Class 12th

14.17 What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO3

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14.17

When D-glucose is heated and treated with HI for a long period of time, then n-hexane is formed, which shows that all the six-carbon atoms are linked in a straight

2. When D-glucose is treated with Br₂ water i.e., bromine water which is a mild oxidising agent, then we get D-gluconic acid as one of the product. This reaction assures the presence of carbonyl group which is available as an aldehydic group.

3. On being treated with HNO₃, D-glucose get oxidised which gives saccharic acid as final Saccharic acid is a di-carboxylic acid. This reaction confirms the presence of alcoholic group (- OH) in the glucose.

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3.33 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes.

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Given -

(i) All the ions are in aqueous state.

Reaction in solution:

AgNO_{3 (s)} + aq → Ag ⁺ + NO³^-

H₂O → H ⁺ + OH ^-

At cathode:

Ag + (aq) + e - →Ag (s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

Ag (s)→ Ag + (aq) + e -

As Ag anode is attacked by NO³^- ions, Ag of the anode will dissolve to form Ag ⁺ ions in the aqueous solution.

(ii) Reaction in solution:

AgNO_{3 (s)} + aq → Ag ⁺ + NO³^-

H₂O óH ⁺ + OH ^-

At cathode:

2Ag ⁺_(aq) + 2e ^- →2Ag_(s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference

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3.32 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe 3+(aq) and I – (aq) (ii) Ag+ (aq) and Cu(s) (iii) Fe3+ (aq) and Br– (aq) (iv) Ag(s) and Fe 3+ (aq) (v) Br 2 (aq) and Fe2+ (aq).

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The electrode reaction is written as,

2Fe³⁺ + 2I ^- → 2Fe²⁺ + I₂

= 0.54V - 0.77V

∴ E⁰_cell = - 0.23 V

It is not feasible, as E⁰_cell is negative, ∴ ?G⁰ is positive.

The electrode reaction is written as,
2Ag⁺ (aq) + Cu(s)→ Cu²⁺ (aq) + Ag(s)

= + 0.80V - 0.34V

∴ E⁰_cell = 0.46V

It is feasible, as E_cell ⁰ is positive, ∴ ?G⁰ is negative.

(iii) The electrode reaction is written as,
2Fe³⁺ (aq) + 2Br^- (aq)→ 2Fe²⁺ (aq) + Br₂

= 0.77V - 1.09V

∴ E⁰_cell = - 0.32 V

It is not feasible, as E⁰_cell is negative, ∴ ?G⁰ is positive.

(iv) The electrode reaction is written as,
Ag(s) + Fe³⁺ (aq) → Fe²⁺ (aq) + Ag⁺ (aq)

= 0.77V - 0.80V

∴ E⁰_cell = -

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14.16 What is the basic structural difference between starch and cellulose?

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14.16

Starch consists of two components – amylase and amylopectin. Amylose is a long linear chain of α–D-(+)-glucose units joined by C₁-C₄glycosidic linkage (α -link).

Amylopectin is a branched-chain polymer of –D-glucose units, in which the chain is formed by C1-C4 glycosidic linkage and the branching occurs by C₁-C₆ glycosidic linkage.

On the other hand, cellulose is the main structural material of tree and other plants. Wood is 50% cellulose, while cotton wool is almost pure cellulose. It is linear chain natural polymers of β-D-glucose units joined by 1, 4-glycosidic linkage (natural linear polymers).

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14.15 What are the hydrolysis products of (i) Sucrose and (ii) Lactose ?

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14.15

Hydrolysis is the process of using water to break down a molecule into two parts. It is usually a type of decomposition reaction where one reactant is water, where water is used to break chemical bonds in the other reactant. It can be considered as reverse of a condensation reaction.

The general formula of a hydrolysis reaction is:

XY + H₂O → XH + YOH

(i) On hydrolysis with dilute acids, sucrose yields an equimolecular mixture of α –D glucose and β–D- fructose.

(ii) The hydrolysis of lactose gives β–D-galactose and β–D-glucose as final products.

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3.31 Three electrolytic cells A, B, C containing solutions of ZnSO4 , AgNO3 and CuSO4 , respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

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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

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3.30 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

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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

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3.29 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2 ? (ii) 1 mol of FeO to Fe2 O3 ?

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

H₂O → H₂ + 1/2O₂

i.e., O²^- →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

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14.14 What is glycogen? How is it different from starch?

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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.

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3.28 How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2?

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