6.Redox Reactions

Qno.1. Choose the most correct option.

A. Oxidation numbers of Cl atoms marked as Clᵃ and Clᵇ in CaOCl₂ (bleaching powder) are

     Clᵃ

    /

Ca

    \

     O — Clᵇ

a. zero in each  

b. -1 in Clᵃ and +1 in Clᵇ

c. +1 in Clᵃ and -1 in Clᵇ

d. 1 in each

B. Which of the following is NOT an example of redox reaction?

a. CuO + H₂ → Cu + H₂O  

b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂  

c. 2K + F₂  → 2KF

d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

C. A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by  

a. A₂(BC₃)₂  

b. A₃(BC₄)₂

c. A₃(B₄C)₂

d. ABC₂

D. The coefficients p, q, r, s in the reaction  

p Cr₂O₇²⁻ + q Fe²⁺ + H⁺ → r Cr³⁺ + s Fe³⁺ + H₂O  

respectively are:  

a. 1, 6, 2, 6  

b. 6, 1, 2, 4

c. 1, 6, 2, 6 

d. 1, 2, 4, 6

E. For the following redox reactions, find the correct statement.  

Sn²⁺ +  2Fe³⁺→ Sn⁴⁺ + 2Fe²⁺

a. Sn²⁺ is undergoing oxidation

b. Fe³⁺ is undergoing oxidation  

c. It is not a redox reaction  

d. Both Sn²⁺ and Fe³⁺ are oxidised  

F. Oxidation number of carbon in H₂CO₃ is  

a. +1 b. +2 c. +3 d. +4  

G. Which is the correct stock notation for manganese dioxide?  

a. Mn(I)O₂   b. Mn(II)O₂  

c. Mn(III)O₂ d. Mn(IV)O₂  

I. Oxidation number of oxygen in superoxide is  

a. -2 b. -1 c. -½ d. 0  

J. Which of the following halogens does always show oxidation state -1?  

a. F  

b. Cl  

c. Br  

d. I  

K. The process SO₂ ⟶ S₂Cl₂ is  

a. Reduction  

b. Oxidation  

c. Neither oxidation nor reduction  

d. Oxidation and reduction.  

Qno.2. Write the formula for the following compounds :  

A. Mercury(II) chloride  

Ans:- HgCl₂

B. Thallium(I) sulphate 

Ans:- Tl₂SO₄ 

C. Tin(IV) oxide  

Ans:- SnO₂

D. Chromium(III) oxide  

Ans:- Cr₂O₃

3. Answer the following questions  

A. In which chemical reaction does carbon exhibit variation of oxidation state from -4 to +4? Write balanced chemical reaction. (1marks)

Ans:- In combustion of methane, carbon exhibits variation from –4 to +4. The reaction is as follows:

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O

In CH₄, the oxidation state of carbon is –4 while in CO₂, the oxidation state of carbon is +4. 

B. In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5?

(2 marks)

Ans:-In the following series of reactions, oxidation number of nitrogen changes from -3 in NH₃ to +5 in HNO₃.

   -3

4NH₃(g)   + 5O₂(g)  ⟶[pt.Rh gauge 

Ammonia   Oxygen

catalyst 500k,9bar] 4NO(g)   + 6H₂O(g)

                                Nitric oxide 

                                        +4

2NO(g).     +  O₂(g) → 2NO₂(g)

Nitric oxide  oxygen    Nitrogen dioxide 

             

                                     +5

3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)

Nitrogen   water        Nitric         Nitric 

dioxide                       acid           oxide 

C. Calculate the oxidation number of underlined atoms.

a. H₂SO₄

Ans:-

Oxidation number of H = +1

Oxidation number of O = -2

H₂SO₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0

∴ 2×(Oxidation number of H) + (Oxidation number of S) + 4×(Oxidation number of O) = 0

∴ 2×(+1) + (Oxidation number of S) + 4×(-2) = 0

∴ Oxidation number of S + 2 - 8 = 0

∴ Oxidation number of S in H₂SO₄ = +6

b.HNO₃     

Ans:-

Oxidation number of H = +1  

Oxidation number of O = –2  

HNO₃ is a neutral molecule.  

∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0  

∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0  

∴ (+1) + (Oxidation number of N) + 3 × (–2) = 0  

∴ Oxidation number of N + 1 – 6 = 0  

∴ Oxidation number of N in HNO₃ = +5

c. H₃PO₃

Ans:-

Oxidation number of O = -2

Oxidation number of H = +1

H₃PO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 3×(Oxidation number of H) + (Oxidation number of P) + 3×(Oxidation number of O) = 0

∴ 3×(+1) + (Oxidation number of P) + 3×(-2) = 0

∴ Oxidation number of P + 3 - 6 = 0

∴ Oxidation number of P in H₃PO₃ = +3

d. K₂C₂O₄    

Ans:-

Oxidation number of K = +1  

Oxidation number of O = –2  

K₂C₂O₄ is a neutral molecule.  

∴ Sum of the oxidation number of all atoms = 0  

∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0  

∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (–2) = 0  

∴ 2 × (Oxidation number of C) + 2 – 8 = 0  

∴ 2 × (Oxidation number of C) = +6  

∴ Oxidation number of C = +6 / 2  

∴ Oxidation number of C in K₂C₂O₄ = +3

e. H₂S₄O₆     

Ans:-

Oxidation number of H = +1  

Oxidation number of O = −2  

H₂S₄O₆ is a neutral molecule.  

∴ Sum of the oxidation numbers of all atoms = 0  

∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0  

∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (−2) = 0  

∴ 2 + 4 × (Oxidation number of S) − 12 = 0  

∴ 4 × (Oxidation number of S) = 12 − 2  

∴ 4 × (Oxidation number of S) = +10  

Oxidation number of S = +10/4  

∴ Oxidation number of S in H₂S₄O₆ = +2.5

f. Cr₂O₇²⁻

Ans:-

Oxidation of O = −2  

Cr₂O₇²⁻ is an ionic species.  

∴ Sum of the oxidation numbers of all atoms = −2  

∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = −2  

∴ 2 × (Oxidation number of Cr) + 7 × (−2) = −2  

∴ 2 × (Oxidation number of Cr) − 14 = −2  

∴ 2 × (Oxidation number of Cr) = −2 + 14  

∴ 2 × (Oxidation number of Cr) = 12  

Oxidation number of Cr = 12/2  

∴ Oxidation number of Cr in Cr₂O₇²⁻ = +6

g. NaH₂PO₄

Ans:-

Oxidation number of Na = +1  

Oxidation number of H = +1  

Oxidation number of O = −2  

NaH₂PO₄ is a neutral molecule  

Sum of the oxidation numbers of all atoms = 0  

(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0  

(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (−2) = 0  

(Oxidation number of P) + 3 − 8 = 0  

Oxidation number of P = 8 − 3  

∴ Oxidation number of P in NaH₂PO₄ = +5

D. Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.

a. 2Cu₂O (s) + Cu₂S (s) ⟶ 6Cu (s) + SO₂ (g)

Ans:-

2Cu₂O(s) + Cu₂S(s) → 6Cu(s)+ SO₂ (g)

a. Write an oxidation number of all the atoms of reactants and products.

2Cu₂O(s) + Cu₂S(s) → 6Cu(s)+ SO₂ (g)

2Cu₂O(s): Cu = +1, O = -2  

Cu₂S(s): Cu = +1, S = -2  

6Cu(s): Cu = 0

b. Identify the species that undergoes a change in oxidation number.

2Cu₂O(s) + Cu₂S(s) → 6Cu(s)+ SO₂ (g)

2Cu₂O(s): Gain of e⁻  

Cu₂S(s): Loss of e⁻  

6Cu(s): Change in oxidation state

c. The oxidation number of S increases from –2 to +4 and that of Cu decreases from +1 to 0. Because the oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.

d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by the gain of electrons, and therefore, Cu is an oxidising agent and itself is reduced.

b. HF (aq) + OH⁻(aq) ⟶ H₂O(l) + F⁻(aq)

Ans:-

a. Write oxidation number of all the atoms of reactants and products.

