Chapter 2 Electrochemistry

NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions for class 12 Chemistry chapter 2 (Electrochemistry) are important for students to understand the connection between electricity and chemical change. This is a really basic area of physical chemistry; it shows students how chemical reactions create electricity (Galvanic cells) and how voltage drives non-spontaneous chemical reactions (Electrolytic cells).

All students who are studying to take either the CBSE board exam or the JEE or NEET should also know the quantitative aspects related to cell potential and conductance. These NCERT solutions help students to break down very complex concepts related to electrode potential and the thermodynamics of redox reactions into smaller, understand and digestible modules so that they will have confidence in solving both Conceptual Problems and Numerical Problems.

1.0Class 12 Chemistry Chapter 2 Electrochemistry: Key Concepts

Class 12 Chemistry Chapter 2 – Electrochemistry explains the relationship between chemical reactions and electrical energy. The chapter covers electrochemical cells, electrode potential, conductance, electrolysis, batteries, and corrosion.

The important concepts in NCERT Class 12 Chemistry Chapter 2 include:

Electrochemical Cells: Understanding the construction and working of electrochemical cells, including the Daniell cell and the role of a salt bridge.
Nernst Equation: Learning how to calculate the cell potential under non-standard conditions using the Nernst equation:
E₍cell₎ = E°₍cell₎ − (RT/nF) ln Q
Gibbs Energy and EMF: Understanding the relationship between Gibbs energy change and the EMF of an electrochemical cell:
ΔG = −nFE₍cell₎
Conductance of Electrolytic Solutions: Studying resistance, resistivity, conductivity (κ), and molar conductivity (Λₘ) of electrolytic solutions.
Kohlrausch’s Law: Learning how to calculate the limiting molar conductivity of weak electrolytes through independent migration of ions.
Electrolysis and Faraday’s Laws: Learning how the amount of a substance deposited or released at an electrode depends on the quantity of electric charge passed through the electrolyte.
Batteries and Fuel Cells: Studying primary cells such as dry cells, secondary cells such as lead-acid batteries, and fuel cells such as the hydrogen-oxygen fuel cell.
Corrosion: Understanding corrosion as an electrochemical process, including the formation of rust on iron.

2.0NCERT Solutions Class 12 Chemistry Chapter 2 Electrochemistry : Detailed Solutions

INTEXT QUESTIONS

How would you determine the standard electrode potential of the system $Mg^{2 +} / Mg$ ? Sol. The standard electrode potential of $Mg^{2 +} ∣ Mg$ can be measured with respect to the standard hydrogen electrode, represented by $Pt (s), H_{2} (g)$ (1 atm) | $H^{+} (aq)$ (1M). A cell, consisting of $Mg ∣ MgSO_{4}, (aq 1 M)$ as the anode and the standard hydrogen electrode as the cathode, is set up. $Mg Mg^{2 +} (aq, 1 M) ∣ H^{+} (aq, 1 M) ∣ H_{2} (g, 1$ bar), $Pt (s)$ 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, $E_{R}^{Θ}$ for the standard hydrogen electrode is zero. Therefore, $E^{Θ} = 0 - E_{L}^{Θ} = - E_{L}^{Θ}$

2. Can you store copper sulphate solution in a zinc pot ? Sol. No, Zinc is more reactive than copper. Zinc reacts with copper sulphate and displaces copper from its salt solution.

$Zn + CuSO_{4} \to ZnSO_{4} + Cu$

Consult the table of the standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions? Sol. $Fe^{2 +} \to Fe^{3 +} + e^{-}; E_{cell}^{o} = + 0.77 V$ Substances which have greater reduction potential than +0.77 V will oxidise $Fe^{2 +}$ , e.g., $Br_{2}, Cl_{2}$ and $F_{2}$ .
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. Sol. $2 H^{+} + 2 e^{-} \to H_{2}$

$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} lo g \frac{P _{H_{2}}}{[ H ^{+} ] ^{2}} E_{cell}^{0} = 0, P_{H_{2}} = 1 atm, [H^{+}] = 1 0^{- pH} = 1 0^{- 10} E_{cell} = - \frac{0.0591}{2} lo g \frac{1}{( 1 0 ^{- 10} ) ^{2}} = - 0.591 V$

