NCERT Chemistry Solutions for Class 12 Chapter 3 – Electrochemistry

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The class 12 chemistry chapter 3 NCERT solutions help students understand the basic concepts of solution and give a detailed explanation of various laws and formulas under the chapter. The NCERT Solution helps students study for the exams without depending on someone to explain. Students may refer to NCERT Solutions Class 12 Chemistry Chapter 3 for free on Extramarks.

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Electrochemistry Class 12 NCERT Solutions Chapter 3 Chemistry

NCERT Solutions Class 12 Chemistry Chapter 3 Electrochemistry explains the pivotal concept of the chapter. Furthermore, the latest term-wise CBSE syllabus 2023-2024 ensures that the content included is appropriate for the students to proceed in their respective streams.

Electrochemistry is the learning of the exchange between electrical and chemical energy. It has critical applications in daily life, stretching from the battery that controls your portable radio devices to the electrorefining that produces the copper pipes carrying your drinking water. These electrochemical processes utilize oxidation and reduction reactions. An oxidation reaction involves the loss of one or more electrons from a chemical species. A reduction reaction is the gain of one or more electrons by a chemical species. Every detail regarding electrochemistry that a student needs is available on the Extramarks website under electrochemistry Class 12 NCERT Solutions Chapter 3 Chemistry.

The Electrochemistry class 12 NCERT Solutions Chapter 3 Chemistry is a comprehensive material that has answers to the exercise present in the NCERT textbook. The NCERT Solutions have been provided in a clear and stepwise process for ease of understanding. For example, at Extramarks, we start with the electrical properties of conductors and their types and continue with their functions, applications, and relationship with temperature. Next, we will explain electrode potential, batteries, fuel cells, and corrosion. So let us start an electrifying tour experience named Electrochemistry with NCERT Solutions Class 12 Chemistry Chapter 3.

NCERT Solutions for Class 12 Chemistry

Chapter 1 – The Solid State

Chapter 2 – Solutions

Chapter 4 – Chemical Kinetics

Chapter 5 – Surface Chemistry

Chapter 6 – General Principles and Processes of Isolation of Elements

Chapter 7 – The p-Block Elements

Chapter 8 – The d and f Block Elements

Chapter 9 – Coordination Compounds

Chapter 10 – Haloalkanes and Haloarenes

Chapter 11 – Alcohols, Phenols and Ethers

Chapter 12 – Aldehydes, Ketones and Carboxylic Acids

Chapter 13 – Amines

Chapter 14 – Biomolecules

Chapter 15 – Polymers

Chapter 16 – Chemistry in Everyday life

Key Topics Covered In NCERT Solutions Class 12 Chemistry Chapter 3

Some of the key topics covered under Electrochemistry Class 12 NCERT Solutions Chapter 3 Chemistry are mentioned in the table below.

Unit	Topic
Ch 3.1	Introduction to an electrochemical cell
Ch 3.2	Half cells and cell potential
Ch 3.3	Primary and secondary cells
Ch 3.4	Types of the electrochemical cell
Ch 3.5	Applications of Electrochemical cells
Ch 3.6	Nernst Equation

A brief of the key topics under class 12 Electrochemistry NCERT Solutions Chapter 3 Chemistry

Introduction to an electrochemical cell

An electrochemical cell is a device that can produce electrical energy from the chemical reaction produced in it. These devices can convert chemical energy into electrical energy or vice versa. A simple example of an electrochemical cell is a standard 1.5V cell used to power many electrical appliances such as tv remotes, toys, watches. Such cells capable of producing an electric current from the chemical reactions are called Galvanic or Voltaic cells. Those that create chemical reactions via electrolysis are known as electrolytic cells. Students quickly understand electrochemical cells by class 12 electrochemistry NCERT Solutions Chemistry Chapter 3.

Detailed diagram of the electrochemical cell

Electrochemical cells normally consist of a cathode and an anode.

Cathode: It represents a positive sign since electrons are absorbed here. A reduction reaction always occurs in the cathode of an electrochemical cell.
Anode: It denotes a negative sign since electrons are released here. Oxidation reaction occurs here, and electrons move out of the anode. Students refer to NCERT Solutions Class 12 Chemistry Chapter 3.

