75 free MCQs on Redox Reactions with worked answers and explanations. Master electron transfer in chemistry: assign oxidation states, balance half-reactions, identify oxidising and reducing agents, and connect redox to everyday reactions.
Take the timed Redox Reactions chapterwise test →Below are 75 practice questions on Redox Reactions, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Redox Reactions notes.
Electron transfer in a redox reaction: zinc loses electrons (oxidised, acts as reducing agent) and copper(II) ions gain those same electrons (reduced, acts as oxidising agent).
Q1.
The oxidation number of oxygen in oxygen difluoride (OF<sub>2</sub>) is:
Answer: C. +2
Why: Fluorine is more electronegative than oxygen, so each F is −1 and oxygen is forced to +2 in OF<sub>2</sub>.
Q2.
In a redox reaction, oxidation involves:
Answer: D. Loss of electrons
Why: Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons.
Q3.
The oxidation number of hydrogen in sodium hydride (NaH) is:
Answer: B. −1
Why: In metal hydrides hydrogen is more electronegative than the metal, so H carries an oxidation number of −1.
Q4.
During a reaction, the reducing agent is itself:
Answer: A. Oxidised
Why: A reducing agent donates electrons and is therefore oxidised in the process.
Q5.
The oxidation number of oxygen in hydrogen peroxide (H<sub>2</sub>O<sub>2</sub>) is:
Answer: C. −1
Why: In peroxides oxygen atoms share a single O–O bond, giving each an oxidation number of −1.
Q6.
Oxidation is defined as:
Answer: B. Loss of electrons
Why: Oxidation is the loss of electrons (or increase in oxidation state). The mnemonic OIL RIG: Oxidation Is Loss (of electrons).
Q7.
Reduction is defined as:
Answer: C. Gain of electrons
Why: Reduction is the gain of electrons (or decrease in oxidation state). In OIL RIG: Reduction Is Gain (of electrons).
Q8.
What is the oxidation state of oxygen in most compounds?
Answer: C. -2
Why: Oxygen has an oxidation state of -2 in most compounds (except peroxides where it is -1, and OF₂ where it is +2).
Q9.
What is the oxidation state of hydrogen in water (H<sub>2</sub>O)?
Answer: B. +1
Why: Hydrogen has an oxidation state of +1 in most compounds, including water.
Q10.
In the reaction 2Mg + O<sub>2</sub> → 2MgO, magnesium is:
Answer: B. Oxidised
Why: Magnesium goes from oxidation state 0 to +2 in MgO; it loses electrons and is oxidised.
Q11.
An oxidising agent:
Answer: B. Accepts electrons and gets reduced
Why: An oxidising agent accepts electrons from another species (which gets oxidised), and in doing so the oxidising agent itself is reduced.
Q12.
What is the oxidation state of a free element (e.g., O<sub>2</sub>, Fe, Cl<sub>2</sub>)?
Answer: C. 0
Why: Any element in its free (uncombined) state has an oxidation state of zero, regardless of the element.
Q13.
The oxidation state of chlorine in HCl is:
Answer: C. -1
Why: In HCl, H is +1 and the molecule is neutral, so Cl must be -1.
Q14.
In the reaction Zn + CuSO<sub>4</sub> → ZnSO<sub>4</sub> + Cu, zinc is:
Answer: C. Oxidised
Why: Zinc changes from oxidation state 0 to +2 in ZnSO₄; it loses electrons and is oxidised. Zinc is the reducing agent.
Q15.
The oxidation state of sulphur in H<sub>2</sub>SO<sub>4</sub> is:
Answer: B. +6
Why: In H₂SO₄: 2(+1) + S + 4(-2) = 0. S = 8 - 2 = +6.
Q16.
Which of the following is a common oxidising agent?
Answer: C. KMnO<sub>4</sub> (potassium permanganate)
Why: KMnO₄ contains Mn in the +7 oxidation state; it readily accepts electrons (Mn goes from +7 to +2 or +4), making it a strong oxidising agent.
Q17.
What is the oxidation state of nitrogen in NH<sub>3</sub>?
