NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry Intext Questions

Chapter 2 Electrochemistry Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download

Page No. 36

2.1. How would you determine the standard electrode potential of the system Mg²⁺ | Mg?

Solution

The standard electrode potential of Mg²⁺ | Mg can be measured with respect to the standard hydrogen electrode, represented by,

Pt_(s), H_2(g) (1 atm) | H⁺_(aq)(1 M).

A cell, consisting of Mg | MgSO₄ (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.

Mg|Mg²⁺(aq,1M) || H⁺(aq,1M) | H₂(g,1bar), Pt(x)

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

E^θ = E_R^θ - E_L^θ

Here, E_R^θ or the standard hydrogen electrode is zero.

∴ E^θ = 0 - E_L^θ

= -E_L^θ

2.2. Can you store copper sulphate solutions in a zinc pot?

Solution

Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution

Zn + CuSO₄ → ZnSO₄ + Cu

Hence, copper sulphate solution cannot be stored in a zinc pot.

 

2.3. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Solution

Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.

Fe²⁺ → Fe³⁺ + e^-1; E^θ = -0.77V

This implies that the substances having higher reduction potentials than +0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are F₂, Cl₂, and O₂.

 

Page No. 41

2.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Solution

For hydrogen electrode,

It is given that pH = 10

∴ [H⁺] = 10^-10 M

Now, using Nernst equation,

= −0.0591 log 10¹⁰

= −0.591 V

2.5. Calculate the emf of the cell in which the following reaction takes place: 

Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s)

Given that E^(-)_(cell) = 1.05 V

Solution

Applying Nernst equation we have:

= 1.05 − 0.02955 log 4 × 10⁴

= 1.05 − 0.02955 (log 10000 + log 4)

= 1.05 − 0.02955 (4 + 0.6021)

= 0.914 V

2.6. The cell in which the following reactions occurs:

2Fe³⁺ (aq) + 2I^– (aq) → 2Fe²⁺ (aq) + I₂ (s) has E^°_cell = 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution

Here, n = 2, E^θ_cell = 0.236 T = 298 K

We know that,

△_rG^θ = -nFE^θ_cell

= −2 × 96487 × 0.236

= −45541.864 J mol⁻¹

= −45.54 kJ mol⁻¹

Again, △_rG^θ = −2.303 RT log K_c

= 7.981

∴K_c = Antilog (7.981)

= 9.57 × 10⁷

Page No. 51

2.7. Why does the conductivity of a solution decrease with dilution?

Solution

The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsibility for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

2.8. Suggest a way to determine the Λ°_m value of water.

Solution

Applying Kohlrausch’s law of independent migration of ions, the Λ°_m value of water can be determined as follows:

Λ°_m(H2O) = λ⁰_H+ + λ⁰_OH-

= (λ⁰_H+ + λ⁰_Cl-) + (λ⁰_Na+ + λ⁰_OH-) – (λ⁰_Na+ + λ⁰_Cl-)

= Λ⁰_m(HCl) + Λ⁰_m(NaOH) - Λ⁰_m(NaCl)

Hence, by knowing the Λ⁰_m values of HCl, NaOH and NaCl, the Λ⁰_m value of water can be determined.

2.9. The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation and dissociation constant Given λ^°(H⁺)=349.6 S cm² mol^-1 and λ^°(HCOO-) = 54.6 S cm² mol^-1.

Solution

C = 0.025 mol L⁻¹

Λ_m = 46.1 Scm²mol^-1

λ⁰(H⁺) = 349.6 Scm²mol^-1

λ⁰(HCOO^-) = 54.6 Scm²mol^-1

Λ⁰_m (HCOOH) = λ⁰(H⁺) + λ⁰(HCOO^-)

= 349.6 + 54.4

= 404.2 Scm²mol^-1

Now, degree of dissociation

= 0.114 (approximately)

Thus, dissociation constant:

= 3.67×10^-4 mol L^-1

2.10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Solution

I = 0.5 A

t = 2 hours = 2 × 60 × 60 s = 7200 s

Thus, Q = It

= 0.5 A × 7200 s

= 3600 C

We know that 96487C = 6.023×10²³ number of electrons

Then,

= 2.25×10²² number of electrons

Hence, 2.25×10²² number of electrons will flow through the wire.

2.11. Suggest a list of metals that are extracted electrolytically.

Solution

Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

2.12. Consider the reaction: Cr₂O₇^2- + 14H⁺ + 6e- → 2Cr³⁺ + 7H₂O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇^2- ?

Solution

The given reaction is as follows:

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Therefore, to reduce 1 mole of Cr₂O₇^2-, the required quantity of electricity will be:

= 6 F

= 6 × 96487 C

= 578922 C

2.13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Solution

A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO₂) as the cathode, and a 38% solution of sulphuric acid (H₂SO₄) as an electrolyte.

When the battery is in use, the following cell reactions take place:

At anode: Pb(s) + SO₄^2-(aq) →PbSO₄(s) + 2e^-

At cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^- → PbSO₄(s) + 2H₂O(l)

The overall cell reaction is given by,

Pbs + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

When a battery is charged, the reverse of all these reactions takes place.

Hence, on charging, PbSO4(s) present at the anode and cathode is converted into Pbs and PbO2(s) respectively.

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

Solution

Methane and methanol can be used as fuels in fuel cells.

2.15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Solution

In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe(s) → Fe²⁺(aq) + 2e^-

Electrons released at the anodic spot move through the metallic object and go to another spot of the object

There, in the presence of H⁺ ions, the electrons reduce oxygen. This spot behaves as the cathode. These H⁺ ions come either from H₂CO_3, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.

The reaction corresponding at the cathode is given by,

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide (Fe₂O₃,xH₂O)i.e., rust.

Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.

O₂(g) + 4H⁺(aq) + 4e^- → 2H₂O(l)

The overall reaction is:

2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l)