Careers360 Logo
NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry

Edited By Shivani Poonia | Updated on Jun 19, 2025 09:00 AM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 15 Jul' 2025 - 15 Jul' 2025

Electrochemistry is an important branch of chemistry that explores between electricity and chemical reactions. It focuses on how chemical energy changes into electrical energy, and vice versa. Have you ever wondered how a battery powers your phone or why metals are coated with a layer of gold? The answers to these questions lie in Electrochemistry.

This Story also Contains
  1. NCERT Solutions Of Class 12 Chemistry Chapter 2 (Intext Questions 2.1 to 2.15)
  2. NCERT Solutions for Class 12 (Exercise Questions)
  3. Class 12 Chemistry NCERT Chapter 2: Higher Order Thinking Skill (HOTS) Questions
  4. Approach to Solve Questions of Chapter 2
  5. Topics of NCERT Class 12 Chemistry Chapter 2
  6. What Extra Should Students Study Beyond NCERT for JEE?
  7. Formulas Of NCERT Class 12 Chemistry Electrochemistry
  8. NCERT Solutions for Class 12 Chemistry
  9. NCERT Exemplar Class 12 Solutions Subject-wise
  10. NCERT Solutions for Class 12 Subject-wise
NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry
NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry

This chapter deals with questions based mainly on electrochemical, galvanic cells, and the Nernst equation to calculate electromotive force potential. NCERT Solutions, designed by subject experts, provide easy-to-understand and comprehensive explanations of every question. NCERT solutions for class 12 help students develop analytical and problem-solving abilities. Electrochemistry is important for both theoretical and practical purposes.


Also Read,


NCERT Solutions Of Class 12 Chemistry Chapter 2 (Intext Questions 2.1 to 2.15)

Page 36

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

To determine the standard electrode potential of the given system we need to use a hydrogen electrode. In the setup, we shall put a hydrogen electrode as cathode and Mg | MgSO4 as an anode.

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

Now we will measure the emf of the cell. This emf will be the standard electrode potential of the magnesium electrode.

E°cell = E° right – E°left
E°left =0 ( The standard hydrogen electrode is always zero)
Hence
Eocell=EoMg|Mg2+

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

Answer:

The standard electrode potential of Zinc is - 0.76 whereas that of Copper is 0.34. So Zinc will reduce copper into the lower state.

It is known that zinc is more reactive than copper. Thus if we will store copper sulphate solution in zinc pot then zinc will displace copper from its solution.

The following reaction will take place:-

Zn+CuSO4ZnSO4+Cu

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

Answer:

The oxidising strength of elements increases as the standard electrode potential increases.

So all the elements having greater standard potential than iron can oxidise it to a higher state.

Few such elements are :- F2 , Cl2, Br2, Ag+1etc.

Page 41

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

Answer:

It is given that pH of the solution is 10,i.e., the hydrogen ion concentration in the solution is 10 -10 M.

H++e12H2

By Nernst equation we have :-

ECell=Ecell RT2Fln1[H+]

So, =00.05911log1[1010]

or = 0.591 V

So the required potential is - 0.591 V.

Question 2.5 Calculate the emf of the cell in which the following reaction takes place: Ni(s)+2Ag+(0.002M)Ni2+(0.160M)+2Ag(s) Given that E(cell)Θ=1.05V

Answer:

Here we can directly apply the Nernst equation :-

ECell=Ecell 0.0591nlog[Ni+2][Ag+]2

Putting the value in this equation :-

=1.05 0.05912log0.160(0.002)2

=1.05 0.02955 log(4×104)

=0.914 V

Hence the required potential is 0.914 V.

Question 2.6 The cell in which the following reaction occurs:
2Fe3+(aq)+2I(aq)2Fe2+(aq)+I2(s) has Ecell0=0.236V at 298K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer:

For finding Gibbs free energy we know the relation :-

ΔGr= nFEcell

= 2×96487×0.236

= 45541.864 J mol1

= 45.54 KJ mol1

Now, for equilibrium constant we will use :-

ΔGr=2.303 RTlog Kc

So, logKc=45.54×1032.303×8.314×298

logKc=7.981

Kc=9.57×107

Page 51

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

Answer:

The conductivity of a solution depends upon the number of ions and the distance between them. In the process of dilution, we don't increase the number of ions in the solution instead we increase the distance between them. So the conductivity of the solution decreases due to dilution.

Question 2.8 Suggest a way to determine the Λm0 value of water.

Answer:

We know :

Λm=ΛmAc12

If we draw a straight line between Λm and c , its slope will be -A and the intercept on the y-axis will be Λm .

In this way, we can obtain the value of limiting molar conductivity.

Question 2.9 The molar conductivity of 0.025molL1 methanoic acid is 46.0Scm2Scm2mol1 Calculate its degree of dissociation and dissociation constant. Given λ0(H+)=349.6Scm2mol1 and λ0(HCOO)=54.6Scm2mol1

Answer:

We know that :-

Λm=λ(H+)+λ(HCOO)

=349.6+54.6

=404.2 Scm2 mol1

For degree of dissociation, we have :-

α=Λm(HCOOH)Λ(HCOOH)

or α=46.1404.2=0.114

For dissociation constant, we have :-

Ka=cα21α

or Ka=0.025×(0.114)210.114

or =3.67×104 mol L1

Page no 54

Question 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?