Answer: C. -3
Why: In NH₃, each H is +1. 3(+1) + N = 0, so N = -3.
Q18.
Rusting of iron is an example of:
Answer: B. Oxidation of iron
Why: In rusting, iron (Fe, oxidation state 0) is oxidised to Fe²⁺/Fe³⁺ in iron oxides/hydroxides by atmospheric oxygen and moisture.
Q19.
In a redox reaction, the substance that gets oxidised is called the:
Answer: B. Reducing agent
Why: The reducing agent is the substance that loses electrons (is oxidised) and thereby reduces another substance.
Q20.
What is the oxidation state of fluorine in all its compounds?
Answer: C. -1
Why: Fluorine is the most electronegative element; it always has an oxidation state of -1 in all its compounds.
Q21.
In the electrolytic decomposition of water: 2H<sub>2</sub>O → 2H<sub>2</sub> + O<sub>2</sub>, oxygen is:
Answer: A. Oxidised
Why: Oxygen in H₂O has oxidation state -2; in O₂ it is 0. It loses electrons and is oxidised (goes to higher, less negative oxidation state).
Q22.
The oxidation state of manganese in KMnO<sub>4</sub> is:
Answer: B. +7
Why: In KMnO₄: K is +1, O is -2 (×4 = -8), so Mn = +7 to balance.
Q23.
A disproportionation reaction is one in which:
Answer: B. The same element is simultaneously oxidised and reduced
Why: In a disproportionation reaction, a single substance acts as both oxidising and reducing agent; the same element changes to two different oxidation states.
Q24.
In the reaction 2H<sub>2</sub>O<sub>2</sub> → 2H<sub>2</sub>O + O<sub>2</sub>, H<sub>2</sub>O<sub>2</sub> undergoes:
Answer: C. Both oxidation and reduction (disproportionation)
Why: In H₂O₂, O is -1. In H₂O, O is -2 (reduction); in O₂, O is 0 (oxidation). The same element (O) is both oxidised and reduced: disproportionation.
Q25.
Which of the following processes is an example of reduction?
Answer: C. Copper(II) oxide reacting with hydrogen to form copper
Why: In CuO + H₂ → Cu + H₂O, Cu goes from +2 to 0 (gains electrons); copper oxide is reduced. (Hydrogen is oxidised from 0 to +1.)
Q26.
The oxidation number of manganese in the manganate ion MnO<sub>4</sub><sup>2−</sup> is:
Answer: D. +6
Why: Mn + 4(−2) = −2, so Mn = +6 in the manganate ion (compared with +7 in permanganate).
Q27.
The average oxidation number of sulphur in sodium thiosulphate Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is:
Answer: A. +2
Why: 2(+1) + 2S + 3(−2) = 0 gives 2S = +4, so the average oxidation number of S is +2.
Q28.
In acidic medium, for MnO<sub>4</sub><sup>−</sup> → Mn<sup>2+</sup>, the number of electrons gained per MnO<sub>4</sub><sup>−</sup> is:
Answer: C. 5
Why: Manganese goes from +7 to +2, a change of 5 units, so 5 electrons are gained.
Q29.
Among the halogens, the strongest oxidising agent is:
Answer: B. F<sub>2</sub>
Why: Fluorine has the highest standard reduction potential among the halogens, making F<sub>2</sub> the strongest oxidising agent.
Q30.
The n-factor of oxalic acid acting as a reductant (C<sub>2</sub>O<sub>4</sub><sup>2−</sup> → 2CO<sub>2</sub>) is:
Answer: D. 2
Why: Each carbon changes from +3 to +4; with two carbons the total change is 2 electrons, so the n-factor is 2.
Q31.
What is the oxidation state of chromium in K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> (potassium dichromate)?
Answer: C. +6
Why: K₂Cr₂O₇: 2(+1) + 2(Cr) + 7(-2) = 0. 2Cr = 14 - 2 = 12. Cr = +6.
Q32.
Balance the redox half-reaction in acidic medium: MnO<sub>4</sub><sup>-</sup> → Mn<sup>2+</sup>. How many H+ ions are needed?
