Ncert Solutions Chemistry Class 12th

Equivalent weight is Ag, E_Ag = 180/1  = 180

Equivalent weight is Cu, E_Cu = 63.5 / 2 = 31.75

Equivalent weight is Zn, E_Zn= 65/2 = 32.5

Using Faraday's second law of electrolysis, to find the mass of Cu and Zn, we use Equation 1,

∴ W_Zn = 0.436 g

To find the time of current flow, using Faraday's first law of electrolysis we get,

M = Z *I *t ⇒ Equation 2

? Z = Equivalent Weight / 96487, Equation 2 becomes,

M = 108 / 96487 X 1.5 X t

t = 1.45 X 96487 / 108X 1.5

t = 864 seconds.

The time of current flow, t = 864 seconds, the mass of Cu is 0.426 g and mass of Zn is 0.436 g

Quantity of electricity passed = 5 A * (20 * 60 sec)
= 6000 C ⇒ Equation 1
The electrode reaction is written as,
Ni²⁺ + 2e → Ni
Thus, the quantity of electricity required = 2F
= 2*96487 C
= 192974 C
? 192974 C of electricity deposits 1 mole of Ni, which is 58.7 g ⇒ Equation 2
Thus, equating equations 1 and 2, we get

192974 C of electricity deposits = 58.7 g
6000 C of electricity will deposit = 58.7 X 6000 / 192974
= 1.825g of Ni
The mass of Ni deposited at the cathode is 1.825g of Ni

(i) The electrode reaction for 1 mole of H₂O is given as,

H₂O → H₂ + 1/2O₂

i.e., O^2- →1/2 O₂ + 2e ^-

∴ The quantity of electricity required = 2F

= 2*96487 C

= 192974 C

The quantity of electricity required in coulomb for the oxidation of 1 mol of H₂O to O₂ is 192974 C

(ii) The electrode reaction for 1 mole of FeO is

FeO + 1/2 O₂ → 1/2 Fe₂O₃

i.e., Fe²⁺ → Fe³⁺ + e ^-

∴ The quantity of electricity required = 1F

= 1*96487 C

= 96487 C

The quantity of electricity required in coulomb for the oxidation of 1 mol of FeO to Fe₂O₃ is 96487 C

14.14

Glycogen is a polysaccharide-type of carbohydrate. In animals, carbohydrates are stored as glycogen.

But starch is a carbohydrate which consists of two components –amylase (15 -20 %) and amylopectin (80 – 85%). However, glycogen is also like amylopectin but branching will take place after every 5 to 6 glucose unit. Also, glycogen is highly branched.

(i) Ca²⁺ + 2e^- → Ca
⇒ Here, 1 mole of Ca, i.e., 40g of Ca requires = 2 F electricity (F if Faraday)
∴ 20g of Ca requires = 20X2/40
= 1 F of electricity
Electricity in terms of Faraday required to produce 20.0 g of Ca from molten CaCl₂ is 1 F of electricity.

(ii) Al³⁺ + 3e^-  → Al
⇒ 1 mole of Al, i.e., 27g of Al requires = 3 F electricity (F if Faraday)
∴ 40.0 g of Al will require = 3/27 X 40
= 4.44 F of electricity
Electricity in terms of Faraday required to produce 40.0 g of Al from molten Al₂O₃ is 4.44 F of electricit

14.13 

The condensation of the hydroxyl group of two monosaccharides to form a link between them is called glycosidic linkage. In other words, it refers to linkage developed between two different monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, α-glucose and β–fructose, are joined together by a glycosidic linkage.

The electrode reaction is given as,  

Al³⁺ (aq) + 3e^- → Al (s)

∴ The quantity of charge required for the reduction of 1 mol of Al³⁺ = 3F

= 3*96487 C

= 289461 C

The electrode reaction is given as,

Cu²⁺ (aq) + 2e^- → Cu (s)

∴ The quantity of charge required for the reduction of 1 mol of Cu²⁺ = 2F

= 2*96487 C

= 192974 C

The electrode reaction is given as, MnO₄→ Mn²⁺

i.e., Mn⁷⁺ + 5e ^- → Mn²⁺

∴ The quantity of charge required for the reduction of 1 mol of Mn⁷⁺ = 5F

= 5*96487 C

= 482435 C

14.12 

i. Ribose, 2-deoxyribose, galactose, and fructose are

Monosaccharides are the simplest units of carbohydrates which cannot be hydrolyzed into simpler compounds.

ii. Maltose and lactose are

A disaccharide is a carbohydrate that is formed when two monosaccharides are joined together and a molecule of water is removed from the structure. Lactose is a disaccharide formed from the combination of galactose and glucose.