HF (aq) + OH⁻(aq) ⟶ H₂O(l) + F⁻(aq)

 HF(aq):  

  H = +1  

  F = –1  

 OH⁻(aq):  

  O = –2  

  H = +1  

  → H₂O(l):  

  H = +1  

  O = –2  

b. Since the oxidation numbers of all the species remain the same, this is NOT a redox reaction.

c. I₂(aq) + 2S₂O₃²⁻(aq) ⟶ S₄O₆²⁻(aq) + 2I⁻(aq)

Ans:-

a. Write oxidation number of all the atoms of reactants and products.

 I₂(aq) + 2S₂O₃²⁻(aq) ⟶ S₄O₆²⁻(aq) + 2I⁻(aq)

 I₂(aq)-0

2S₂O₃²⁻(aq) - +2,-2

S₄O₆²⁻(aq) - +2.5,-2

2I⁻(aq) - -1

b. Identify the species that undergoes a change in oxidation number.

Gain of e⁻

 I₂(aq) + 2S₂O₃²⁻(aq) ⟶ S₄O₆²⁻(aq)

Loss of e⁻.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to –1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.

d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidized. On the other hand, the oxidation number of I decreases by a gain of electrons, and therefore, I is an oxidizing agent and itself is reduced.

E. What is oxidation? Which one of the following pairs of species is in its oxidized state?

Ans:-Oxidation is a process that involves the loss of one or more electrons from a substance. 

a. Mg / Mg²⁺  

Ans:-

Here, Mg loses two electrons to form Mg²⁺ ion.

Mg(s) → Mg²⁺(aq) + 2e⁻

Hence, Mg/Mg²⁺ is an oxidized state

b. Cu / Cu²⁺

Ans:-

Here, Cu loses two electrons to form Cu²⁺ ion.

Cu(s) → Cu²⁺(aq) + 2e⁻

Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O₃  

Ans:-

Here, O loses two electrons to form O²⁻ ion.

 O₂(g) + 2e⁻ →2O²⁻(aq)

Hence, O₂,O²⁻ is in reduced state.

d. Cl₂ / Cl⁻

Ans:-

Here, each Cl gains one electron to form Cl⁻ ions.

Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)

Hence, Cl₂/Cl⁻ is in a reduced state. 

F. Justify the following reaction as redox reaction.

2 Na(s) + S(s) ⟶ Na₂ S(s)

Find out the oxidizing and reducing agents.(3marks)

Ans:-

i. Redox reaction can be described as electron transfer as shown below:

 2Na(s) + S(s) ⟶ 2Na⁺ + S²⁻

ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S²⁻.

This can be represented as shown in the adjacent reaction:

       

     Loss of e⁻ 

            ↓

  ┌───────────────┐

2Na(s) + S(s) → 2Na⁺ + S²⁻

           ↑

      Gain of e⁻

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electron to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S²⁻ in Na₂S.

iv. Each step represents a half reaction which involves electron transfer (loss or gain).

v. Sum of these two half reactions or the overall reaction is a redox reaction.

vi. Oxidising agent is an electron acceptor and hence, S is an orxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

G. Provide the stock notation for the following compounds: HAuCl₄, Ti₂O₃, FeO₃, Fe₂O₃, MnO and CuO.[3 marks]

Ans:-

      +1+3-1

i. HAuCl₄ → HAu(III)Cl₄  

ii. +1–2

  Tl₂O ⟶ Tl₂(I)O

      +2–2  

iii.  FeO → Fe(II)O  

iv.  +3–2

      Fe₂O₃→Fe₂(III)O₃  

    +2–2  

v.  MnO → Mn(II)O         vi. +2–2

  CuO ⟶ Cu(II)O

                  H. Assign oxidation number to each atom in the following species.

a. Cr(OH)₄⁻

Ans:- 

Oxidation number of O = -2  

Oxidation number of H = +1  

Cr(OH)₄⁻ is an ionic species.