Answer: B. 8
Why: MnO₄⁻ → Mn²⁺: Mn changes from +7 to +2 (gains 5e⁻). Balance O by adding 4H₂O on the right; balance H by adding 8H⁺ on the left; add 5e⁻ on the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Q33.
In the reaction: Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> + Fe<sup>2+</sup> → Cr<sup>3+</sup> + Fe<sup>3+</sup> (acidic), how many Fe<sup>2+</sup> ions are oxidised per one Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> ion?
Answer: D. 6
Why: Each Cr goes from +6 to +3 (gains 3e⁻); two Cr atoms gain 6e⁻ total. Each Fe²⁺ loses 1e⁻. To balance: 6 Fe²⁺ are oxidised per Cr₂O₇²⁻ reduced.
Q34.
What is the oxidation state of phosphorus in H<sub>3</sub>PO<sub>4</sub>?
Answer: B. +5
Why: In H₃PO₄: 3(+1) + P + 4(-2) = 0. P = 8 - 3 = +5.
Q35.
When Cu reacts with dilute H<sub>2</sub>SO<sub>4</sub>, no reaction occurs. But with concentrated H<sub>2</sub>SO<sub>4</sub>, the reaction gives:
Answer: B. CuSO<sub>4</sub> + SO<sub>2</sub> + H<sub>2</sub>O
Why: Concentrated H₂SO₄ is an oxidising acid: Cu + 2H₂SO₄(conc.) → CuSO₄ + SO₂↑ + 2H₂O. Dilute H₂SO₄ cannot oxidise copper as Cu is below H in the electrochemical series.
Q36.
In the electrochemical series, which of the following is the strongest reducing agent?
Answer: C. Li
Why: Standard reduction potential: Li⁺/Li = -3.04 V (most negative), meaning Li has the greatest tendency to lose electrons and is the strongest reducing agent.
Q37.
Balance the redox reaction in basic medium: MnO<sub>4</sub><sup>-</sup> + I- → MnO<sub>2</sub> + I<sub>2</sub>. The balanced equation is:
Answer: C. 2MnO<sub>4</sub><sup>-</sup> + 3I- + 2H<sub>2</sub>O → 2MnO<sub>2</sub> + I<sub>2</sub> + 4OH-... (check)
Why: In basic medium: Mn goes from +7 to +4 (gains 3e⁻); I goes from -1 to 0 (loses 1e⁻). Multiply to equalise: 2 MnO₄⁻ (6e⁻ gained) and 3 I⁻ (but I₂ has 2 atoms, so need even numbers). Balanced: 2MnO₄⁻ + 3I⁻ + 2H₂O → 2MnO₂ + (3/2)I₂... multiply by 2 for whole numbers.
Q38.
In the reaction: Fe + dilute H<sub>2</sub>SO<sub>4</sub> → FeSO<sub>4</sub> + H<sub>2</sub>, the element that acts as the reducing agent is:
Answer: C. Fe
Why: Iron goes from 0 to +2 (loses electrons); it is oxidised and is the reducing agent. H⁺ is reduced to H₂.
Q39.
What is the oxidation state of iron in Fe<sub>3</sub>O<sub>4</sub>?
Answer: C. Mixed: 1/3 Fe<sup>2+</sup> and 2/3 Fe<sup>3+</sup> (average +8/3)
Why: Fe₃O₄ is a mixed oxide: FeO·Fe₂O₃. It contains one Fe²⁺ and two Fe³⁺ per formula unit. Average oxidation state = (2 + 3 + 3)/3 = 8/3 ≈ +2.67.
Q40.
The reaction Cl<sub>2</sub> + 2NaOH → NaCl + NaOCl + H<sub>2</sub>O is an example of:
Answer: B. Disproportionation of Cl<sub>2</sub>
Why: Cl₂ (oxidation state 0) produces NaCl (Cl is -1, reduced) and NaOCl (Cl is +1, oxidised). The same element undergoes both oxidation and reduction: disproportionation.
Q41.
Which of the following can act as both an oxidising agent and a reducing agent?