The cathode must be represented on the right side in an electrochemical cell, whereas the anode on the left side.

Half cells and cell potential

Electrochemical cells connect two half cells, each consisting of an electrode dipped in the electrolyte. This same electrolyte can be applied to half cells. A salt bridge is connected with these half cells, which come up with a platform to form ionic contact between them, not allowing them to fuse. A good example of a salt bridge is a filter paper dipped in potassium nitrate solution Or NaCl solutions. Students may refer to NCERT Solutions Class 12 Chemistry

One of the half cells of the electrochemical cell loses electrons due to oxidation, and the other gains one electron in a reduction process. As a result, an equilibrium reaction occurs in both the half cells, and once the equilibrium reaches net voltage becomes zero, the cell stops producing electricity.

The tendency of an electrode in contact with an electrolyte to gain or lose an electron is explained by its electrode potential. Therefore, the value of these potentials can predict the overall cell potential. Commonly, the electrode potentials are measured with the help of the standard hydrogen electrode as a reference electrode (known electrode potential), given in NCERT Solutions Class 12 Chemistry Chapter 3.

Primary and secondary cells

Primary cells are used as “use and throw” galvanic cells. These electrochemical reactions that take place in this cell are irreversible. Therefore, the reactants are utilised to create electrical energy, and the cell is not producing an electric current once the reactants are consumed completely
Secondary cells are also called rechargeable batteries. The electrochemical reaction that takes place in these cells is reversible. The cell can be used as a Galvanic cell and an electrolyte cell. Students can understand further by referring to NCERT Solutions Class 12 Chemistry Chapter 3.

Types of the electrochemical cell

The electrochemical cells are of two types, Galvanic cells, and Electrolytic cells.

Galvanic cell: In a galvanic cell, a spontaneous reaction occurs, the chemical energy is converted into electrical energy. It is also known as the Voltaic cell or Daniel cell.
Electrolytic cell: The nonspontaneous redox reaction is carried out by electrical energy in an electrolytic cell.

This will help understand the brief knowledge of electrochemical cells under NCERT Solutions Class 12 Chemistry Chapter 3.

Applications of Electrochemical cells

Some of the applications include

Electrolytic cells are applied in the electrorefining of many non-ferrous metals. These are also utilisedin the electrowinning of these metals.
The production of high-purity metals such as Lead, Zinc, Aluminum, and Copper involves using electrolytic cells.
Metallic sodium can be extracted from molten NaCl by placing it in an electrolytic cell and passing an electric current.
Fuel cells are an essential class of electrochemical cells that serve as a source of clean energy in several remote locations.

The application of electrochemical cells is more defined in the NCERT Solutions Class 12 Chemistry Chapter 3.

Nernst Equation

The Nernst equation is named by” German physicist Walther Nernst”. This equation guarantees cell potential under nonstandard conditions, relates the measured potential to the reaction quotient, and permits the exact measurement of equilibrium constants.

Let us consider an electrochemical reaction shown in the following type

aA + bB ⇾ cC + dD

The equation can be written as follows.

Ecell = E–cell – RT/ nF lnQ

= E–cell – RT/ nF ln [C]c [D]d

—————-

[A]a [B]b

In the case of a Daniel cell, the Nernst equation is as follows

Ecell = E–cell – RT/ nF lnQ

E–cell – RT/ 2F ln [Zn2+]

—————-

[Cu2+]

This equation implies that the value increases with the increase in the concentration of Copper ions and decreases the concentration of Zn ions. Students may refer to NCERT Solutions Class 12 Chemistry Chapter 3 for a more detailed explanation.

Fuel cells: Galvanic cells that are designed to convert the energy of fuel combustion like hydrogen, methane, methanol, etc., directly into electrical energy are called fuel cells.

Corrosion: Metals react with atmospheric oxygen and produce metal oxides that are basic in nature because they react with water to form bases.

In the case of rusting of iron, the iron reacts with the oxygen present in air and moisture and develops rust (hydrated iron (III) oxide). More details of corrosion properties are shown in NCERT class 12 chapter 3 Electrochemistry.

4Fe + 3O2 + 2 H2O → 2Fe2O3

Students may refer to Extramarks NCERT Solutions Class 12 Chemistry Chapter 3 for more details on fuel cells and corrosion.

