Redox Questions

Redox reactions are oxidation-reduction chemical reactions in which the oxidation states of the reactants change. The term ‘redox’ is an abbreviation for reduction-oxidation. All redox reactions can be divided into two types of processes: reduction and oxidation.

The redox reaction always involves simultaneous oxidation and reduction reactions. In a chemical reaction, the substance being reduced is known as the oxidising agent, while the substance being oxidised is known as the reducing agent.

Redox Chemistry Questions with Solutions

Q1. For the redox reaction, MnO₄^– +C₂O₄^2– + H⁺ → Mn²⁺ + CO₂ +H₂O

The correct coefficients of the reactants for the balanced equation are-

1. 16, 5, 2
2. 2, 5, 16
3. 2, 16, 5
4. 5, 16, 2

Correct Answer: (b) 2, 5, 16

Q2. Which oxidation states of phosphorus can be found in the disproportionation reaction below?

P₂H₄ → PH₃ + P₄H₂

1. P₄²⁺
2. P³⁺
3. P₂^4–
4. P^2–

Correct Answer: (c)

Q3. What is the oxidation number of alkali metals in its compounds?

Answer. +1.

Q4. What is the fundamental principle of balancing redox reactions using the ion-electron method?

Answer. The number of electrons lost during oxidation is equal to the number of electrons gained during reduction.

Q5. Would you use an oxidizing agent or reducing agent in order for the following reactions to occur?

1. ClO₃^– → ClO₂
2. SO₄^2– → S^2–
3. Mn²⁺ → MnO₂
4. Zn → ZnCl₂

Answer.

1. ClO₃^– → ClO₂ reducing agent
2. SO₄^2– → S^2– reducing agent
3. Mn²⁺ → MnO₂ oxidizing agent
4. Zn → ZnCl₂ oxidizing agent

Q6. How would you know whether a redox reaction is taking place in an acidic/alkaline or neutral medium?

Answer. When H⁺ or any acid appears on either side of a chemical equation, the reactions occur in an acidic medium. The solution is basic if OH^– or any base appears on either side of a chemical equation. If there is no H⁺, OH^–, acid, or base in the chemical equation, the solution is neutral.

Q7. Justify the reaction:

2Cu₂O(S) + Cu₂S (s) → 6Cu (s) + SO₂ (g) is a redox reaction. Identify the species oxidized, and reduced, which acts as an oxidant and which acts as a reductant.

Answer.

Copper is reduced from + 1 oxidation state to 0 oxidation state in this reaction, and sulphur is oxidised from – 2 to +4 state. As a result, the reaction is a redox reaction, additionally, Cu₂O aids sulphur in Cu₂S and Cu₂O in decreasing its oxidation number. Thus, sulphur in Cu₂S is a reducing agent.

Q8. Can we store copper sulphate solution in a gold vessel?

Given E°_{Cu2+| Cu}= +0.34 V and E°_{Au3+| Au} = +1.50V

Answer. If the following redox reactions occur, we can store CuSO4 solution in a gold vessel.

2Au + 3Cu²⁺ → 2Au³⁺ + 3Cu

The cell corresponding to the preceding redox reaction can be represented as follows:

Au | Au³⁺ || Cu²⁺|Cu

Since the reduction potential of Au³⁺∣Au is higher than Cu²⁺∣Cu. So, the copper ion will not be reduced. Therefore, copper sulphate solution can be stored in a gold vessel. Thus, the CuSO₄ solution can be stored in a gold vessel.

Q9. What are the highest and lowest oxidation numbers of N?

Answer. N has a maximum oxidation number of +5 because it has five electrons in the valence shell (2s²2p³) and a minimum oxidation number of -3 because it can accept three more electrons to achieve the nearest inert gas (Ne) configuration.

Q10. Consider the following galvanic cell.

Cd | Cd²⁺ (1M) || H⁺(1M) | H₂ (g/atm)

(i) Write the overall cell reaction

(ii) What do the double vertical lines denote?

Answer:

(i) The anodic reaction is

Cd(s) → Cd²⁺ (aq) + 2e^–

The Cathodic reaction is

2H⁺(aq) + 2e^– → H₂(g)

The overall reaction is

Cd(s) + 2H⁺(aq) → Cd²⁺(aq) + H₂(g)

(ii) The salt bridge that connects the oxidation and reduction half cells is represented by the double vertical lines.

Q11. Write the following redox reaction in the oxidation & reduction half-reaction.

1. 2K(s) + Cl₂(g) → 2KCl(s)
2. 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)

Answer:

1. 2K(s) + Cl₂(g) → 2KCl(s)

K(s) → K⁺(aq) + e^– (Oxidation)

Cl₂(g) + 2e^– → 2Cl^– (Reduction)

1. 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)

Al(s) → Al³⁺ (aq) + 3e^– (Oxidation)

Cu²⁺ + 2e^– → Cu(s) (Reduction)

Q12. HNO₃ acts only as an oxidant whereas HNO₂ acts both as an oxidant and reductant. Why?

Answer.

The oxidation number of N in HNO₃ = +5

The oxidation number of N in HNO₂ = +3

The maximum oxidation number that N can show is = + 5

(∴ It has only 5 valance electrons 2s²2p³)

The Oxidation number of N in HNO₃ is maximum and it can only decrease. Therefore HNO₃ can act only as an oxidant. The minimum Oxidation number of N is -3.