∴ Sum of the oxidation numbers of all atoms = -1

∴ Oxidation number of Cr  + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = -1  

∴ Oxidation number of Cr + 4 × (-2)  + 4 × (+1) = -1

∴ Oxidation number of Cr - 8 + 4 = -1  

∴ Oxidation number of Cr - 4 = -1  

∴ Oxidation number of Cr = -1 + 4  

∴ Oxidation number of Cr in Cr(OH)₄⁻ = +3

b. Na₂S₂O₃ 

Ans:-

Oxidation number of Na = +1  

Oxidation number of O = –2  

Na₂S₂O₃ is a neutral molecule.  

∴ Sum of the oxidation numbers of all atoms = 0  

∴  2 × (Oxidation number of Na)  

   + 2 × (Oxidation number of S)  

   + 3 × (Oxidation number of O) = 0  

∴ 2 × (+1) + 2 × (Oxidation number of S)  

   + 3 × (–2) = 0  

∴ 2 × (Oxidation number of S) + 2 – 6 = 0  

∴ 2 × (Oxidation number of S) = +4  

∴ Oxidation number of S = +4 / 2  

                = +2  

∴ Oxidation number of S in Na₂S₂O₃ = +2

c. H₃BO₃

Ans:-

Oxidation number of H = +1  

Oxidation number of O = –2  

H₃BO₃ is a neutral molecule.  

∴ Sum of the oxidation numbers of all atoms = 0  

∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0  

∴ 3 × (+1) + (Oxidation number of B) + 3 × (–2) = 0  

∴ Oxidation number of B + 3 – 6 = 0  

∴ Oxidation number of B in H₃BO₃ = +3  

I. Which of the following redox couple is stronger oxidizing agent?

a. Cl₂ (E° = 1.36 V) and Br₂ (E° = 1.09 V)

b. MnO₄⁻ (E° = 1.51 V) and Cr₂O₇²⁻ (E° = 1.33 V).   [2marks]

Ans:-

i. Cl₂ has larger positive value of E⁰ than Br₂. Thus, Cl₂ is stronger oxidizing agent than Br₂.

ii. MnO₄⁻ has larger positive value of E⁰ than Cr₂O₇²⁻. Thus, MnO₄⁻ is stronger oxidizing agent than Cr₂O₇²⁻.

J. Which of the following redox couple is stronger reducing agent?

a. Li (E° = – 3.05 V) and Mg (E° = – 2.36 V)

b. Zn (E° = – 0.76 V) and Fe (E° = – 0.44 V).     [2marks]

Ans:-

i. Li has larger negative value of E⁰ than Mg. Thus, Li is stronger reducing agent than Mg.

ii. Zn has larger negative value of E⁰ than Fe. Thus, Zn is stronger reducing agent than Fe.

Qno.4. Balance the reactions/equations:

A. Balance the following reactions by oxidation number method [3 marks]

a. Cr₂O₇²⁻ (aq) + SO₃²⁻ (aq) ⟶ Cr³⁺ (aq) + SO₄²⁻ (aq) (acidic)

Ans:-

i. Cr₂O₇²⁻ (aq) + SO₃²⁻ (aq) ⟶ Cr³⁺ (aq) + SO₄²⁻ (aq) (acidic)

Step 1: Write skeletal equation and balance the elements other than O and H.

Cr₂O₇²⁻ (aq) + SO₃²⁻ (aq)  ⟶  2Cr³⁺ (aq) + SO₄²⁻ (aq)

Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

Cr₂      O₇²⁻ (aq) + S  O₃²⁻ (aq) ⟶ 2Cr³⁺ (aq) 

↑          ↑                 ↑   ↑                    ↑ 

2×(+6) 7×(–2).     +4 3×(–2).         +3

+ S   O₄²⁻ (aq)

    ↑   ↑

   +6  4+(-2)

Increase in oxidation number:   SO₃²⁻ → 

                                                        ↑            ↑ 

                                                       +4.        +6  (Increase per atom = 2)

Decrease in oxidation number: Cr₂O₇²⁻ →

                                                       ↑

                                                     +6

                                                    2Cr³⁺(aq)

                                                       ↑

                                                     +3

(Decrease per atom = 3)

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) ⟶ 2Cr³⁺ (aq) + SO₄²⁻ (aq) 

Step 3: Balance 'O' atoms by adding 4H2O to the right-hand side.

