Ncert Solutions Chemistry Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (ii, iii)

At equilibrium, ?G $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

-nFE $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{\mathrm{n}\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

For Daniel cell, n=2

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

At equilibrium, E $°$ = 1.1V

1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$

(i) Since ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  this option is incorrect

(ii) As derived 1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

(iii) As derived _{${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$} so this option is correct

(iv) As ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  _,so this option is incorrect

This is a Multiple Choice Questions as classified in NCERT Exemplar

Correct options: A and D

Activation Energy is the amount of energy released when reactant molecules collide and form an activated complex. When energy is released, the complex decomposes into a product.

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

Correct options: A, C and D

The pressure in this reaction is extremely high, and it becomes independent of the ammonia concentration. The metal surface becomes saturated with gas molecules when the rate of reaction=Rate constant. The rate of a zero-order reaction is independent of the concentration of reactants in the reaction.

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

Correct options: A and D

A complex reaction is one that does not happen in a single step.

The total number of molecules involved in the slowest step of the reaction determines the overall reaction rate.

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)

Correct options: A and B

This is a Multiple Choice Questions as classified in NCERT Exemplar

Unimolecular reaction: Unimolecular means relating to a single molecular reaction that describes decomposition of a single reactant.

If product A→ (B)

rate = k [A]

Example Na₂O₄ (g)→2No₂ (g)

Order = 1

Molecularity = 1

The slowest step is the rate determining which determines the Order and Molecularity.

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.