NCERT Solutions Class 12 Chemistry Chapter 3 is explained in detail by the subject matter experts at Extramarks. In addition to chapter 3, students can access NCERT Solution for all other Chemistry chapters of class 12. Furthermore, students can click on the links provided below to access the study material of different classes.

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Electrochemistry NCERT Solutions Class 12 Chemistry Chapter 3 Exercise & Answer Solutions

Students may refer to the Electrochemistry NCERT Solutions Class 12 Chemistry Chapter 3 Electrochemistry on Extramarks. The exercise and answer solutions are explained in detailed to help students understand the various concepts mentioned in the chapter. Every minute detail that a student may need to understand electrochemistry is mentioned in the NCERT solutions.

In addition to exercise and answer solutions, at Extramarks we provide sample question papers, past year question papers, important questions, revision notes, and more. Students may find all information under one channel, making it easier for them to study, especially during an examination.

Q.1 Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn

Ans.

The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Al, Zn, Fe, Cu

Q.2 Given the standard electrode potentials,
K⁺/K = −2.93V, Ag⁺/Ag = 0.80V,
Hg²⁺/Hg = 0.79V
Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = − 0.74V
Arrange these metals in their increasing order of reducing power.

Ans.

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K⁺/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

Q.3 Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Ans.

The galvanic cell in which the given reaction takes place is depicted as:

${Zn}_{(s)} ∥ {Zn}_{(aq)}^{2 +} ∥ {Ag}_{(aq)}^{+} ∥ {Ag}_{(s)}$

(i) Zn electrode (anode) is negatively charged.

(ii) Metal ions are carriers of current in the internal circuit and electrons act as current carriers in the external circuit,hence current will flow from silver to zinc.

(iii) The reaction taking place at the anode is given by,

${Zn}_{(s)} \to {Zn}_{(aq)}^{2 +} + 2_{e}^{_}$

The reaction taking place at the cathode is given by,

${Ag}_{(aq)}^{+} +_{e}^{-} \to {Ag}_{(s)}$

Q.4 Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd
(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Calculate the ∆rG^θ and equilibrium constant of the reactions.

Ans.

Q.5 Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s) | Mg²⁺(0.001M) || Cu²⁺ (0.0001 M) | Cu(s)
(ii) Fe(s) | Fe²⁺ (0.001M) || H⁺ (1M)|H₂(g)(1bar) | Pt(s)
(iii) Sn(s) | Sn²⁺ (0.050 M) || H⁺ (0.020 M) | H₂(g) (1 bar) | Pt(s)
(iv) Pt(s) | Br₂(l) | Br^– (0.010 M) || H⁺ (0.030 M) | H₂(g) (1 bar) | Pt(s).

Ans.

$\begin{array}{l} (i) For the given reaction, the Nernst equation can be given as : \\ E_{cell} = E_{cell}^{\emptyset} - \frac{0 .0591}{n} \log \frac{[{Mg}^{2 +}]}{[{Cu}^{2 +}]} \\ = {0 .34 - (- 2 .36)} - \frac{0 .0591}{2} \log \frac{. 001}{. 0001} \\ = 2 .7 - \frac{0 .0591}{2} \log 10 \\ = 2 .7 - 0 .02955 \\ = 2 .67 V (approximately) \\ (ii) For the given reaction, the Nernst equation can be given as : \\ E_{cell} = E_{cell}^{\emptyset} - \frac{0 .0591}{n} \log \frac{[{Fe}^{2 +}]}{{[H^{+}]}^{2}} \\ = {0 - (- 0 .44)} - \frac{0 .0591}{2} \log \frac{0 .001}{1^{2}} \\ = 0 .44 - 0 .02955 (- 3) \\ = 0 .52865 V \\ = 0 .53 V (approximately) \\ (iii) For the given reaction, the Nernst equation can be given as : \\ E_{cell} = E_{cell}^{\emptyset} - \frac{0 .0591}{n} \log \frac{[{Sn}^{2 +}]}{{[H^{+}]}^{2}} \\ = {0 - (- 0 .14)} - \frac{0 .0591}{2} \log \frac{0 .050}{{(0 .020)}^{2}} \\ = 0 .14 - 0 .0295 x \log 125 \\ = 0 .14 - 0 .062 \\ = 0 .078 V \\ = 0 .08 V (approximately) \\ (iv) For the given reaction, the Nernst equation can be given as : \\ E_{cell} = E_{cell}^{\emptyset} - \frac{0 .0591}{n} \log \frac{1}{{[{Br}^{-}]}^{2} {[H^{+}]}^{2}} \\ = (0 - 1 .09) - \frac{0 .0591}{2} \log \frac{1}{{(0 .010)}^{2} {(0 .030)}^{2}} \\ = - 1 .09 - 0 .02955 \times \log \frac{1}{0 .00000009} \\ = - 1 .09 - 0 .02955 \times \log \frac{1}{9 \times 10^{- 8}} \\ = - 1 .09 - 0 .02955 \times \log (1 .11 \times 10^{7}) \\ = - 1 .09 - 0 .02955 (0 .0453 + 7) \\ = - 1 .09 - 0 .208 \\ = - 1 .298 \end{array}$