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) ⟶ 2Cr³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H+ on the left-hand side.

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺(aq) ⟶ 2Cr³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)

Step 5: Check two sides for balance of atoms and charges.

Hence, balanced equation: 

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺(aq)⟶ 2Cr³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)

b. MnO₄⁻ (aq) + Br⁻ (aq) ⟶ MnO₂ (s) + BrO₃⁻ (aq) (basic)

Ans:-

MnO₄⁻(aq) + Br⁻(aq) ⟶ MnO₂(s) + BrO₃⁻(aq) (basic)

Step 1: Write skeletal equation and balance the elements other than O and H.

MnO₄⁻(aq) + Br⁻(aq) ⟶ MnO₂(s) + BrO₃⁻(aq)

Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

MnO₄⁻(aq) + Br⁻(aq) ⟶ MnO₂(s) + BrO₃⁻(aq)

MnO₄⁻(aq): Mn = +7, 4 × (-2), -1  

MnO₂(s): Mn = +4, 2 × (-2)  

BrO₃⁻(aq): Br = +5, 3 × (-2)

Br⁻(aq): -1

Increase in oxidation number:  

Br⁻ ⟶ BrO₃⁻  

↑         ↑

-1  +5  

(Increase per atom = 6)

Decrease in oxidation number:  

MnO₄⁻ ⟶ MnO₂  

↑                ↑

+7   +4  

(Decrease per atom = 3)

To make the net increase and decrease equal, we must take 2 atoms of Mn.

2MnO₄⁻(aq) + Br⁻(aq) ⟶ 2MnO₂(s) + BrO₃⁻(aq)

Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.

2MnO₄⁻(aq) + Br⁻(aq) ⟶ 2MnO₂(s) + BrO₃⁻(aq) + H₂O(l)

Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

2MnO₄⁻(aq) + Br⁻(aq) + 2H⁺(aq) ⟶ 2MnO₂(s) + BrO₃⁻(aq) + H₂O(l)

Add OH⁻ ions equal to the number of H⁺ ions on both sides of the equation.

2MnO₄⁻(aq) + Br⁻(aq) + 2H⁺(aq) + 2OH⁻(aq) ⟶ 2MnO₂(s) + BrO₃⁻(aq) + H₂O(l) + 2OH⁻(aq)

The H⁺ and OH⁻ ions appearing on the same side of the reaction are combined to give H₂O molecules.

2MnO₄⁻(aq) + Br⁻(aq) + 2H₂O(l) ⟶ 2MnO₂(s) + BrO₃⁻(aq) + H₂O(l) + 2OH⁻(aq)

2MnO₄⁻(aq) + Br⁻(aq) + H₂O(l) ⟶ 2MnO₂(s) + BrO₃⁻(aq) + 2OH⁻(aq)

Step 5: Check two sides for balance of atoms and charges.

Hence, balanced equation:

2MnO₄⁻(aq) + Br⁻(aq) + H₂O(l) ⟶ 2MnO₂(s) + BrO₃⁻(aq) + 2OH⁻(aq)

c. H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + SO₂ (g) + H₂O (l) (acidic)

Ans:-

H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + SO₂ (g) + H₂O (l) (acidic)

Step 1: Write skeletal equation and balance the elements other than O and H.

H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + SO₂ (g) + H₂O (l) (acidic)   

Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.

Increase in oxidation number:    

C → CO2

↑             ↑

0            +4

(Increase per atom = 4)

Decrease in oxidation number:   H2SO4  → SO2

  ↑                     ↑

    +6                   +4

(Decrease per atom =2)

To make the net increase and decrease equal, we must take 2 atoms of S.

2H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + 2SO₂ (g) + H₂O (l) 

Step 3: Balance 'O' atoms by adding H2O to the right-hand side.

2H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + 2SO₂ (g) + H₂O (l) + H₂O (l)    

Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both sides.

2H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + 2SO₂ (g) + 2H₂O (l)   

Step 5: Check two sides for balance of atoms and charges.