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

$Ni (s) + 2 Ag^{+} (0.002 M) \to Ni^{2 +} (0.160 M) + 2 Ag (s); Given that E_{cell}^{\circ} = 1.05 V$

Sol. Applying Nernst equation we have:

$E_{(cell)} = E_{cell}^{Θ} - \frac{0.0591}{n} lo g \frac{[ Ni ^{2 +} ]}{[ Ag ^{+} ] ^{2}} = 1.05 - \frac{0.0591}{2} lo g \frac{( 0.160 )}{( 0.002 ) ^{2}} = 1.05 - 0.02955 lo g \frac{0.16}{0.000004} = 1.05 - 0.14 = 0.91 V$

The cell in which the following reaction occurs :

$2 Fe^{3 +} (aq.) + 2 I^{-} (aq.) \to 2 Fe^{2 +} (aq.) + I_{2} (s) has E_{cell}^{\circ} = 0.236 V at 298 K$

Calculate the standard Gibbs energy and equilibrium constant of the cell reaction. Sol. $Δ_{r} G^{o} = - nFE_{cell}^{\circ}$

$= - 2 \times 96500 \times 0.236 joule mol^{- 1} = - 45.55 kJ mol^{- 1} Δ_{r} G^{\circ} = - 2.303 RT log K_{c} (R = 8.314 J K^{- 1} mol^{- 1}) - 45.55 = - 2.303 \times 8.314 \times 298 lo g K_{c} or lo g K_{c} = 7.981$

$\Rightarrow K_{c} = antilog (7.981)$ $\Rightarrow K_{c} = 9.6 \times 1 0^{7}$

7. Why does the conductivity of a solution decrease with dilution? Sol. The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) per unit volume decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

Suggest a way to determine the $Λ_{m}^{0}$ value of water. Sol. Applying Kohlrausch's law of independent migration of ions, the value of water can be determined as follows :

$Λ_{m (H_{2} O)}^{0} = Λ_{H^{+}}^{0} + Λ_{OH^{-}}^{0} = (Λ_{H^{+}}^{0} + Λ_{Cl^{-}}^{0}) + (Λ_{Na}^{0} + Λ_{OH^{-}}^{0}) - (Λ_{Na}^{0} + Λ_{Cl^{-}}^{0}) ∴ Λ_{m (H_{2} O)}^{0} = Λ_{m (HCl)}^{0} + Λ_{m (NaOH)}^{0} - Λ_{m (NaCl)}^{0}$

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

9. The molar conductivity of $0.025 mol / L$ methanoic acid is $46.1 S cm^{2} / mol$ . Calculate the degree of dissociation and dissociation constant. Given: $λ_{H^{+}}^{o} = 349.6 Scm^{2} mol^{- 1}$ and $λ_{HCOO^{-}}^{o} = 54.6 Scm^{2} mol^{- 1}$ Sol. $λ_{HCOOH}^{o} = λ_{H^{+}}^{o} + λ_{HCOO^{-}}^{o} = 349.6 + 54.6 = 404.2$

$α = \frac{λ _{m}^{c}}{λ _{m}^{o}} = \frac{46.1}{404.2} = 0.114; K_{a} = \frac{C α ^{2}}{1 - α} = \frac{0.025 \times 0.114 \times 0.114}{1 - 0.114} = 3.67 \times 1 0^{- 4}$

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ? Sol. $I = 0.5 A, t = 2$ hours $= 2 \times 60 \times 60 s = 7200 s$

$Q = I t = 0.5 A \times 7200 s = 3600 C ∵ 96500 coulomb are equivalent of 6.023 \times 1 0^{23} number of electrons ∴ 3600 coulomb are equivalent to \frac{6.023 \times 1 0 ^{23} \times 3600}{96500} = 2.246 \times 1 0^{22} electrons$

Suggest a list of metals that are extracted electrolytically? Sol. Alkali metals such as Na, K etc., alkaline earth metals such as Mg, Ca etc. and aluminium.
Consider the reaction,

$Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O$

What is the quantity of electricity in coulombs needed to reduce one mol of $Cr_{2} O_{7}^{2 -}$ ? Sol. To reduce 1 mole of $Cr_{2} O_{7}^{2 -}$ , the required quantity of electricity will be :