Q.6 In the button cells widely used in watches and other devices the following reaction takes place.Zn(s)+ Ag2O(s)+ H2O(l)→Zn2+(aq)+ 2Ag(s)+ 2OH−(aq)Determine=ΔrGϕ and Eϕ for the reaction.

Ans.

$\begin{array}{l} {Zn}_{(s)} \to {Zn}_{(aq)}^{2 +} + 2_{e}^{-}; E^{\emptyset} = 0 .76 V \\ {Ag}_{2} O_{(s)} + H_{2} O_{(l)} + 2_{e}^{-} \to 2 {Ag}_{(s)} + 2 {OH}_{(aq)}^{^{-}}; E^{\emptyset} = 0 .344 V \\ ____________________________________________________ \\ {Zn}_{(s)} + {Ag}_{2} O_{(s)} + H_{2} O_{(l)} \to {Zn}_{(aq)}^{2 +} + 2 {Ag}_{(s)} + 2 {OH}_{(aq)}^{-}; E^{\emptyset} = 1 .104 V \\ ∴ E^{\emptyset} = 1 .104 V \\ We know that, \\ Δ_{r} G^{\emptyset} = - {nFE}^{\emptyset} \\ = - 2 \times 96487 \times 1 .04 \\ = - 213043 .296 J \\ = - 213 .04 kJ \end{array}$

Q.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans.

Conductivity of a solution is defined as the conductance of an electrolyte in which electrodes are unit distance apart and have area of cross section between them as unity. If ρ is resistivity, then we can write:

$κ = \frac{1}{ρ}$

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

$i . e ., G = k \frac{a}{l} = k . 1 = k$

(Since a = 1, l = 1)

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity:

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

$\land_{m} = k \frac{A}{l}$

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

$∴ \land_{m} = kV$

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

The variation of ∧_m with √C for strong and weak electrolytes is shown in the following plot:

Q.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.

Ans.

$\begin{array}{l} \begin{array}{l} Given, \\ κ = 0 .0248 S {cm}^{- 1} \\ c = 0 .20 M \end{array} \\ ∴ Molar conductivity, \land_{m} = \frac{k \times 1000}{c} \\ = \frac{0 .0248 \times 1000}{0 .2} \\ = 124 S c m^{2} {mol}^{- 1} \end{array}$

Q.9 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M 0.001 0.010 0.020 0.050 0.100

10² × κ/S m^-1 1.237 11.85 23.15 55.53 106.74

Calculate ∧_m for all concentration and draw a plot between ∧_m and c½. Find the value of ∧_m⁰.

Ans.