Hence, balanced equation: 

2H₂SO₄ (aq) + C (s) ⟶ CO₂ (g) + 2SO₂ (g) + 2H₂O (l)   

d. Bi(OH)₃ (s) + Sn(OH)₃⁻ (aq) ⟶ Bi (s) + Sn(OH)₆²⁻ (aq) (basic)

Ans:-

Bi(OH)₃ (g) + Sn(OH)₃⁻ (aq) ⟶ Bi (s) + Sn(OH)₆²⁻ (aq)

Step 1: Write skeletal equation and balance the elements other than O and H.

Bi(OH)₃ (s) + Sn(OH)₃⁻ (aq) ⟶ Bi (s) + Sn(OH)₆²⁻ (aq)

Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

Bi(OH)₃ (s) + Sn(OH)₃⁻ (aq) ⟶ Bi (s) + Sn(OH)₆²⁻ (aq)

Bi : +3

(OH)₃: 3×(-1)

Sn: +2

(OH)₃⁻ : 3×(-1)

Bi:0 

Sn:+4

(OH)₆²⁻: 6×(-1)

Increase in oxidation number: 

Sn(OH)3 → Sn(OH)62–

    ↑                   ↑

   +2     +4

(increase per atom = 2)  

Decrease in oxidation number: 

Bi(OH)3 → Bi

↑                        ↑

+3      0

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) ⟶ 2Bi (s) + 3Sn(OH)₆²⁻ (aq)

Step 3: Balance ‘O’ atoms by adding 3H2O to the left-hand side.

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) +3H₂O(l) ⟶ 2Bi (s) + 3Sn(OH)₆²⁻ (aq)

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3H+ on the right-hand side.

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) +3H₂O(l) ⟶ 2Bi (s) + 3Sn(OH)₆²⁻ (aq) + 3H⁺(aq)

Add OH⁻ ions equal to the number of H+ ions on both sides of the equation.

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) +3H₂O(l) + OH⁻(aq) ⟶ 2Bi (s) + 3Sn(OH)₆²⁻ (aq) + 3H⁺(aq) + 3OH⁻(aq)

The H+ and OH^- ions appearing on the same side of the reaction are combined to give H2O molecules.

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) +3H₂O(l) + OH⁻(aq) ⟶ 4Bi (s) + 3Sn(OH)₆²⁻ (aq) + 3H₂O(l)

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) + OH⁻(aq) ⟶ 2Bi (s) + 3Sn(OH)₆²⁻ (aq) 

Step 5: Check two sides for balance of atoms and charges.

Hence, balanced equation: 2Bi(OH)3(s) + 

2Bi(OH)₃ (s) + 3Sn(OH)₃⁻ (aq) + OH⁻(aq) ⟶ 2Bi (s) + 3Sn(OH)₆²⁻ (aq) 

B. Balance the following redox equation by half reaction method. [3 marks each ]

a. H₂C₂O₄ (aq) + MnO₄⁻ (aq) ⟶ CO₂ (g) + Mn²⁺ (aq) (acidic)

Ans:-

i. H₂C₂O₄ (aq) + MnO₄⁻ (aq) ⟶ CO₂ (g) + Mn²⁺ (aq)

Step 1: write unbalanced equation for the redox reaction.Assign oxidation number to all the atoms in reactants and products.Divide the equation into two half equations.

H₂C₂O₄ (aq) + MnO₄⁻ (aq) ⟶ CO₂ (g) + Mn²⁺ (aq)

H₂-2× (+1)

C₂-+3

O₄-4×(-2)

Mn-+7

O₄⁻-4×(+2)

C-+4

O₂-2×(+2)

Mn²⁺-+2

+3 to +4- Loss of e⁻

+7 to +2- gain of e⁻

Oxidation half reaction: H₂C₂O₄(aq) → CO₂(g)

Reduction half reaction: MnO₄⁻(aq) → Mn²⁺(aq)

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Oxidation: H₂C₂O₄(aq) → 2CO₂(g)

Reduction: MnO₄⁻(aq) → Mn²⁺(aq) + 4H₂O(l)

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Oxidation: H₂C₂O₄(aq) → 2CO₂(g) + 2H⁺(aq)

Reduction: MnO₄⁻(aq) + 8H⁺(aq) → Mn²⁺(aq) + 4H₂O(l)

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Oxidation: H₂C₂O₄(aq) → 2CO₂(g) + 2H⁺(aq) + 2e⁻

Reduction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in half equations. Then add two half equations.