$6 F = 6 \times 96487 coulomb = 578922 C$

Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. Sol. lead storage battery consists of a lead anode, a grid of lead packed with lead oxide $(PbO_{2})$ as the cathode, and a 38% solution of sulphuric acid $(H_{2} SO_{4})$ as an electrolyte. When the battery is in use, the following cell reactions take place: At Anode :

$Pb (s) + SO_{4}^{2 -} (aq.) ⟶ PbSO_{4} (s) + 2 e^{-} At Cathode : PbO_{2} (s) + 4 H^{+} (aq.) + SO_{4}^{- 2} (aq.) + 2 e^{-} ⟶ PbSO_{4} (s) + 2 H_{2} O (ℓ)$

Net cell reaction : $Pb (s) + PbO_{2} (s)$

$+ 2 H_{2} SO_{4} (aq.) ⟶ 2 PbSO_{4} (s) + 2 H_{2} O (ℓ)$

When a battery is charged, the reverse of all these reactions takes place. Hence, on charging, $PbSO_{4} (s)$ present at the anode and cathode is converted into $Pb (s)$ and $PbO_{2} (s)$ respectively.

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

Sol. Methane and methanol.

3.0NCERT EXERCISE

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 Sol. The following is the order in which the given metals displace each other from the solution of their salts. Mg, Al, Zn, Fe, Cu
Given the standard electrode potentials,

$K^{+} / K = - 2.93 V, Ag^{+} / Ag = 0.80 V Hg^{2 +} / Hg = 0.79 V Mg^{2 +} / Mg = - 2.37 V, Cr^{3 +} / Cr = - 0.74 V$

Arrange these metals in their increasing order of reducing power. Sol. The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of $K^{+} / K < Mg^{2 +} / Mg < Cr^{3 +} / Cr < Hg^{2 +} / Hg < Ag^{+} / Ag$ . Hence, the reducing power of the given metals increase in the following order :

$Ag < Hg < Cr < Mg < K$

Depict the galvanic cell in which the reaction $Zn (s) + 2 Ag^{+} (aq) ⟶ Zn^{2 +} (aq) + 2 Ag (s)$ 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. Sol. The galvanic cell in which the given reaction takes place is depicted as :

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

(i) Zn electrode (anode) is negatively charged. (ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc. (iii) The reaction taking place at the anode is given by,

$Zn_{(s)} ⟶ Zn_{(aq)}^{2 +} + 2 e^{-}$

The reaction taking place at the cathode is given by,

$2 Ag_{(aq)}^{+} + 2 e^{-} ⟶ 2 Ag_{(s)}$

Calculate the standard cell potentials of galvanic cell in which the following reactions take place : (i) $2 Cr (s) + 3 Cd^{2 +} (aq) ⟶ 2 Cr^{3 +} (aq) + 3 Cd (s)$

$E_{Cr^{3 +} / Cr}^{⊖} = - 0.74 V; E_{Cd^{2 +} / Cd}^{⊖} = - 0.40 V$

(ii) $Fe^{2 +} (aq) + Ag^{+} (aq) ⟶ Fe^{3 +} (aq) + Ag (s)$

$E_{Fe^{+ 3} / Fe^{+ 2}}^{Θ} = 0.77 V; E_{Ag^{+} / Ag}^{Θ} = 0.80 V$

Calculate the $Δ_{r} G^{⊖}$ and equilibrium constant of the reactions.

Sol.

(i) The galvanic cell of the given reaction is depicted as :

$Cr (s) Cr^{3 +} (aq) ∥ Cd^{2 +} (aq) Cd (s)$

Now, the standard cell potential is

$E_{cell}^{Θ} = E_{R}^{Θ} - E_{L}^{Θ} = - 0.40 - (- 0.74) = + 0.34 V Δ_{r} G^{Θ} = - nFE_{cell}^{Θ}$

In the given equation,

$n = 6; F = 96500 C mol^{- 1}; E_{cell}^{⊖} = + 0.34 V$

Then, $Δ_{r} G^{⊖} = - 6 \times 96500 C mol^{- 1} \times 0.34 V$

$= - 196.86 KJ mol^{- 1}$

Again,

$Δ_{r} G^{⊖} = - RT ln K_{c} = - 2.303 RT lo g K_{c}$

$\Rightarrow lo g K_{c} = - \frac{Δ _{r} G}{2.303 RT} = \frac{- 196.86 \times 1 0 ^{3}}{2.303 \times 8.314 \times 298} = 34.50 ∴ K_{c} = antilog (34.50) = 3.16 \times 1 0^{34}$

