the way to +6 in sulphate.
3. The stronger oxidant, bromine, oxidises thiosulphate further, so bromine is the stronger oxidant.

Teacher's Note:
1. The extent of oxidation of the same reductant shows the strength of the oxidant.
2. Both halogens are reduced in these reactions, not oxidised.

Answer: (a) The oxidation number of hydrogen is always +1

Explanation:
1. In metal hydrides such as NaH and \(\ce{CaH2}\), hydrogen is -1.
2. So "always +1" is not a correct rule.
3. The other three statements are correct.

Teacher's Note:
1. Hydrogen is +1 with non-metals and -1 with metals in hydrides.
2. Fluorine, being the most electronegative, is always -1.

Answer: (b) \(\ce{NH4NO3}\)

Explanation:
1. In \(\ce{NH4+}\) nitrogen is -3 and in \(\ce{NO3-}\) it is +5.
2. In \(\ce{NH2OH}\) N is -1 and in \(\ce{N2H4}\) it is -2, both with identical nitrogen atoms.
3. \(\ce{N3H}\) has an average of -1/3. Only \(\ce{NH4NO3}\) shows two distinct states.

Teacher's Note:
1. Treat ionic compounds as separate ions when counting oxidation states.
2. Identical atoms in a molecule normally share one state unless they are in different environments.

Answer: (b) Zinc is a reductant and the hydrogen ion is an oxidant

Explanation:
1. Zinc goes from 0 to +2, so it is oxidised and acts as the reductant.
2. \(\ce{H^+}\) goes from +1 to 0, so it is reduced and acts as the oxidant.
3. Chlorine stays at -1.

Teacher's Note:
1. The species that loses electrons is the reducing agent.
2. Chloride is only a spectator in this reaction.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Fluorine has the highest standard reduction potential of the halogens, so it is the strongest oxidant.
2. It is also the most electronegative element, which gives its strong tendency to gain electrons.
3. So A and R are true and R explains A.

Teacher's Note:
1. Strong tendency to gain electrons makes a species a strong oxidising agent.
2. Electronegativity measures that tendency in a bond.

Answer: (c) A is true but R is false

Explanation:
1. Permanganate oxidises iodide to iodine, so A is true.
2. Manganese is reduced from +7 (in \(\ce{MnO4^-}\)) to a lower state such as +2, not from +2 to +7.
3. Hence R is false.

Teacher's Note:
1. Oxidising agents are themselves reduced, so their oxidation number falls.
2. Read the direction of change in a Reason statement carefully.

Answer: (c) reduces permanganate to \(\ce{Mn^{2+}}\)

Explanation:
1. Chloride ions are oxidised to chlorine by permanganate.
2. This uses up some permanganate, so more titrant is consumed than oxalic acid alone requires.
3. So HCl itself reduces permanganate to \(\ce{Mn^{2+}}\) and causes error.

Teacher's Note:
1. Use only an acid that cannot be oxidised, such as dilute sulphuric acid.
2. Extra H+ would not affect the end point as long as acid is in excess.

Answer: (d) an acid as well as an oxidant

Explanation:
1. Sulphur is reduced from +6 in \(\ce{H2SO4}\) to +4 in \(\ce{SO2}\), so the acid is an oxidant.
2. Some sulphate also remains in \(\ce{Ag2SO4}\), so part of the acid behaves as an acid giving the salt.
3. So it acts as both an acid and an oxidant.

Teacher's Note:
1. Check which part of the acid ends up unchanged in the product salt.
2. Part of the sulphate is a spectator and the rest is reduced.

Answer: (c) M

Explanation:
1. Fe changes from +2 in \(\ce{FeSO4}\) to +3 in \(\ce{Fe2(SO4)3}\).
2. The change per formula unit of \(\ce{FeSO4}\) is 1 electron.
3. Equivalent weight = M/1 = M.

Teacher's Note:
1. Equivalent weight = molar mass divided by electrons transferred per formula unit.
2. Do not count the 2 Fe atoms of the product when working per \(\ce{FeSO4}\).

Answer: (c) +5, 0, 0, +5

Explanation:
1. Each terminal sulphur is bonded to three oxygens (in a sulphonate group) and one sulphur.
2. Terminal S: 3(-2) + x = -1 (ion charge shared), giving x = +5.
3. The two middle S atoms bond only to sulphur, so each is 0. The sum is +10, matching 12 - 2.

Teacher's Note:
1. Treat S-S bonds as sharing equally, contributing zero.
2. The average oxidation state, +2.5, hides the real +5, 0, 0, +5 pattern.