Oxidation: 5H₂C₂O₄(aq) → 10CO₂(g) + 10H⁺(aq) + 10e⁻

Reduction: 2MnO₄⁻(aq) + 16H⁺(aq) + 10e⁻ → 2Mn²⁺(aq) + 8H₂O(l)

Add two half equations:

5H₂C₂O₄(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) → 10CO₂(g) + 2Mn²⁺(aq) + 8H₂O(l)

The equation is balanced in terms of number of atoms and charges.

Hence, balanced equation: 5H₂C₂O₄(aq) + 2MnO₄⁻(aq) + 6H⁺(aq) → 10CO₂ + 2Mn²⁺(aq) + 8H₂O(l)

b. Bi(OH)₃ (s) + SnO₂²⁻ (aq) ⟶ SnO₃²⁻ (aq) + Bi (s) (basic)

Ans:-

Bi(OH)₃ (s) + SnO₂²⁻ (aq) ⟶ SnO₃²⁻ (aq) + Bi (s)

Step 1: write unbalanced equation for the redox reaction.Assign oxidation number to all the atoms in reactants and products.Divide the equation into two half equations.

Bi(OH)₃ (s) + SnO₂²⁻ (aq) ⟶ SnO₃²⁻ (aq) + Bi (s)

Bi-+3

(OH)₃-3×(-1)

Sn-+2

O₂²⁻-2×(-2)

Sn-+4

O₃²⁻-3×(-2)

Bi-0

+3 to 0 - Loss of e⁻ 

+2 to +4- Gain of e⁻ 

Oxidation half reaction - SnO₂²⁻ (aq)→SnO₃²⁻ (aq)

Reduction half reaction - Bi(OH)₃ (s) →Bi (s)

Step 2:Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1 H₂O to left side of oxidation half equation and 3 H₂O to the right side of reduction half equation.

- Oxidation: SnO₂²⁻(aq) + H₂O(l) → SnO₃²⁻(aq)

- Reduction: Bi(OH)₃(s) → Bi(s) + 3H₂O(l)

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2 H⁺ ions to the right side of oxidation half equation and 3 H⁺ ions to the left side of reduction half equation.

- Oxidation: SnO₂²⁻(aq) + H₂O(l) → SnO₃²⁻(aq) + 2H⁺(aq)

- Reduction: Bi(OH)₃(s) + 3H⁺(aq) → Bi(s) + 3H₂O(l)

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

- Oxidation: SnO₂²⁻(aq) + H₂O(l) → SnO₃²⁻(aq) + 2H⁺(aq) + 2e⁻

- Reduction: Bi(OH)₃(s) + 3H⁺(aq) + 3e⁻ → Bi(s) + 3H₂O(l)

Step 5:Multiply oxidation half equation by 3, reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equations.

- Oxidation: 3SnO₂²⁻ + 3H₂O(l) → 3SnO₃²⁻ + 6H⁺(aq) + 6e⁻

- Reduction: 2Bi(OH)₃(s) + 6H⁺(aq) + 6e⁻ → 2Bi(s) + 6H₂O(l)

Add two half equations:

2Bi(OH)₃(s) + 3SnO₂²⁻ → 3SnO₃²⁻ + 2Bi(s) + 3H₂O(l)

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH⁻ ions. The equation is balanced in terms of number of atoms and the charges.

Hence, balanced equation: 2Bi(OH)₃(s) + 3SnO₂²⁻(aq) → 3SnO₃²⁻(aq) + 2Bi(s) + 3H₂O(l)

Qno.5. Complete  the following table:

Assign oxidation number to the underlined species and write Stock notation of compound. [½ mark each]

Ans:-