Thus HNO₂ in which the oxidation number of N is +3 can decrease as well as increase. Thus HNO₂ can act as an oxidant as well as a reductant.

Q13. Identify the species being oxidized and reduced in each of the following reactions:

1. Cr⁺ + Sn⁴⁺ → Cr³⁺ + Sn²⁺
2. 3Hg²⁺ + Fe (s) → 3Hg₂ + 2Fe³⁺
3. 2As (s) + 3Cl₂ (g) → 2AsCl₃

Answer.

1. Cr⁺ + Sn⁴⁺ → Cr³⁺ + Sn²⁺
   Cr⁺: oxidized, Sn⁴⁺: reduced
2. 3Hg²⁺ + Fe (s) → 3Hg₂ + 2Fe³⁺
   Hg²⁺: reduced, Fe: oxidized
3. 2As (s) + 3Cl₂ (g) → 2AsCl₃
   As: oxidized, Cl₂: reduced

Q14. Calculate the oxidation number of underlined elements in the followings:

1. Na₂B₄O₇
2. H₄P₂O₇
3. CaO₂
4. NaBH₄
5. H₂S₂O₇

Answer:

1. Na₂B₄O₇
   Let’s assume the oxidation number of B is x.
   Oxidation number of H = +1
   Oxidation number of O = -2
   Then we have: 2 + 4(x) – 14 = 0
   ⇒4x = 12
   ⇒x = 3
   Hence, the oxidation number of B is +3.
2. H₄P₂O₇
   Let’s assume the oxidation number of P is x.
   Oxidation number of H = +1
   Oxidation number of O = -2
   Then we have: 4(+1) + 2(x) + 7 (-2) = 0
   ⇒ 4 + 2x – 14 = 0
   ⇒ 2x – 10 = 0
   ⇒ 2x = +10
   ⇒ x = +5
   Hence, the Oxidation number of P is +5.
3. CaO₂
   Let’s assume the oxidation number of O is x.
   The oxidation number of Ca = +2
   Then we have: 1(+2) + 2(x) = 0
   ⇒ 2 + 2x = 0
   ⇒ 2x = -2
   ⇒ x = -1
   Hence, the Oxidation number of O is -1
4. NaBH₄
   Let’s assume the oxidation number of B is x.
   Oxidation number of Na = +1
   Oxidation number of H = -1
   Then we have: 1(+1) + 1(x) + 4(-1) = 0
   ⇒ 1 + x -4 = 0
   ⇒ x – 3 = 0
   ⇒ x = +3
   Hence, the Oxidation number of B is +3.
5. H₂S₂O₇
   Let’s assume the oxidation number of S is x.
   Oxidation number of O = -2
   Oxidation number of H = +1
   Then we have: 2(+1) + 2(x) + 7(-2) = 0
   ⇒ 2 + 2x – 14 = 0
   ⇒ 2x – 12 = 0
   ⇒ x = +6
   Hence, the Oxidation number of S is +6.

Q15. Balance the following equation by the ion-electron method

Zn (s) + NO₃^– → Zn²⁺ (aq)+ NH₄ (aq) + H₂O (In acid solution)

Answer. Oxidation half-reaction-

Zn → Zn²⁺ (aq)

Zn (s) → Zn²⁺ + 2e^– (To balance charge) [equation 1]

Reduction half-reaction

NO₃^– (aq) → NH₄⁺ (aq)

NO₃^– (aq) → NH₄⁺ + 3H₂O (l) (To balance O atom)

NO₃^– (aq) +10H⁺ (aq)→ NH₄⁺(aq) +3H₂O (l) (To balance H atom)

NO₃^– (aq) +10H⁺ (aq) + 8e^–→ NH₄⁺(aq) +3H₂O (l)

Multiply eq. 1 by 4 to equalise the number of electrons in both the half reactions-

4Zn (s) → 4Zn²⁺ + 8e^–

Add both the half reaction-

4Zn (s) + NO₃^– (aq) +10H⁺ (aq) → 4Zn²⁺ + NH₄⁺(aq) +3H₂O (l)

Practise Questions on Redox Chemistry

Q1. Determine the oxidation number of the elements in each of the following compounds:

1. H₂CO₃
2. N₂
3. Zn(OH)₄^2–
4. NO₂^–
5. LiH

Q2. Write the balanced half-reactions of the following reactions:

1. NiO₂ + 2H₂O + Fe → Ni(OH)₂ + Fe(OH)₂ in basic solution
2. CO₂ + 2N₂HOH → CO + N₂ + 3H₂O in basic solution

Q3. Balance the following equation in an acidic medium by the oxidation number method.

MnO₄^– + (COOH)₂ → Mn²⁺ (aq) + CO₂ + H₂O

Q4. How many grams of K₂Cr₂O₇ is required to oxidize Fe²⁺ present in 15.2 gm of FeSO₄ to Fe³⁺ if the reaction is carried out in an acidic medium.

Q5. How many millimoles of potassium dichromate are required to oxidize 24 cm³ of 0.5 M mohr’s salt solution in an acidic medium.

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