Chemistry NCERT Exemplar Solutions Class 12th Chapter Three

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

In the electrolysis of sulphuric acid, the following reactions occur

2SO₄²⁻ (aq)→S₂O₈²⁻ (aq) + 2e⁻ E? _Cell =1.96V

2H₂O (l)→O₂ (g)+4H⁺ (aq) + 4e⁻ E? _Cell =1.23V

The reaction will lower the value of E? _Cell is preferred at anode so the second reaction is feasible.

H⁺ + e⁻ → $\frac{1}{2}$ H₂ E? _Cell = 0.00V

At the cathode, reduction of water occurs. Therefore, in dilute sulphuric acid solution, hydrogen will be reduced at the cathode.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (ii, iv)

The lesser the E? value of the redox couple higher is the reducing power

For Cu²⁺ + 2e⁻→Cu; E? =0.34V

For 2H⁺ + 2e⁻→H₂ E? =0.00V

Since the second redox couple has less standard reduction potential than the first so it can be concluded that the redox couple is a stronger oxidizing agent than H⁺/H₂ and copper cannot displace H₂ from acid.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (iv)

The electrolysis of aqueous sodium chloride is as follows

NaCl→Na⁺ + Cl⁻

H₂O→H⁺ + OH⁻

At the cathode, the reaction goes this way,

H₂O + e⁻ → $\frac{1}{2}$ H₂ + OH⁻

At the anode, the reaction goes this way,

Cl⁻ (aq)→ $\mathrm{}\frac{1}{2}$ Cl₂ (g)+ e⁻ E? _Cell =1.36V

2H₂O (g)→O₂ (g) + 4H⁺ (aq) + 4e⁻ E? _Cell =1.23V

At the anode, the reaction with a lower E? value will be preferred and oxidation of oxygen is a slow process and requires high voltage so the first reaction will take place.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (ii)

Given: Salt ammonium hydroxide

Apply Kohlrausch Law 

Kohlrausch law of states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte

Λ⁰_{m (NaOH) =}  Λ⁰_{m (NH4}⁺₎ + Λ⁰_{m (OH}^-₎ + Λ⁰_{m (Na}⁺₎ + Λ⁰_{m (OH}^-₎ - Λ⁰_{m (Na}⁺₎- Λ⁰_{m (Cl}^-₎

Λ⁰_{m (NH4OH)}  = Λ⁰_{m (NH4OH)}  + Λ⁰_{m (NaOH)} – Λ⁰ _(NaCl)

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (i)

In the lead storage battery, a reverse reaction occurs and lead sulfate is formed at the anode, and at the cathode, it is converted into lead and lead oxide respectively. The reaction is as follows

At the cathode, the reaction goes this way,

PbSO₄ (s)+2e⁻→Pb (s)+SO₄²⁻ (aq), Reduction process

At the anode, the reaction goes this way,

PbSO₄ (s)+2H₂O→PbO₂ (s)+SO₄²⁻+2H⁺ +2e, oxidation process

The overall reaction is as follows;

2PbSO₄ (s)+2H₂O→Pb (s)+PbO₂ (s)+4H+ (aq)+2SO₄²⁻ (aq)

Therefore, at anode lead sulfate is reduced to lead.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (iv)

The cell constant is described as the ratio of the length of the object and the area of cross-section. It can be written as follow

G= $\frac{\mathrm{l}}{\mathrm{A}}$

As l and A remains constant for an object it can be inferred that the cell constant for a conductivity cell remains constant for a cell.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Correct Option: (iii)

 Ans: Given: One mole of aluminium

Calculate the number of electrons required to convert Al₂O₃ to Al

It is the charge required to obtain one mole of aluminium from Al₂O₃

The dissociation of aluminium oxide is as follows;

Al₂O₃→2Al³⁺+3O²⁻

Al³⁺+3e⁻→Al

Therefore 3Fcharge is required to obtain one mole of aluminum from Al₂O_3.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (i)

According to the given table, has the most negative value among the given species. So

Cr³⁺ is the most stable oxidized species.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (iv)

is the highest positive value among the given species. So Mn²⁺ is the most stable ion in its reduced form.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (iii)

According to the electrochemical series, the lower the reduction potential higher is the reducing power.

Therefore, the order of reducing power is Mn²⁺ < Cl < Cr³⁺ < Cr