(ii) $E_{Fe^{3 +} / Fe^{2 +}}^{Θ} = 0.77 V; E_{Ag^{+} / Ag}^{Θ} = 0.80 V$ The galvanic cell of the given reaction is depicted as :

$Fe^{2 +} (aq) Fe^{3 +} (aq) ∥ Ag^{+} (aq) Ag (s)$

Now, the standard cell potential is

$E_{cell}^{⊖} = E_{R}^{⊖} - E_{L}^{⊖} = 0.80 - 0.77 = 0.03 V$

Here, $n = 1; Δ_{r} G^{⊖} = - nFE_{cell}^{⊖}$

$= - 1 \times 96500 C mol^{- 1} \times 0.03 V = - 2.895 kJ mol^{- 1}$

Again, $Δ_{r} G^{⊖} = - 2.303 RT Iog K_{c}$

$\Rightarrow lo g K_{c} = \frac{- Δ _{r} G ^{⊖}}{2.303 RT} = - \frac{- 2.895 \times 1 0 ^{3}}{2.303 \times 8.314 \times 298} = 0.5073 ∴ K_{c} = antilog (0.5073) = 3.2 (approximately)$

Write the Nernst equation and emf of the following cells at 298K : (i) $Mg (s) Mg^{2 +} (0.001 M) Cu^{2 +} (0.0001 M) Cu (s)$ (ii) $Fe (s) Fe^{2 +} (0.001 M) ∣ H^{+} (1 M) ∣ H_{2} (g) ($ lbar $) ∣ Pt (s)$ (iii) $Sn (s) Sn^{2 +} (0.050 M) ∣ H^{+} (0.020 M) ∣ H_{2} (g) ($ lbar $) ∣ Pt (s)$ (iv) $Pt (s) ∣ Br_{2} (l) ∣ Br^{-} (0.010 M) ∣∣ H^{+} (0.030 M) ∣ H_{2} (g) (1 bar) ∣ Pt (s)$

Sol.

(i) For the given reaction, the Nernst equation can be given as :

$E_{cell} = E_{cell}^{⊖} - \frac{0.0591}{n} lo g \frac{[ Mg ^{2 +} ]}{[ Cu ^{2 +} ]} = {0.34 - (- 2.36)} - \frac{0.0591}{2} lo g \frac{.001}{.0001} = 2.67 V (approx)$

(ii) For the given reaction, the Nernst equation can be given as :

$E_{cell} = E_{cell}^{⊖} - \frac{0.0591}{n} lo g \frac{[ Fe ^{2 +} ]}{[ H ^{+} ] ^{2}} = {0 - (- 0.44)} - \frac{0.0591}{2} lo g \frac{0.001}{1 ^{2}} = 0.53 V$

(approx) (iii) For the given reaction, the Nernst equation can be given as :

$E_{cell} = E_{cell}^{Θ} - \frac{0.0591}{n} lo g \frac{[ Sn ^{2 +} ]}{[ H ^{+} ] ^{2}} = {0 - (- 0.14)} - \frac{0.0591}{2} lo g \frac{0.050}{( 0.020 ) ^{2}} = 0.08 V (approx)$

(iv) For the given reaction, the Nernst equation can be given as :

$E_{cell} = E_{cell}^{Θ} - \frac{0.0591}{n} lo g \frac{1}{[ Br ^{-} ] ^{2} [ H ^{+} ] ^{2}} = (0 - 1.09) - \frac{0.0591}{2} lo g \frac{1}{( 0.010 ) ^{2} ( 0.030 ) ^{2}} = 1.09 - 0.02955 \times lo g \frac{1}{0.00000009} = 1.09 - 0.02955 \times lo g \frac{1}{9 \times 1 0 ^{- 8}} = 1.09 - 0.02955 \times lo g (1.11 \times 1 0^{7}) = - 1.09 - 0.02955 (0.0453 + 7) = - 1.298 V$

In the button cells widely used in watches and other devices the following reaction takes place :