Answer: C. H<sub>2</sub>O<sub>2</sub>
Why: H₂O₂ can oxidise (reducing to H₂O, O changes from -1 to -2) or reduce (oxidising to O₂, O changes from -1 to 0), acting as both an oxidising and a reducing agent depending on the conditions.
Q42.
The oxidation state of nitrogen in NH<sub>4</sub><sup>+</sup> ion is:
Answer: C. -3
Why: In NH₄⁺: overall charge is +1. 4(+1) + N = +1. N = +1 - 4 = -3.
Q43.
In the ion-electron method of balancing, when balancing the following half-reaction in acidic medium: Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> → Cr<sup>3+</sup>, how many electrons are transferred?
Answer: C. 6
Why: Each Cr goes from +6 to +3 (gains 3e⁻). There are 2 Cr atoms in Cr₂O₇²⁻, so total electrons gained = 2 × 3 = 6.
Q44.
In the reaction: 2F<sub>2</sub> + 2H<sub>2</sub>O → 4HF + O<sub>2</sub>, fluorine is:
Answer: B. Reduced from 0 to -1
Why: F₂ contains F at 0; in HF, F is -1. F₂ gains electrons (reduction) and is the oxidising agent. Oxygen in H₂O is oxidised from -2 to 0 in O₂.
Q45.
What is the change in oxidation state of sulphur when H<sub>2</sub>S is converted to SO<sub>2</sub>?
Answer: C. -2 to +4 (change of +6)
Why: In H₂S, S is -2. In SO₂, S is +4. The change in oxidation state is +4 - (-2) = +6; sulphur is oxidised (loses 6 electrons).
Q46.
The standard electrode potential for the H+/H<sub>2</sub> couple is:
Answer: C. 0.00 V
Why: The Standard Hydrogen Electrode (SHE) is the reference electrode with E° = 0.00 V by definition. All other electrode potentials are measured relative to the SHE.
Q47.
In the reaction: Zn + 2HCl → ZnCl<sub>2</sub> + H<sub>2</sub>, identify the oxidising agent:
Answer: C. HCl (specifically H+)
Why: H⁺ (from HCl) is reduced: 2H⁺ + 2e⁻ → H₂. Since H⁺ accepts electrons, it is the oxidising agent. Zn is oxidised (reducing agent).
Q48.
What is the oxidation state of carbon in glucose (C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>)?
Answer: A. 0
Why: Average oxidation state of C in C₆H₁₂O₆: C₆H₁₂O₆ = 0 charge. 6C + 12(+1) + 6(-2) = 0. 6C = 0. Average oxidation state of C = 0.
Q49.
Bleaching powder decolourises coloured substances because:
Answer: B. The Ca(OCl)Cl releases nascent oxygen which oxidises the colouring matter
Why: Bleaching powder [Ca(OCl)Cl] releases hypochlorite (OCl⁻) in water; this produces nascent oxygen which oxidises and destroys the coloured compounds, causing decolourisation.
Q50.
The method of balancing redox reactions that involves calculating the change in oxidation numbers and multiplying coefficients to make total increase = total decrease is called:
Answer: C. Oxidation number method
Why: The oxidation number (or electron transfer) method directly tracks changes in oxidation states and adjusts stoichiometric coefficients so that the total electron gain equals the total electron loss.
Q51.
In the balanced acidic-medium reaction MnO<sub>4</sub><sup>−</sup> + Fe<sup>2+</sup> + H<sup>+</sup> → Mn<sup>2+</sup> + Fe<sup>3+</sup> + H<sub>2</sub>O, the mole ratio of MnO<sub>4</sub><sup>−</sup> to Fe<sup>2+</sup> is:
Answer: A. 1 : 5
Why: MnO<sub>4</sub><sup>−</sup> gains 5 electrons while each Fe<sup>2+</sup> loses 1; balancing electrons needs 5 Fe<sup>2+</sup> per MnO<sub>4</sub><sup>−</sup>.
Q52.
The oxidation number of chromium in chromium pentoxide CrO<sub>5</sub> (containing two peroxide linkages) is:
Answer: A. +6
Why: CrO<sub>5</sub> has two O<sub>2</sub><sup>2−</sup> (−1 each O) and one O<sup>2−</sup>; Cr + 4(−1) + (−2) = 0 gives Cr = +6.