Given,
κ = 1.237 × 10-2 S m^-1, c = 0.001 M
Then, κ = 1.237 × 10^-4 S cm^-1, c½ = 0.0316 M^1/2
∴∧_m = κ/C
=(1.237 x 10^-4 Scm^-1 x 1000cm³/ 0.001 mol L^-1 x L
=123.7 Scm² L^-1
Given,
κ = 11.85 × 10-2 S m-1, c = 0.010M
Then, κ = 11.85 × 10-4 S cm-1, c½ = 0.1 M^1/2
∴∧_m = κ/C
=118.5 Scm² L^-1
Given,
κ = 23.15 × 10-2 S m-1, c = 0.020 M
Then, κ = 23.15 × 10^-4 S cm^-1, c^1/2 = 0.1414 M^1/2
∴∧_m= κ/CL
=115.8 Scm² L^-1
Given,
K = 55.53 × 10-2 S m^-1, c = 0.050 M
Then, κ = 55.53 × 10^-4S cm^-1, c^1/2 = 0.2236 M^1/2
∴∧_m= κ/C
=111.1 1 Scm² L^-1
Given,
κ = 106.74 × 10^-2 S m^-1, c = 0.100 M
Then, κ = 106.74 × 10^-4 S cm^-1, c^1/2 = 0.3162 M^1/2
∴∧_m= κ/C
=106.74 Scm² L^-1
Now, we have the following data:

C^1/2 ⁄M^1/2 0.0316 0.1 0.1414 0.2236 0.3162

∧_m (Scm²mol^-1) 123.7 118.5 115.8 111.1 106.74

Q.10 Conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if ∧_m⁰ for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?

Ans.

$\begin{array}{l} {Given, κ = 7.896 × 10}^{–5} {Scm}^{–1} \\ {c = 0.00241 mol L}^{–1} \\ Then, molar conductivity, Λ_{m} = \frac{K}{C} \\ = \frac{{7.896 × 10}^{–5} {Scm}^{–1}}{{0.00241 mol L}^{–1}} × \frac{{1000 cm}^{3}}{L} \\ = 32 .76 {Scm}^{2} {mol}^{- 1} \\ Again, Λ_{m}^{0} = 390 .5 {Scm}^{2} {mol}^{- 1} \\ α = \frac{Λ_{m}}{Λ_{m}^{0}} = \frac{{32.76 Scm}^{2} {mol}^{–1}}{{390.5 Scm}^{2} {mol}^{–1}} \\ Now, \\ = 0 .084 \\ Dissociation constant, K_{a} = \frac{{cα}^{2}}{(1 - α)} \\ = \frac{({0.00241 molL}^{–1}) {(0.084)}^{2}}{(1 – 0.084)} \\ = 1 .86 \times 10^{- 5} {molL}^{- 1} \end{array}$

Q.11

Ans.

$\begin{array}{l} {(i) Al}^{3 +} {+3}_{e}^{–} \to Al \\ ∴ Required charge = 3F \\ = 3 × 96487 C \\ = 289461C \\ (ii) {Cu}^{2 +} {+2}_{e}^{–} \to Cu \\ ∴ Required charge = 2 F = 2 × 96487 C \\ = 192974C \\ {(iii) MnO}_{4}^{-} \to {Mn}^{2 +} \\ i .e {., Mn}^{7 +} {+ 5}_{e}^{-} \to {Mn}^{2 +} \\ Required charge = 5 F \\ = 5 × 96487 C \\ = 482435C \end{array}$

Q.12 How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl₂.
(ii) 40.0 g of Al from molten Al₂O₃.

Ans.

$\begin{array}{l} (i) According to the question, \\ {Ca}^{2 +} + 2e – \to Ca \\ 40g \\ Electricity required to produce 40g of calcium = 2F \\ Electricity required to produce 20g of calcium = \frac{2 × 20}{40} F \\ Therefore, electricity required to produce 20g of calcium = 1F \\ (ii) According to the question, \\ {Al}^{3 +} {+ 3}_{e}^{-} \to Al \\ 27g \\ Electricity required to produce 27g of Al = 3F \\ = \frac{3 \times 40}{27} F \\ Therefore, electricity required to produce 40g of Al = 4.44 F. \end{array}$

Q.13

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 H₂SO₄with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Ans.