$Zn (s) + Ag_{2} O (s) + H_{2} O (l) ⟶ Zn^{2 +} (aq) + 2 Ag (s) + 2 OH^{-} (aq)$

Determine $Δ_{r} G^{⊖}$ and $E^{Θ}$ for the reaction. Sol. $Zn_{(s)} \to Zn^{2 +}_{(aq)} + 2 e^{⊖}; E^{⊖} = - 0.76 V$ $Ag_{2} O_{(s)} + H_{2} O_{(l)} + 2 e^{⊖} \to 2 Ag_{(g)} + 2 OH_{(aq)}^{⊖}; E^{⊖} = 0.344 V$

$Zn_{(s)} + Ag_{2} O_{(s)} + H_{2} O_{(l)} \to Zn_{(aq)}^{2 +} + 2 Ag_{(s)} + 2 OH_{(aq)}^{⊖}; E^{⊖} = 1.104 V Δ_{r} G = - nFE^{⊖} = - 2 \times 96500 \times 1.104 = - 213072 J = - 213.07 kJ$

Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration. Sol. 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. The conductivity of a solution is defined as the conductance of a solution of 1 cm of length and area of cross section $1 cm^{2}$ . The inverse of resistivity is called conductivity.

$i.e., G = κ \frac{A}{ℓ} = κ .1 = κ (Since A = 1, = 1)$

Conductivity always decrease 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 is 1.

$Λ_{m} = κ \frac{A}{ℓ} = κ$

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

$∴ Λ_{m} = κ V$

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of an electrolyte also increases on dilution. The variation of $Λ_{m}$ with $c$ for strong and weak electrolytes is shown in the following plot:

NCERT Solution Chapter 2 Electrochemistry Ques 7

8. The conductivity of 0.20 M solution of KCl at 298 K is $0.0248 Scm^{- 1}$ . Calculate its molar conductivity. Sol. Given, $κ = 0.0248 S cm^{- 1}; c = 0.20 M$

$∴ Molar conductivity (Λ_{m}) = \frac{κ \times 1000}{c} = \frac{0.0248 \times 1000}{0.2} = 124 Scm^{2} mol^{- 1}$

The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is $1500Ω$ . What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is $0.146 \times 1 0^{- 3} S cm^{- 1}$ . Sol. Given, Conductivity, $κ = 0.146 \times 1 0^{- 3} S cm^{- 1}$ Resistance, $R = 1500Ω$

$∴ Cell constant = κ \times R = 0.146 \times 1 0^{- 3} \times 1500 = 0.219 cm^{- 1}$

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

Concentration/ M	$10^{2} \times κ / S m^{- 1}$
0.001	1.237
0.01	11.85
0.02	23.15
0.05	55.53
0.1	106.74

Calculate $Λ_{m}$ for all concentrations and draw a plot between $Λ_{m}$ and $C \frac{1}{2}$ . Find the value of $Λ_{m}^{0}$ . Sol. Given,

$κ = 1.237 \times 1 0^{- 2} S m^{- 1}, c = 0.001 M Then, κ = 1.237 \times 1 0^{- 4} S cm^{- 1}, c^{1/2} = 0.0316 M^{1/2} ∴ Λ_{m} = \frac{κ}{c} = \frac{1.237 \times 1 0 ^{- 4} S cm ^{- 1}}{0.001 molL ^{- 1}} \times \frac{1000 cm ^{3}}{L} = 123.7 S cm^{2} mol^{- 1}$

Given,

$κ = 1185 \times 1 0^{- 2} S m^{- 1}, c = 0.010 M Then, κ = 11.85 \times 1 0^{- 4} S cm^{- 1}, c^{1/2} = 0.1 M^{1} ∴ Λ_{m} = \frac{κ}{c} = \frac{11.85 \times 1 0 ^{- 4} Scm ^{- 1}}{0.010 mol L ^{- 1}} \times \frac{1000 cm ^{3}}{L} = 118.5 S cm^{2} mol^{- 1}$

Given, $κ = 23.15 \times 1 0^{- 2} S m^{- 1}, c = 0.020 M$ Then, $k = 23.15 \times 1 0^{- 4} S cm^{- 1}, c^{1/2} = 0.1414 M^{1/2}$