Q53.
The equivalent weight of KMnO<sub>4</sub> (molar mass ≈ 158 g/mol) acting as an oxidant in acidic medium is:
Answer: B. 31.6
Why: In acidic medium Mn goes +7 to +2 (n-factor 5), so equivalent weight = 158/5 = 31.6 g.
Q54.
Chlorine exhibits an oxidation state of +5 in which of the following ions?
Answer: B. ClO<sub>3</sub><sup>−</sup>
Why: In ClO<sub>3</sub><sup>−</sup>: Cl + 3(−2) = −1 gives Cl = +5 (ClO<sup>−</sup> is +1, ClO<sub>2</sub><sup>−</sup> is +3, ClO<sub>4</sub><sup>−</sup> is +7).
Q55.
For a spontaneous redox reaction, the standard cell potential E°<sub>cell</sub> and ΔG° must be, respectively:
Answer: C. Positive and negative
Why: ΔG° = −nFE°<sub>cell</sub>; spontaneity requires ΔG° < 0, which needs E°<sub>cell</sub> > 0.
Q56.
Balance the following redox reaction in acidic medium: MnO<sub>4</sub><sup>-</sup> + C<sub>2</sub>O<sub>4</sub><sup>2-</sup> → Mn<sup>2+</sup> + CO<sub>2</sub>. What are the stoichiometric coefficients (MnO<sub>4</sub><sup>-</sup> : C<sub>2</sub>O<sub>4</sub><sup>2-</sup>)?
Answer: B. 2 : 5
Why: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (reduction, 5e⁻ per Mn). C₂O₄²⁻ → 2CO₂ + 2e⁻ (oxidation, 2e⁻ per oxalate). LCM(5,2) = 10. Multiply: 2 MnO₄⁻ (10e⁻ gained) and 5 C₂O₄²⁻ (10e⁻ lost). Ratio = 2 : 5.
Q57.
The disproportionation of H<sub>3</sub>PO<sub>3</sub> (phosphorous acid) on heating gives H<sub>3</sub>PO<sub>4</sub> and:
Answer: A. PH<sub>3</sub>
Why: H₃PO₃ has P in +3 oxidation state. On heating: 4H₃PO₃ → 3H₃PO₄ + PH₃. P goes from +3 to +5 (in H₃PO₄, oxidised) and +3 to -3 (in PH₃, reduced). This is disproportionation.
Q58.
What is the oxidation state of sulphur in peroxomonosulphuric acid (Caro acid, H<sub>2</sub>SO<sub>5</sub>)?
Answer: A. +6 (one S-OH and one S-O-O-H group making it a peroxoacid)
Why: H₂SO₅ is peroxomonosulphuric acid (also called Caro acid). The peroxy O-O linkage contributes -1 per O (not -2). 2(+1) + S + 4(-2) + 1(-1) = 0, giving S = +6. The S is +6 but one O is the peroxide type (-1).
Q59.
In the reaction: I<sub>2</sub> + 2S<sub>2</sub>O<sub>3</sub><sup>2-</sup> → 2I- + S<sub>4</sub>O<sub>6</sub><sup>2-</sup>, the oxidation state of sulphur in S<sub>4</sub>O<sub>6</sub><sup>2-</sup> is:
Answer: B. +2.5
Why: In S₄O₆²⁻: 4S + 6(-2) = -2. 4S = +10. S = +2.5. In S₂O₃²⁻: 2S + 3(-2) = -2; 2S = +4; S = +2. Sulphur is oxidised from +2 to +2.5.
Q60.
The comproportionation reaction (reverse of disproportionation) has the general form: A(higher OS) + A(lower OS) → A(intermediate OS). Which of the following is an example?
Answer: B. Cu<sup>2+</sup> + Cu → 2Cu+
Why: Cu²⁺ (OS +2) + Cu (OS 0) → 2Cu⁺ (OS +1). The higher and lower oxidation state of the same element combine to give an intermediate oxidation state. This is comproportionation.
Q61.
