Ncert Solutions Chemistry Class 12th

Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Anode: Lead (Pb)

Cathode: a grid of lead packed with lead oxide (PbO₂)

Electrolyte: 38% solution of sulphuric acid (H₂SO₄)

The cell reactions are as follows :

Pb (s) + SO₂^-4 (aq) ⇒ PbSO₄ (s) + 2e^- (anode)

PbO₂ (s) + SO₂^-4 (aq) + 4H⁺ (aq) +2e^-⇒ PbSO₄ (s) +2H₂O (l) (cathode)

Pb (s) + PbO₂ (s) +2H₂SO₄ (aq)⇒ 2PbSO₄ (s) +2H₂O (l)

(overall cell reaction)

On charging, all these reactions will be reversed.

14.2

Lactose is a disaccharide carbohydrate (made up of two monosaccharide units) composed of β-D-galactose and β-D-glucose units. Hydrolysis breaks the glycosidic bond converting sucrose into β-D- galactose and β-D-glucose.

NOTE: But however, this reaction is so slow that it takes years for the solution of sucrose to undergo negligible change. Hence an enzyme called sucrase is added to proceed rapidly.

Cr₂O₇^2– + 14H⁺ + 6e^–⇒ 2Cr³⁺ + 7H₂O

A 3.12 

Cr₂O₇^2– + 14H⁺ + 6e^–⇒ 2Cr³⁺ + 7H₂O

For reducing one mole of Cr₂O₇^2–, 6 mole of electrons are required. Hence, 6 Faraday charges is needed. Hence, 6F = 6*96487 = 578922 C. Thus, the quantity of electricity is needed is 578922 C.

Metals with greater reactivity can be extracted electrolytically. Sodium, potassium, calcium, lithium, magnesium, aluminium which are present in the top of the reactivity series are extracted electrolytically. 

Current I = 0.5A
Time t = 2hrs = 2*60*60 = 7200 seconds
Charge Q = I * t
Q = 0.5*7200 = 3600 C
Charge carried by 1 mole of electrons (6.023*10²³electrons) is equal to 96487C.
No of electrons = 6.023*10²³ * 3600/96487
No of electrons = 2.25*10²² electrons

C = 0.025 mol L^-1
A_m = 46.1 Scm² mol L^-1
λ⁰ (H⁺) = 349.6 Scm² mol L^-1
λ⁰ (HCOO^-) = 54.6 Scm² mol L^-1
Λ⁰m (HCOOH) = Λ⁰ (H⁺) + Λ⁰ (HCOO^-) 
= 349.6 + 54.6
= 404.2 S cm² mol L^-1

^{Now, the degree of dissociation:}

The conductivity of a solution depends on the amount of ions present per volume of the solution. When diluted, the concentration of the ions decreases which implies that the number of ions per volume decreases thus, in turn, conductivity decreases.