$\begin{array}{l} (i) At cathode:The following reduction reactions compete to take place at the cathode . \\ {Ag}_{(aq)}^{+} + e^{-} \to {Ag}_{(s)}; E^{°} = 0.80 V \\ H_{(aq)}^{+} + e^{-} \to \frac{1}{2} H_{2 (g)}; E^{°} = 0.00 V \\ The reaction with a higher value of E^{°} takes place at the cathode . \end{array}$ $\begin{array}{l} Therefore,deposition of silver will take place at the cathode . \\ At anode: \\ The Ag anode is attacked by {NO}_{3}^{–} ions .Therefore,the silver electrode \\ at the anode dissolves in the solution to from Ag . \\ (ii) At cathode:The following reduction reactions compete to take place at the cathode \\ {Ag}_{(aq)}^{+} + e^{-} \to {Ag}_{(s)}; E^{°} = 0.80 V \\ H_{(aq)}^{+} + e^{-} \to \frac{1}{2} H_{2 (g)}; E^{°} = 0.00 V \\ The reaction with a higher value of E^{°} takes place at the cathode . \end{array}$ $\begin{array}{l} At cathode : \\ The following oxidation reactions are possible at the anode . \\ {Cl}_{(aq)}^{-} \to \frac{1}{2} {Cl}_{2 (g)} + e^{- 1}; E^{°} = 1 .36 V \\ 2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H_{(aq)}^{+} + 4 e^{-}; E^{°} = 1 .23 V \\ At the anode, there action with a lower value of E^{°} is preferred . \end{array}$ $\begin{array}{l} But due to the overpotential of oxygen, \\ {Cl}^{–} {gets oxidised at the anode to produce Cl}_{2} gas . \end{array}$

Q.14 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe³⁺(aq) and I^– (aq)
(ii) Ag⁺ (aq) and Cu(s)
(iii) Fe³⁺ (aq) and Br^‑ (aq)
(iv) Ag(s) and Fe³⁺ (aq)
(v) Br₂ (aq) and Fe²⁺ (aq)

Ans.

$\begin{array}{l} (i) {Fe}^{3 +} (aq) + e^{-} \to {Fe}^{2 +} (aq) \times 2; E ° = + 0 .77 V \\ 2 I^{-} (aq) \to I_{2} (s) + 2 e^{-} E ° = - 0 .54 V \\ \overset{- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -}{2 {Fe}^{3 +} (aq) + 2 I^{-} (aq) \to 2 {Fe}^{2 +} (aq) + I_{2} (s); E ° = + 0 .23 V} \\ {Since E}^{°} for the overall reaction is positive, \\ {the reaction between Fe}^{3 +} {(aq)}_{} {and I}^{-} (aq) is feasible . \\ (ii) {Ag}^{+} (aq) + e^{-} \to Ag (s) \times 2; E ° = + 0 .80 V \\ Cu (s) \to {Cu}^{2 +} (aq) + 2 e^{-} E ° = - 0 .34 V \\ \overset{- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -}{\begin{array}{l} 2 {Ag}^{+} (aq) + Cu (s) \to 2 Ag (s) + {Cu}^{2 +} (aq); E ° = + 0 .46 V \end{array}} \\ Since E ° for the overall reaction is positive, \\ {the reaction between Ag}^{+} {(aq)}^{} and Cu (s) is feasible . \\ (iii) {Fe}^{3 +} (aq) + e^{-} \to {Fe}^{2 +} (aq) \times 2; E ° = + 0 .77 V \\ 2 {Br}^{-} (aq) \to {Br}_{2} (l) + 2 e^{-} E ° = - 1 .09 V \\ \overset{- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -}{\begin{array}{l} 2 {Fe}^{3 +} (aq) + 2 {Br}^{-} (aq) \to 2 {Fe}^{2 +} (aq) + {Br}_{2} (l); E ° = - 0 .32 V \end{array}} \\ Since E ° for the overall reaction is negative, \\ {the reaction between Fe}^{3 +} (aq) {and Br}^{-} (aq) is not feasible . \\ (iv) Ag (s) \to {Ag}^{+} (aq) + e^{-} E ° = - 0 .80 V \\ {Fe}^{3 +} (aq) + e^{-} \to {Fe}^{2 +} (aq) E ° = + 0 .77 V \\ \overset{- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -}{Ag (s) + {Fe}^{3 +} (aq) \to {Ag}^{+} (aq) + {Fe}^{2 +} (aq) E ° = - 0 .03 V} \\ Since E^{°} for the overall reaction is negative, the reaction between Ag (s) and Fe 3 + (aq) is not feasible . \end{array}$ $\begin{array}{l} (i v) A g (s) \to A g^{+} (a q) + e^{-} E ° = - 0.80 V \\ F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q) E ° = + 0.77 V \\ \overset{- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -}{A g (s) + F e^{3 +} (a q) \to A g^{+} (a q) + F e^{2 +} (a q) E ° = - 0.03 V} \\ Since E ° for the over all reaction is positive, the reaction between {Br}_{2} (aq) and {Fe}^{2 +} (aq) is feasible . \end{array}$

Q.15 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?