In the permanganate titration of ferrous ions in acidic medium, the end point is determined by:
Answer: B. Persistent pale pink colour of excess KMnO<sub>4</sub> (self-indicator)
Why: KMnO₄ is itself a self-indicator: MnO₄⁻ is deep purple; Mn²⁺ (product) is nearly colourless. At the end point, a single excess drop of KMnO₄ imparts a persistent pale pink/purple colour.
Q62.
What is the n-factor (equivalents per mole) of KMnO<sub>4</sub> in neutral/faintly alkaline medium?
Answer: B. 3
Why: In acidic medium, Mn goes from +7 to +2 (n-factor = 5). In neutral or faintly alkaline medium, MnO₄⁻ is reduced to MnO₂ (Mn +4); n-factor = 7 - 4 = 3.
Q63.
Assign the oxidation state of each element in [Fe(CN)<sub>6</sub>]4- and determine the oxidation state of iron:
Answer: A. Fe = +2 (CN- is -1 each; 6(-1) + Fe = -4; Fe = +2)
Why: In [Fe(CN)₆]⁴⁻: each CN⁻ has a charge of -1; 6 CN⁻ contribute -6. Overall charge = -4. So Fe + (-6) = -4; Fe = +2. Ferrocyanide contains Fe(II).
Q64.
Balance the redox reaction in basic medium: Cr<sup>3+</sup> + H<sub>2</sub>O<sub>2</sub> → CrO<sub>4</sub><sup>2-</sup> + H<sub>2</sub>O. What is the oxidising agent?
Answer: B. H<sub>2</sub>O<sub>2</sub> (H<sub>2</sub>O<sub>2</sub> oxidises Cr<sup>3+</sup> to CrO<sub>4</sub><sup>2-</sup>)
Why: Cr goes from +3 to +6 (loses 3e⁻ per Cr; oxidised); H₂O₂ accepts electrons (O goes from -1 to -2): H₂O₂ is the oxidising agent. In basic medium the reaction is: 2Cr³⁺ + 3H₂O₂ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O.
Q65.
The equivalent weight of K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> in acidic medium (where Cr goes from +6 to +3) is:
Answer: B. Molar mass / 6
Why: In K₂Cr₂O₇: each Cr goes from +6 to +3 (change of 3 per Cr); two Cr atoms change, so total electron transfer per mole = 6 (n-factor = 6). Equivalent weight = Molar mass / n-factor = M/6.
Q66.
In the reaction of F<sub>2</sub> with water: F<sub>2</sub> + H<sub>2</sub>O → HF + HOF, fluorine:
Answer: A. Undergoes disproportionation (one F atom is 0→-1 and the other is 0→+1 in HOF)
Why: One F₂ molecule gives HF (F = -1, reduced) and HOF (F = +1 in HOF; O is -2, H is +1, so F = +1, oxidised). This is disproportionation: F goes from 0 to both -1 and +1 simultaneously.
Q67.
In calculating n-factor for Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> in its reaction with I<sub>2</sub> (where products include Na<sub>2</sub>S<sub>4</sub>O<sub>6</sub>), the n-factor of Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is:
Answer: B. 1
Why: In S₂O₃²⁻: S is +2. In S₄O₆²⁻: S is +2.5. Each Na₂S₂O₃ loses 0.5 electrons (per S₂O₃²⁻ unit, 2 S atoms each go from +2 to +2.5 = 1 electron lost total per mole). n-factor = 1.
Q68.
What happens to the oxidising power of KMnO<sub>4</sub> as the pH of the medium decreases (more acidic)?
Answer: C. Increases
Why: The reduction potential of the MnO₄⁻/Mn²⁺ couple increases with acidity (Nernst equation; H⁺ is a reactant). More acidic conditions make KMnO₄ a stronger oxidising agent.
Q69.
The oxidation state of nitrogen in NH<sub>2</sub>OH (hydroxylamine) is:
Answer: B. -1
Why: In NH₂OH: H is +1 (two H on N, one on O), O is -2. Summing: N + 2(+1) + (-2) + (+1) = 0. N + 1 = 0. N = -1. Hydroxylamine has nitrogen in the -1 oxidation state, intermediate between NH₃ (-3) and N₂ (0).
