CBSE Class 12 Chemistry Notes Chapter 3 Electrochemistry

Chapter 3 Electrochemistry Class 12 Chemistry Notes for CBSE Students

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

Ans: To determine the electrode potential of mg+2/mg1 we use hydrogen electrode (anode)
The standard hydrogen electrode potential is zero.
 ∴Eo cell=Eo right+Eo left
Eocell=Eomg+2/mg−1Eon2​/n1​ 
Eomg+2/mg=Eocell

3.2 Can you store copper Sulphate solutions in a zinc pot?

Ans: Zinc will replace the copper from its salt solution. When the copper Sulphate solution is stored in a zinc pot, copper is replaced by zinc, in the copper Sulphate solution. So copper Sulphate solution cannot be stored in a zinc pot.

3.3 Consult the table of standard electrode potentials and suggest three substances that can Oxidise Ferrous Ions under suitable conditions.

Ans: The metals having higher oxidation potential than iron can oxidize iron.
∴Ag++e−⟶AgEo=0.80 V
Br2​+2e−⟶Br⊝Eo=1.06 V
Cl2​+2e−⟶Cl⊝Eo=1.36 V

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

Ans: pH = 10
Using formula [H⁺] = 10^-pH
So [H⁺]   = 10^-10 M
Electrode reaction will be
H⁺ + e^– → ½ H²
Use the formula
Value of E⁰ = 0
E^cell = E⁰-(0.059/1) log [H₂] /[H⁺]
E^cell =-(0.059/1) log (1/10^-10)
E^cell =-0.0591 (10) = -0.591

3.5 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) V = 1.05 V

Ans: Given that [Ag+] = 0.002 M

[Ni2+] =0.160 M {n=10}

Apply the Nernst equation:

3.6 The cell in which the following reaction occurs:
2Fe 2I 2Fe I aq aq aq 2 s + − + + → + has 0 Ecell = 0.236 V at 298 K
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Ans: Given that
n = 2
T = 298 K
E° cell =  0.236 V
We have the formula
∆rG^θ = – nFE° cell   
Plug the values, we get
∆rG^θ = −2 × 96487 × 0.236
∆rG^θ = −45541.86 J mol⁻¹
Divide by 100o to convert in KJ
∆rG^θ = −45.54 kJ mol⁻¹  
Use second formula of ∆rG^θ
∆rG^θ =−2.303RT log K_c
Plug the values we get
 −45541.86 J mol⁻¹ = –2.303 × 8.314 × 298  log K_c
Solve it we get
Log K_c  = 7.98
Take antilog both side, we get
K_c = Antilog (7.98) = 9.6 × 107

3.7 Why does the conductivity of a solution decrease with dilution?

Ans= The conductivity of a solution is the conductance of ions presenting a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

3.8 Suggest a way to determine the Λ° m value of water.

Ans: According to kohlrausch’s law at infinite dilution each ion makes a definite contribution towards equivalent conductance of electrolyte irrespective of the nature of ion with which it is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contributions from its constituent ions
Applying Kohlrausch’s law of independent migration of ions, the ∧m 0 value of water can be determined as follows:

∧m(H2O) 0 = λH− 0 + λOH− 0 = (λH− 0 + λCl − 0 ) + (λNa+ 0 + λOH− 0 ) − (λNa + 0 + λCl − 0 ) =∧m(HCl) 0 +∧m(NaOH) 0 −∧m(NaCl) 0
Hence, by knowing the ∧m 0 values of HCl,NaOH and NaCl, the ∧m 0 value of water can be determined.

3.9 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 0 (H⁺ ) = 349.6 S cm2 mol^–1 and λ 0 (HCOO^– ) = 54.6 S cm2 mol^–1 .

Ans: Given that
λ0(H⁺)= 349.6 S cm2 mol^–1
λ0(HCOO^–) = 54.6 S cm2 mol^–1
Concentration ,C = 0.025 mol L^-1
λ(HCOOH) = 46.1 S cm2 mol⁻¹
use formula
λ°(HCOOH) = λ0(H⁺) + λ0(HCOO^–)
plug the values we get
λ°(HCOOH) = 0.349.6 + 54.6
                   =404.2 S cm2 mol⁻¹
Formula of degree of dissociation:
ά = λ°(HCOOH)/ λ°(HCOOH)
ά = 46.1 / 404.2
ά = 0.114
Formula of dissociation constant:
K = (c ά2)/(1 – ά)
Plug the values we get

K=[0/025x(0.114)²]/(1-0.114)

K = 3.67 × 10–4 mol per liter

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

Ans: Given:
current (I) = 0.5 ampere
time (t) = 2 hours
= 2 × 60 × 60 s = 7200 s
Find charge (Q) = ?
Formula: charge (Q) = current(I) × time(t)in second
= 0.5 A × 7200 s = 3600 C
We know that 1 F = 96500 C charge = 6.023 × 1023 number of electrons.
Then, 3600 C = 6.023×1023×3600 96500
number of electrons = 2.28 × 1022 number of electrons
Hence, 2.28 × 1022 number of electrons will flow through the wire.

3.11 Suggest a list of metals that are extracted electrolytically.

ANS= Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert Electrodes . The products of electrolysis depend on the different oxidizing and reducing species present in the electrolytic cell and their standard electrode potentials. Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these do not seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur. Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

3.12 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–?

Ans: mol of Cr2O7 2−= 1
The given reaction is as follows:
Cr2O7 2− + 14H + + 6e − → 2Cr 3+ + 7H2O
We know:
Therefore, to reduce 1 mole of Cr2O7 2− need 6 mole of electron,
Charge on one mole of electron = 1F 1 F = 96500
the required quantity of electricity will be: = 6 F
(because 6 mole electrons are using there) = 6 × 96500 C = 579000 C

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

Ans: In lead storage battery we have,
anode – porous lead
cathode- lead dioxide
The reactions that are taking place during the recharge of lead storage battery is
2PbSO4​+2H2​O→Pb(s)+PbO2​+4H++2SO42−

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

Ans: Galvanic cells that are designed to convert, the energy of combustion of fuels, hydrogen, methane, methanol are directly into electrical energy are called fuel cells.
Concept insight:
Cells that are designed to convert the energy of combustion of fuels directly into electrical energy are called fuel cells.

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

Ans: In process of rusting, water molecules present at the layer of iron react with oxides and get dissociated and give H+ ions
 H₂O + CO₂ → H₂CO₃
H₂CO₃ ↔ 2H⁺ + CO₃^2–
In present of H⁺ ions iron convert in Fe²⁺ so this part is act as anode
Reaction at anode:
Fe(s) → Fe²⁺(aq) + 2e^–
Electron released at anodic spot move through the metals and go to another spot on the metal and reduce oxygen this spot act as cathode
Reaction at Cathode:
O₂(g) +  4H⁺(aq) + 4e⁺( aq) → 2H₂O(l)
The overall reaction
2Fe(s)  + O₂(g) + 4H⁺(aq) → 2H₂O(l)  + Fe²⁺(aq)
Hence, rusting of iron act as electrochemical cell.