Answer: (b) Oxidation number of N in \(\ce{NH4+}\) changes from -3 to 0 and that in \(\ce{NO2-}\) from +3 to 0

Explanation:
1. In \(\ce{NH4+}\) nitrogen is -3 and in \(\ce{NO2-}\) it is +3.
2. Both end as \(\ce{N2}\) with oxidation number 0.
3. So -3 goes to 0 (oxidation) and +3 goes to 0 (reduction), a comproportionation.

Teacher's Note:
1. Assign oxidation numbers ion by ion.
2. Here the same element is oxidised and reduced to the same product.

Answer: (c) 0.5

Explanation:
1. Per mole of \(\ce{FeC2O4}\), Fe goes from +2 to +3 (1 electron) and oxalate to \(\ce{CO2}\) (2 electrons), a total of 3 electrons.
2. One \(\ce{Cr2O7^{2-}}\) accepts 6 electrons.
3. x = 3/6 = 0.5.

Teacher's Note:
1. Add the electrons lost by both Fe and oxalate.
2. Dichromate takes 6 electrons per ion.

Answer: (c) 0.6 moles

Explanation:
1. Ferrous oxalate loses 1 electron at Fe and 2 at oxalate, 3 electrons per mole.
2. Each \(\ce{MnO4^-}\) gains 5 electrons in acid.
3. Moles of permanganate = 3/5 = 0.6.

Teacher's Note:
1. Equate the electrons lost and gained.
2. Permanganate takes 5 electrons in acidic medium.

Answer: (b) 40 mL

Explanation:
1. Cr goes from +3 to +6 (3 electrons): 20 × 0.2 = 4 mmol, so 12 mmol electrons are lost.
2. Cl goes from +1 to -1 (2 electrons per NaOCl), so NaOCl needed = 6 mmol.
3. Volume = 6/0.15 = 40 mL.

Teacher's Note:
1. Equate the total electrons given and taken.
2. Millimoles divided by molarity gives volume in mL.

Answer: (d) 10

Explanation:
1. Milli-equivalents of permanganate = 10 × (1/0.56) = 17.86.
2. Normality of \(\ce{H2O2}\) = 17.86/10 = 1.786 N.
3. Volume strength = 5.6 × N = 5.6 × 1.786 = 10.

Teacher's Note:
1. Volume strength of \(\ce{H2O2}\) equals 5.6 times its normality.
2. Equivalents of oxidant and reductant are equal at the end point.

Answer: (c) \(\ce{S4O6^{2-}}\)

Explanation:
1. V in \(\ce{VO4^{3-}}\) is +5, Mn in \(\ce{Mn2O3}\) is +3 and Cl in \(\ce{Cl2O7}\) is +7.
2. In \(\ce{S4O6^{2-}}\), 4x - 12 = -2, so x = +2.5.
3. Only tetrathionate has a fractional average oxidation state.

Teacher's Note:
1. Fractional oxidation numbers are average values over non-equivalent atoms.
2. Tetrathionate has sulphur at +5 and 0.

Answer: (b) 20%

Explanation:
1. Thiosulphate used = 90 × 0.2 = 18 mmol.
2. 2 BaCrO4 give 1 \(\ce{Cr2O7^{2-}}\), which gives 3 \(\ce{I2}\), which needs 6 thiosulphate, so 1 mol BaCrO4 corresponds to 3 mol thiosulphate.
3. Ba = 18/3 = 6 mmol = 1.464 g of \(\ce{BaCl2.2H2O}\), so purity = 1.464/7.32 = 20%.

Teacher's Note:
1. Follow the chain Ba to chromate to dichromate to iodine to thiosulphate.
2. Use the mole ratio 1 : 3 for barium to thiosulphate.

Answer: (a) 15

Explanation:
1. Three Mn atoms split into a atoms of Mn(+7) and b atoms of Mn(+2), with a + b = 3.
2. Conservation of oxidation number: 7a + 2b = 8, so a = 0.4 and b = 2.6.
3. Electrons lost (oxidation) = 0.4 × (7 - 8/3) = 26/15, so equivalent mass = M/(26/15) = (15/26)M and x = 15.

Teacher's Note:
1. In disproportionation the equivalent mass uses only the oxidation half (or only the reduction half).
2. First find how many atoms go to each state.

Answer: (b) 3

Explanation:
1. \(\ce{2Cu^{2+} + 4I^- -> 2CuI + I2}\). Moles of \(\ce{Cu^{2+}}\) = 5 mmol and \(\ce{I^-}\) = 5 mmol, so iodide is limiting.
2. 5 mmol \(\ce{I^-}\) need 2.5 mmol \(\ce{Cu^{2+}}\), and only half of the iodide is oxidised, 2.5 mmol.
3. Each \(\ce{Cu^{2+}}\) gains one electron: electrons = \(2.5 \times 10^{-3}\) mol, so x = 3.

Teacher's Note:
1. Identify the limiting reagent using the balanced equation.
2. Part of the iodide is held in CuI and is not oxidised.