Q70.
During iodometric back-titration, Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is used to titrate liberated I<sub>2</sub>. The reaction is: 2S<sub>2</sub>O<sub>3</sub><sup>2-</sup> + I<sub>2</sub> → S<sub>4</sub>O<sub>6</sub><sup>2-</sup> + 2I-. What is the role of Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> here?
Answer: B. Reducing agent
Why: S₂O₃²⁻ loses electrons (S goes from +2 to +2.5 on average in S₄O₆²⁻); I₂ gains electrons (reduced to I⁻). Therefore Na₂S₂O₃ is the reducing agent.
Q71.
The reaction: 5H<sub>2</sub>O<sub>2</sub> + 2KMnO<sub>4</sub> + 3H<sub>2</sub>SO<sub>4</sub> → 2MnSO<sub>4</sub> + K<sub>2</sub>SO<sub>4</sub> + 5O<sub>2</sub> + 8H<sub>2</sub>O shows that H<sub>2</sub>O<sub>2</sub> is acting as:
Answer: B. Reducing agent (H<sub>2</sub>O<sub>2</sub> is oxidised: O from -1 to 0 in O<sub>2</sub>)
Why: In this reaction, H₂O₂ is the reducing agent: O in H₂O₂ goes from -1 to 0 in O₂ (loses electrons, oxidised). KMnO₄ is the oxidising agent: Mn goes from +7 to +2 (gains electrons, reduced).
Q72.
In the reaction: 3Cl<sub>2</sub> + 6NaOH (hot and conc.) → 5NaCl + NaClO<sub>3</sub> + 3H<sub>2</sub>O, which of the following statements is correct?
Answer: B. Cl disproportionates: 0 → -1 (in NaCl) and 0 → +5 (in NaClO<sub>3</sub>)
Why: Cl₂ (0) in hot concentrated NaOH disproportionates more extensively: Cl goes to -1 in NaCl (reduced) and +5 in NaClO₃ (oxidised). This contrasts with cold NaOH giving NaOCl (+1).
Q73.
A sample of iron ore weighing 1.00 g is dissolved in acid and all iron converted to Fe<sup>2+</sup>. The solution requires 25.0 mL of 0.0200 M KMnO<sub>4</sub> for titration. The percentage of Fe in the ore is approximately:
Answer: A. 14.0%
Why: Moles KMnO₄ = 0.025 × 0.02 = 5 × 10⁻⁴ mol. From MnO₄⁻ + 5Fe²⁺: moles Fe²⁺ = 5 × 5 × 10⁻⁴ = 2.5 × 10⁻³ mol. Mass Fe = 2.5 × 10⁻³ × 56 = 0.14 g. %Fe = (0.14/1.00) × 100 = 14.0%.
Q74.
In photo-oxidation of silver halides during photography, the reaction Ag+ + e- → Ag involves:
Answer: B. Reduction of Ag+ to Ag<sup>0</sup> by light-generated electrons
Why: When light hits silver halide crystals, it excites electrons; these electrons reduce Ag⁺ ions to Ag⁰ atoms, forming a latent image. Ag⁺ is reduced (gains one electron) to metallic silver.
Q75.
Using the standard electrode potentials E°(Fe<sup>3+</sup>/Fe<sup>2+</sup>) = +0.77 V and E°(I<sub>2</sub>/I-) = +0.54 V, predict whether Fe<sup>3+</sup> will oxidise I- ions:
Answer: B. Yes, because E°(Fe<sup>3+</sup>/Fe<sup>2+</sup>) > E°(I<sub>2</sub>/I-) so the cell EMF is positive and the reaction is spontaneous
Why: E°(cell) = E°(cathode) - E°(anode) = E°(Fe<sup>3+</sup>/Fe<sup>2+</sup>) - E°(I<sub>2</sub>/I-) = 0.77 - 0.54 = +0.23 V. Since E°(cell) > 0, the reaction Fe<sup>3+</sup> + I⁻ → Fe<sup>2+</sup> + I₂ is spontaneous. Fe<sup>3+</sup> oxidises I⁻ to I₂.