$∴ Λ_{m} = \frac{κ}{c} = \frac{23.15 \times 1 0 ^{- 4} Scm ^{- 1}}{0.020 mol L ^{- 1}} \times \frac{1000 cm ^{3}}{L} = 115.8 cm^{2} mol^{- 1}$

Given, $k = 55.53 \times 1 0^{- 2} S m^{- 1}, c = 0.050 M$ Then, $k = 55.53 \times 1 0^{- 4} S cm^{- 1}, c^{1/2} = 0.2236 M^{1/2}$

$∴ Λ_{m} = \frac{κ}{c} = \frac{55.53 \times 1 0 ^{- 4} Scm ^{- 1}}{0.050 mol L ^{- 1}} \times \frac{1000 cm ^{2}}{L} = 111.11 S cm^{2} mol^{- 1}$

Given, $κ = 106.74 \times 1 0^{- 2} Sm^{- 1}, c = 0.100 M$ Then, $κ = 106.74 \times 1 0^{- 4} S cm^{- 1}, c^{1/2} = 0.3162 M^{1/2}$

$∴ Λ_{m} = \frac{κ}{c} = \frac{106.74 \times 1 0 ^{- 4}}{0.100 molL ^{- 1}} \times \frac{1000 cm ^{3}}{L} = 106.74 S cm^{2} mol^{- 1}$

Now, we have the following data.

$C^{1/2}$	$λ_{m} (S cm^{2} mol^{- 2})$
0.0316	123.7
0.1	118.5
0.1414	115.8
0.2236	111.1
0.3162	106.74

NCERT Chapter 2 Electrochemistry Ques 10

Since the line interrupts $Λ_{m}$ at $124.0 S cm^{2} mol^{- 1}, Λ_{m}^{0} = 124.0 S cm^{2} mol^{- 1}$

Conductivity of 0.00241 M acetic acid is $7.896 \times 1 0^{- 5} S cm^{- 1}$ . Calculate its molar conductivity. If $Λ_{m}^{0}$ for acetic acid is $390.5 S cm^{2} mol^{- 1}$ , what is its dissociation constant? Sol. Given, $κ = 7.896 \times 1 0^{- 5} S cm^{- 1}$ $c = 0.00241 mol L^{- 1}$ Then, molar conductivity,

$Λ_{m} = \frac{κ}{c} = \frac{7.896 \times 1 0 ^{- 5} Scm ^{- 1}}{0.00241 mol L ^{- 1}} \times \frac{1000 cm ^{3}}{L} = 32.76 S cm^{2} mol^{- 1}$

Again, $Λ_{m}^{0} = 390.5 S cm^{2} mol^{- 1}$ Now, $α = \frac{Λ _{m}}{Λ _{m}^{0}} = \frac{32.76 S cm ^{2} mol ^{- 1}}{390.5 S cm ^{2} mol ^{- 1}} = 0.084$ ∴ Dissociation constant,

$K_{a} = \frac{c ^{2}}{( 1 - α )} = \frac{( 0.00241 mol L ^{- 1} ) ( 0.084 ) ^{2}}{( 1 - 0.084 )} = 1.86 \times 1 0^{- 5} mol L^{- 1}$

How much charge is required for the following reductions : (i) 1 mol of $Al^{3 +}$ to Al . (ii) 1 mol of $Cu^{2 +}$ to Cu . (iii) 1 mol of $MnO_{4}^{-}$ to $Mn^{2 +}$ Sol. (i) $Al^{3 +} + 3 e^{-} ⟶ Al$ ∴ Required charge $= 3 F$ (ii) $Cu^{2 +} + 2 e^{-} ⟶ Cu$ ∴ Required charge $= 2 F$ (iii) $MnO_{4}^{-} ⟶ Mn^{2 +}$ i.e., $Mn^{7 +} + 5 e^{-} ⟶ Mn^{2 +}$ ∴ Required charge $= 5 F$
How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten $CaCl_{2}$ . (ii) 40.0 g of Al from molten $Al_{2} O_{3}$ . Sol. (i) According to the question, $Ca^{2 +} + 2 e^{-} ⟶ Ca$ Electricity required to produce 40 g of calcium $= 2 F$ Therefore, electricity required to produce 20 g of calcium $= \frac{2 \times 20}{40} F = 1 F$ (ii) According to the question, $Al^{3 +} + 3 e^{-} ⟶ Al$ Electricity required to produce 27 g of $Al = 3 F$ Therefore, electricity required to produce 40 g of $Al = \frac{3 \times 40}{27} F$ $= 4.44 F$
How much electricity is required in coulomb for the oxidation of (i) 1 mol of $H_{2} O$ to $O_{2}$ . (ii) 1 mol of FeO to $Fe_{2} O_{3}$ .

