Questions & explanations
1. For the spontaneous cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which of the following is correct?
- Zinc electrode acts as cathode.
- Copper electrode is the anode.
- Electrons flow from zinc to copper in the external circuit.
- Reduction occurs at the zinc electrode.
Answer: Electrons flow from zinc to copper in the external circuit.
In this galvanic cell, zinc undergoes oxidation (Zn → Zn²⁺ + 2e⁻) at the anode, so electrons leave the zinc electrode. Copper ions are reduced (Cu²⁺ + 2e⁻ → Cu) at the cathode, so electrons enter the copper electrode. Thus, electrons flow from zinc (anode) to copper (cathode). Options A and B incorrectly swap the roles of anode and cathode; option D is false because reduction occurs at the copper electrode, not the zinc electrode.
2. When the following equation is balanced by the oxidation‑number method in acidic medium, what is the coefficient of H⁺? MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (unbalanced)
- A) 4
- B) 8
- C) 5
- D) 1
Answer: B) 8
Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (increase of 1). Reduction: MnO₄⁻ (Mn +7) + 5e⁻ → Mn²⁺ (Mn +2) (decrease of 5). Multiply oxidation by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻. Add to reduction: MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺. Balance O by adding H₂O: MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O? Actually left has 4 O, right none, so need 4H₂O on right, then balance H by adding 8H⁺ on left: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Thus coefficient of H⁺ is 8.
3. Which of the following reactions is an example of comproportionation?
- A) 2H2O2 → 2H2O + O2
- B) Cl2 + 2NaOH → NaCl + NaOCl + H2O
- C) 2FeCl3 + Fe → 3FeCl2
- D) 2KClO3 → 2KCl + 3O2
Answer: C) 2FeCl3 + Fe → 3FeCl2
Comproportionation is a reaction in which two different oxidation states of the same element combine to form a single intermediate oxidation state. Here Fe is present as Fe³⁺ (in FeCl3, oxidation number +3) and Fe⁰ (metallic Fe) and both produce Fe²⁺ (in FeCl2, oxidation number +2), so it is comproportionation. Options A and B are disproportionation; D is a decomposition.
4. Balance the following redox reaction in basic medium:
Cr(OH)₃ + IO₃⁻ → CrO₄²⁻ + I⁻
The coefficient of OH⁻ in the balanced equation is:
- 2
- 4
- 6
- 8
Answer: 4
The half-reactions are: Cr(OH)₃ → CrO₄²⁻ (oxidation, loss of 3e⁻) and IO₃⁻ → I⁻ (reduction, gain of 6e⁻). Balancing: Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻; IO₃⁻ + 3H₂O + 6e⁻ → I⁻ + 6OH⁻. Multiplying the first by 2 and adding gives 2Cr(OH)₃ + IO₃⁻ + 4OH⁻ → 2CrO₄²⁻ + I⁻ + 5H₂O. Thus, the coefficient of OH⁻ is 4. Options 2, 6, and 8 correspond to incorrect balancing.
5. The equivalent weight of K₂Cr₂O₇ in acidic medium is: (Molar mass = 294 g mol⁻¹)
- 24.5 g eq⁻¹
- 49 g eq⁻¹
- 98 g eq⁻¹
- 147 g eq⁻¹
Answer: 49 g eq⁻¹
In acidic medium, Cr₂O₇²⁻ is reduced to Cr³⁺; each Cr atom changes from +6 to +3 (gain of 3 electrons), so the dichromate ion accepts 6 electrons (n = 6). Equivalent weight = Molar mass / n = 294/6 = 49 g eq⁻¹. Options A, C, and D arise from using n = 12, n = 3, or n = 2, which are incorrect for this reduction.
6. The reaction Cl2 + 2NaOH → NaCl + NaOCl + H2O is an example of:
- A) Combination reaction
- B) Decomposition reaction
- C) Displacement reaction
- D) Disproportionation reaction
Answer: D) Disproportionation reaction
In this reaction, chlorine (oxidation number 0) is simultaneously reduced to Cl⁻ (oxidation number -1) in NaCl and oxidized to ClO⁻ (oxidation number +1) in NaOCl. Such a reaction where the same element undergoes both oxidation and reduction is called disproportionation.
7. Consider the reaction: 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4. Which statement is correct?
- A) FeCl3 is reduced because the oxidation number of Fe decreases from +3 to +2.
- B) SnCl2 is reduced because the oxidation number of Sn increases from +2 to +4.
- C) FeCl3 is oxidized because the oxidation number of Fe increases.
- D) SnCl2 is oxidized because the oxidation number of Sn decreases.
Answer: A) FeCl3 is reduced because the oxidation number of Fe decreases from +3 to +2.
In terms of oxidation number change, reduction is a decrease in oxidation number. Fe in FeCl3 has oxidation number +3 and in FeCl2 it is +2, a decrease, so FeCl3 is reduced. Sn in SnCl2 goes from +2 to +4 (increase), so it is oxidized. Thus only option A is correct.
8. In the reaction 2KMnO4 + 10FeSO4 + 8H2SO4 → 2MnSO4 + 5Fe2(SO4)3 + K2SO4 + 8H2O, which species acts as the reducing agent?
- A) KMnO4
- B) FeSO4
- C) H2SO4
- D) MnSO4
Answer: B) FeSO4
The reducing agent is the species that gets oxidized (loses electrons). Fe in FeSO4 has oxidation number +2, which increases to +3 in Fe2(SO4)3, indicating oxidation. Thus FeSO4 is the reducing agent. KMnO4 (Mn from +7 to +2) is the oxidizing agent.
9. Balance the following redox reaction in acidic medium by the ion‑electron method: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺. What is the coefficient of H₂O in the balanced equation?
- A) 7
- B) 6
- C) 14
- D) 3
Answer: A) 7
Half‑reactions: Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 6: 6Fe²⁺ → 6Fe³⁺ + 6e⁻). Adding: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺. Hence the coefficient of H₂O is 7.
10. In terms of electron transfer, which statement correctly describes oxidation and reduction in a chemical reaction?
- A) Oxidation is the gain of electrons, reduction is the loss of electrons.
- B) Oxidation is the loss of electrons, reduction is the gain of electrons.
- C) Both oxidation and reduction involve gain of electrons.
- D) Both oxidation and reduction involve loss of electrons.
Answer: B) Oxidation is the loss of electrons, reduction is the gain of electrons.
Oxidation is defined as the loss of electrons, while reduction is the gain of electrons. This is often remembered by the mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain). Hence option B is correct.
11. What is the oxidation number of chromium in K2Cr2O7?
- A) +5
- B) +6
- C) +7
- D) +4
Answer: B) +6
Using standard rules: K is +1, O is -2. Let Cr be x. Total charge = 0: 2(+1) + 2x + 7(-2) = 0 → 2 + 2x - 14 = 0 → 2x = 12 → x = +6. So the oxidation number of Cr is +6.