Ans.

$\begin{array}{l} According to the reaction : \\ {Ag}^{+} (aq) + e^{-} \to Ag (s) \\ 108 g \\ i . e ., 10 8g of Ag is deposited by 96487 C . \\ = \frac{96487 \times 1 .45}{108} = 1295 .43 C \\ Given, \\ Current = 1 . 5A \\ = \frac{1295 .43}{1 .5} s \\ Time \\ = 863 . 6s \\ = 864s \\ = 14.40 min \\ Again, \\ {Cu}^{2 +} (aq) + 2 e^{-} \to Cu (s) \\ 63 .5 g \\ i . e ., 2 \times 96487 C of charge deposit = 63 . 5g of Cu \\ Therefore, 1295 . 43C of charge will deposit \\ = \frac{63 .5 x 1295 .43}{2 x 96487} g \\ = 0 . 426g of Cu \\ {Zn}_{(aq)}^{2 +} + 2_{e}^{-} \to {Zn}_{(s)} \\ 65 . 4 g \\ i . e ., 2 \times 96487C of charge deposit = 65 . 4g of Zn \\ Therefore, 1295 . 43C of charge will deposit \\ = \frac{65 .4 x 1295 .43}{2 x 96487} g \\ = 0 . 439g of Zn \end{array}$

Q.16 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?

Ans.

$\begin{array}{l} Given, \\ Current = 5 A \\ Time = 20 \times 60 = 1200 s \\ ∴ Charge = Current \times Time \\ = 5 \times 1200 \\ = 6000 C \\ According to the reaction, \\ {Ni}^{2 +} (aq) + 2 e^{-} \to Ni (s) \\ 58 .7 g \\ Nickel deposited by 2 \times 96487 C = 58 .71 g \\ Therefore, nickel deposited by 6000 C = \frac{58 .71 \times 6000}{2 \times 96487} g \\ = 1 .825 g \\ Hence, 1 .825 g of nickel will be deposited at the cathode . \end{array}$

Q.17 How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H₂O to O₂.
(ii) 1 mol of FeO to Fe₂O₃

Ans.

$\begin{array}{l} (i) According to the question, \\ H_{2} O \to H_{2} + \frac{1}{2} O_{2} \\ Now, we can write : \\ O^{2 -} \to \frac{1}{2} O_{2} + 2_{e}^{-} \\ Electricity required for the oxidation of \\ {1 mol of H}_{2} {O to O}_{2} = 2F \\ 2 \times 96487C \\ = 192974C \\ (ii) According to the question, \\ {Fe}^{2 +} \to {Fe}^{3 +} + e^{- 1} \\ Electricity required for the oxidation of \\ {1mol of FeO to Fe}_{2} O_{3} = 1F = 96487C \end{array}$

Q.18 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^–3S cm^–1.

Ans.

Given,

Conductivity, κ = 0.146 × 10^–3 S cm^–1

Resistance, R = 1500

Cell constant = κ × R

= 0.146 × 10^–3 × 1500

= 0.219 cm^–1

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Related Chapters

Chapter 1 - The Solid State

Chapter 2 - Solutions

Chapter 4 - Chemical Kinetics

Chapter 5 - Surface Chemistry

Chapter 6 - General Principles and Processes of Isolation of Elements

Chapter 7 - The p-Block Elements

Chapter 8 - The d and f Block Elements

Chapter 9 - Coordination Compounds

Chapter 10 - Haloalkanes and Haloarenes

Chapter 11 - Alcohols, Phenols and Ethers

Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

Chapter 13 - Amines

Chapter 14 - Biomolecules

Chapter 15 - Polymers

Chapter 16 - Chemistry in Everyday life

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NCERT Chemistry Solutions for Class 12 Chapter 3 – Electrochemistry