Sol.

(i) According to the question,

$H_{2} O ⟶ H_{2} + \frac{1}{2} O_{2}$

Now, we can write :

$O^{2 -} ⟶ \frac{1}{2} O_{2} + 2 e^{-}$

Electricity required for the oxidation of

$1 mol of H_{2} O to O_{2} = 2 F = 2 \times 96500 C = 193000 C$

(ii) According to the question,

$Fe^{2 +} ⟶ Fe^{3 +} + e^{- 1}$

Electricity required for the oxidation of 1 mol of FeO to $Fe_{2} O_{3} = 1 F = 96500 C$

15. A solution of $Ni (NO_{3})_{2}$ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? Sol. Given

$Current = 5 A; Time = 20 \times 60 = 1200 s ∴ Charge = current \times time = 5 \times 1200 = 6000 C$

According to the reaction,

$Ni^{2 +}_{(aq)} + 2 e^{-} ⟶ Ni_{(s)}$

Nickel deposited by

$2 \times 96500 C = 58.71 g$

Therefore, nickel deposited by 6000 C

$= \frac{58.71 \times 6000}{2 \times 96500} g = 1.825 g$

Hence, 1.825 g of nickel will be deposited at the cathode.

16. Three electrolytic cells A, B, C containing solutions of $ZnSO_{4}, AgNO_{3}$ and $CuSO_{4}$ , 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? Sol. According to the reaction:

$Ag_{(aq)}^{+} + e^{-} ⟶ 108 g Ag_{(s)}$

i.e., 108 g of Ag is deposited by 96500 C . Therefore, 1.45 g of Ag will be deposited by

$= \frac{96500 \times 1.45}{108} C = 1295.60 C$

Given, Current $= 1.5 A$

$∴ Time = \frac{1295.60}{1.5} s = 863.73 s = 864 s = 14.40 min Again, Cu_{(aq)}^{2 +} + 2 e^{-} ⟶ 63.5 g Cu_{(s})$

i.e., $2 \times 96500 C$ of charge deposit $= 63.5 g$ of Cu Therefore, 1295.60 C of charge will deposit

$= \frac{63.5 \times 1295.60}{2 \times 96500} g = 0.426 g of Cu$

Given, $Zn_{(aq)}^{2 +} + 2 e^{-} ⟶ 65.4 g Zn_{(s)}$ i.e., $2 \times 96500 C$ of charge deposit $= 65.4 g$ of Zn Therefore, 1295.60 C of charge will deposit

$= \frac{65.4 \times 1295.60}{2 \times 96500} g = 0.439 g of Zn$

Using the standard electrode potentials given in electrochemical series predict if the reaction between the following is feasible : (i) $Fe^{3 +} (aq)$ and $I^{-} (aq)$ (ii) $Ag^{+} (aq)$ and $Cu (s)$ (iii) $Fe^{3 +} (aq)$ and $Br^{-} (aq)$ (iv) $Ag (s)$ and $Fe^{3 +} (aq)$ (v) $Br_{2} (aq)$ and $Fe^{2 +} (aq)$ .

Sol.

$Fe_{(aq)}^{3 +} + e ⊖ Fe_{(aq)}^{2 +} 2; E = + 0.77 V$

(i)

$2 1^{⊖}_{(aq)} 2 Fe^{3 +}_{(aq)} + 2 I_{(aq)}^{⊖} ⟶ ⟶ I_{2 (s)} + 2 e; E = - 0.54 V 2 Fe^{2 +}_{(aq)} + I_{2 (s)}; E = + 0.23 V$

Since $E^{⊖}$ for the overall reaction is positive, the reaction between $Fe_{(aq)}^{3 +}$ and $I_{(aq)}^{-}$ is feasible.

