Explore topic-wise InterviewSolutions in .

(d) I & IV is correct 

(c) 1.10V

Anodic oxidation: (Reverse the given reaction)

(E_ox⁰) = 0.76V cathodic reduction

E_cell⁰ = (E_ox⁰) + (E⁰_red) = 0.76 + 0.34 = 1.1V

i.   b c d a

(c) Galvanisation

The value of cell emf of Mercury button cell is 1.35 V

i.   c a d b

(a) HgO mixed with graphit

(c) Both a & b

(d) NH₄CI + ZnCI₂

(a) Porous graphite

(c) Lead storage battery

(b) Pacemaker. In pacemaker mercury button cell is used whereas in other three, lead storage battery is used up

(d) R is the odd one, and other three represents Ohm’s law

(b) NaCl_(solid)

In solid state, NaCl does not dissociate into ions so it does not conduct electricity

(b) ΔG° = – nFE°_{​​​​​​cell​}

i.   c d a b

(c) 6.25 x 10^-4 Sm² mol^-1 

ii.  c a d b

iii.  d c a b

iv.  c a d b

(a) Y will oxidize X and not Z

2H₂O→ 2O₂ +4 H⁺ + 4e^- 

4 F of electricity produced O₂ = 32 

 1 F of electricity produced O₂ =32/4 = 8  

15.8 grams of copper

(a) (i) only

Hg(NO₃)₂ = 200.6g

(d) Electronic watch. in this mercury button cell is used whereas in others Li-ion battery is used up.

(a) O₂ (g) + 2H₂O (l) + 4e^- → 4OH^- (aq) 

Oxidation at anode: 

2H₂ (g) + 4OH^- (aq) → 4H₂O (1) + 4e^- 

1 mole of hydrogen gas produces 2 moles of electrons at 25°C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres

∴no. of moles of hydrogen gas produced = \(\frac{1mole}{22.4litres}\) x 44.8 litres = 2 moles of hydrogen

∴2 of moles of hydrogen produces 4 moles of electro 

i.e., 4F charge. We know that Q = It

I = \(\frac{Q}{t}\,=\,\frac{4F}{10mins}\,=\,\frac{4\times96500}{10x60s}\)

I = 643.33 A

Electro deposition of copper 

Cu²⁺_(aq) + 2e^- → Cu_(s) 

2F charge is required to deposit 1 mole of copper i.e., 63.5 g 

If the entire current produced in the fuel cell i.e., 4 F is utilised for electrolysis, then 2 x 63.5 i.e., 127.0 g copper will be deposited at cathode.

In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively. 

Oxidation occurs at the anode:

2H_2(g) + 4OH^-_(aq) → 4H₂O_(l) + 4e^- 

Reduction occurs at the cathode O₂(g) + 2 H₂O(l) + 4e^- → 4 OH^- (aq) 

The overall reaction is 2H₂ (g) + O₂ (g) → 2H₂O(l) 

The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H₂ – O₂ fuel cell. Other fuel cell like propane – O₂ and methane O₂ have also been developed.

In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. Al, Zn and Mg are used as sacrificial anodes.

1. The watch cells are made up of mercury. This mercury will pollute the water. Water contaminated with mercury leads to accumulation of mercury in the body of fishes and other aquatic life. 

2. It helps in keeping the environment safe from pollution due to mercury.

(a) Faraday's first law of electrolysis states that the amount of chemical reaction which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed, through, the electrolyte (solution or melt).

(b) According to Kohlrausch law of independent migration of ions limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

(a) Both A and R are correct

Cell constant is the ratio of the distance between the electrodes (l) and the area of cross-section (A). It is denoted by 1. Its unit is cm^-1 . Its SI unit is m^-1

(b) (iii) only

(a) Oxidised

One Faraday is defined as the charge of one mole of electron.

Charge of one electron = 1.6 x 10^-19 C

Charge of 1 mole of electrons = 6.023 x 10²³ x 1.602 x 10^-19 C

= 6.023 x 10²³ x 1.602 x 10^-19 C 

= 96488 C 

i.e., IF ≃ 96500C

(c) anode to cathode through the external circuit

The maximum work that can be obtained from a galvanic cell is – nFE

Galvanic Cell

1. It is a device in which a spontaneous chemical reaction generates an electric current. 

2. It converts chemical energy into electrical energy. It is commonly known as Battery. 

3. e.g., Daniel cell, Dry cell. 

4. A salt bridge is used in this.

Electrolytic cell 

1. It is a device in which an electric current from an external source drives a non spontaneous reaction 

2. It converts electrical energy into chemical energy. 

3. e.g., Electrolysis of molten NaCI. 

4. Na salt bridge is used.

(c) 0.66 g

2Cl^- → Cl₂ + 2e^- 

Q = It. 

Amount of current passed 

= 1 x 30 x 60 = 1800C 

The amount of Cl₂ liberated by passing 1800 coulomb of electric charge

= \(\frac{1\times1800\times71}{2\times96500}\)

= 0.66g

I = 0.15 amperes 

t = 150 mins 

= t = 150 x 60sec 

= t = 9000sec 

Q = It 

= Q = 0.15 x 9000 coulombs 

= Q = 1350 coulombs 

Hence, 135 coulombs of electricity deposit is equal to 0.783g of metal.

96500 coulombs of electricity \(\frac{0.783\times96500}{1350}\) = 55.97 gm of metal 

Hence equivalent mass of the metal is 55.97

(d) 1.74 x 10^-12

(b) 1.107 Volts

Cells that converts the energy of combustion of fuels directly into electrical energy.

(i) Mercury cell,

(ii) Fuel cell,

(iii) Lead storage cell,

(iv) Dry cell.

Fuel cells are the cells which converts energy of combustion of fuel directly into electricity.

Example H₂ - O₂ fuel cell.

(a) Electrons flow from Zn to Ag plate.

(b) Zn as anode and Ag acts as cathode

(c) Cell will stop functioning

(d) Concentration of Zn²⁺  ions will increase and that of Ag⁺ ion will decrease.

(e) No change

As corrosion is a phenomenon of oxidation of iron considering the oxidation potentials of all the elements is essential. Element with higher oxidation potential than Fe will oxidise faster than iron preventing corrosion in iron. 

Oxidation potential of Fe = 0.44 V 

Oxidation potential of A = 2.37 V 

Oxidation potential of B = 0.14 V 

As A has higher oxidation potential than iron, it can be used for coating the surface of iron.

A is better than B because its E° value is more negative.