$Ag_{(aq)}^{+} + e ⊖ Ag_{(s)}] 2; E = + 0.80 V$

$Cu_{(s)} ⟶ Cu^{2 +}_{(aq)} + 2 e; E = - 0.34 V$

(ii)

$2 Ag_{(aq)}^{+} + Cu_{(s)} ⟶ 2 Ag_{(s)} + Cu_{(aq)}^{2 +}; E = + 0.46 V$

Since $E^{⊖}$ for the overall reaction is positive, the reaction between $Ag_{(aq)}^{+}$ and $Cu_{(s)}$ is feasible.

$Fe^{3 +}_{(aq)} + e ⊖ Fe^{2 +}_{(aq)}] 2; E = + 0.77 V$

(iii)

$2 Br_{(aq)}^{⊖} ⟶ Br_{2 (l)} + 2 e^{⊖}; E = - 1.09 V 2 Fe ^{3 +} _{(aq)} + 2 Br _{(aq)}^{⊖} ⟶ 2 Fe^{2 +}_{(aq)} + Br_{2 (l)}; E = - 0.32 V$

Since $E^{⊖}$ for the overall reaction is negative, the reaction between $Fe_{(aq)}^{3 +}$ and $Br_{(aq)}$ is not feasible.

$Ag_{(s)} ⟶ Ag_{(aq)}^{+} + e^{⊖}; E = - 0.80 V$

(iv) $Fe^{3 +}_{(aq)} + e ⊖ Fe^{2 +}_{(aq)} 2 Ag + Fe^{3 +}_{(aq)} ⟶ Ag^{+}_{(aq)} + Fe^{2 +}_{(aq)}; E = - 0.03 V$

Since $E^{Θ} E$ for the overall reaction is negative, the reaction between $Ag_{(s)}$ and $Fe_{(aq)}^{3 +}$ is not feasible.

$Br_{2 (aq)} + 2 e ⊖ 2 Br_{(aq)}^{⊖}; E = + 1.09 V$

(v) (v)

$[Fe^{2 +}_{(aq)} Br_{2 (aq)} + 2 Fe^{2 +}_{(aq)} ⟶ Fe^{3 +}_{(aq)} + e^{⊖} 2; E = - 0.77 V ⟶ 2 Br_{(aq)}^{⊖} + 2 Fe^{3 +}_{(aq)}; E = + 0.32 V$

Since $E^{⊖}$ for the overall reaction is positive, the reaction between $Br_{2 (aq)}$ and $Fe_{(aq)}^{2 +}$ is feasible.

18. Predict the products of electrolysis in each to the following : (i) An aqueous solution of $AgNO_{3}$ with silver electrodes. (ii) An aqueous solution of $AgNO_{3}$ with platinum electrodes. (iii) A dilute solution of $H_{2} SO_{4}$ with platinum electrodes. (iv) An aqueous solution of $CuCl_{2}$ with platinum electrodes.

Sol.

	Cathode	Anode
$AgNO_{3}$	Ag	$Ag^{+}$
$AgNO_{3}$	Ag	$O_{2}$
$H_{2} SO_{4}$	$H_{2}$	$O_{2}$
$CuCl_{2}$	Cu	$Cl_{2}$

4.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

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Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 3	Chemical Kinetics
Chapter 4	D- and F-Block Elements
Chapter 5	Coordination Compounds
Chapter 6	Haloalkanes and Haloarenes
Chapter 7	Alcohols, Phenols and Ethers
Chapter 8	Aldehydes, Ketones and Carboxylic Acids
Chapter 9	Amines
Chapter 10	Biomolecules

5.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 2

Derive step by step how the Equilibrium Constant relates to the Cell Potential - clear and concise.
Solve using Nernst Equation correctly with Unit Errors (Coulombs, Faradays, and voltage).
Why does Conductivity decrease with dilution but Molar Conductivity increase? (Common Trick question in Exam).
How Fuel Cells Work and Why They are A Considered "Green" Future of Energy, will be illustrated through examples.
The structure of the answers, to enable maximum scoring in descriptive exam questions.

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NCERT Solutions Class 12 Chemistry Chapter 2 - Electrochemistry

1.0Class 12 Chemistry Chapter 2 Electrochemistry: Key Concepts

2.0NCERT Solutions Class 12 Chemistry Chapter 2 Electrochemistry : Detailed Solutions

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