Answer:

Firstly we will find total charge flown through the wire then we will calculate number of electrons.

We are given :- I = 0.5 A, Time = 2 hours = 7200 seconds.

We have, Q = I.t

= (0.5)7200 = 3600 C.

Now we will convert charge into number of electrons.

We know that 96487 C=6.023×1023 No. of electrons

So toal number of electrons :

=360096487×6.023×1023

or =2.25×1022 no. of electrons will flow through wire.

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

Answer:

Metals like Na, Mg, Al, etc. are produced on a large scale by electrochemical reduction of their respective cations or by the process electrolysis because there are no suitable reducing agents available for this purpose.

Question 2.12 Consider the reaction: Cr2O72+14H++6e2cr3++7H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72 ?

Answer:

It is clear from the given reaction that reduction of 1 mol of Cr2O72- will be

= 6 F (as 6 electrons are required to balance the reaction; Charge required = nF)

=6×96500

=579000 C

Thus 578922 C charge is required for reduction of 1 mol of Cr2O72- .

Page no 58

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

Answer:

The lead storage battery can be recharged by reversing the direction of current passing through it.

For recharging PbSO4 is converted into Pb at the anode and into PbO2 at the cathode.

The chemical reactions are as follows:-

1643888556447

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

Answer:

The two materials are methane and methanol that can be used as fuels in fuel cells.

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

Answer:

The chemistry of corrosion is quite complex but it can be understood by considering it as an electrochemical phenomenon. Consider a particular spot on an object where corrosion takes place. At here oxidation takes place and this spot behaves as the anode. The released electrons at anodic spot go through the metal and go to another spot on the metal and reduction of oxygen takes place in the presence of H+. This spot behaves as a cathode with the reaction. In this way this analogy is possible.


NCERT Solutions for Class 12 (Exercise Questions)

Question 2.1 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.

Answer:

The order in which metals displace each other from the solution of their salts can be given with the help of their standard electrode potential. Since magnesium has the least standard electrode potential so it is the most strong reducing agent. So the required order we get is:-

Mg>Al>Zn>Fe>Cu

Question 2.2 Given the standard electrode potentials,

K+/K=2.93V,Ag+/Ag=0.80V,

Hg2+/Hg=0.79V

Mg2+/Mg=2.37V,Cr3+/Cr=0.74V

Arrange these metals in their increasing order of reducing power.

Answer:

Elements with reducing power or reducing agents have least/minimum standard electrode potential i.e., reducing power increases with a decrease in standard electrode potential. So the result obtained is:-

K > Mg > Cr > Hg > Ag

Question 2.3 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?

Answer:

The galvanic cell of the given reaction is depicted below:-

Zn (s) | Zn+2 (aq) || Ag+ (aq) | Ag (s)

Clearly Zn electrode is negatively charged.

Question 2.3 Depict the galvanic cell in which the reaction

Zn(s)+2Ag+(aq)Zn2+(aq)+2Ag(s) takes place.

(ii) The carriers of the current in the cell.

Answer:

The carriers of current in the cell are ions. and Current flows from silver to zinc in the external circuit.

Question 2.3 Depict the galvanic cell in which the reaction Zn(s)+2Ag2+(aq)Zn2+(aq)+2Ag(s) takes place.

(iii) Individual reaction at each electrode.

Answer:

The reaction taking place at both cathode and anode are shown below :-

(i) Cathode reaction :-

Ag(aq)++eAg(s)

(ii) Anode reaction :-

Zn(s) Zn(aq)2+ + 2e

Question 2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(i) 2Cr(s)+3Cd2+(aq)2Cr3+(aq)+3Cd

Calculate the ΔrG0 and equilibrium constant of the reactions.

Answer:

The galvanic cell of the given reaction is shown below:-

Cr(s)|Cr(aq)3+Cd(aq)2+|Cd(s)

The standard electrode potential of Cr and Cd can be found in table of standard electrode potential.

So, we get :

E=ER  EL

=0.40  (0.74)

= 0.34 V

Now

ΔGr = nFEcell

Putting values :

ΔGr = 6×96487×0.34

=196.83 KJ mol1

Now for finding equlilibrium constant we have :

log k=ΔGr2.303×R×T

or log k=34.496

or K=3.13×1034

Question 2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(ii). Fe2+(aq)+Ag+(aq)Fe3+(aq)+Ag(s)

Calculate the ΔrGe and equilibrium constant of the reactions.

Answer:

The galvanic cell of the given reaction is shown below :-

Fe(aq)2+|Fe(aq)3+||Ag(aq)+|Ag(s)

We can know about the electrode potential of Fe and Ag with the help of table of standard electrode potential.

We have : E(cell)=EREL

or =0.800.77

or =0.03 V

Now consider : ΔGr= nFE(cell)

or =1×96487×0.03

=2.89 KJ mol1

Now for equilibrium constant :

log K= ΔGr2.303×RT

or = 2894.612.303×8.314×298

or = 0.5073

Thus k  3.2

Question 2.5 Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s)|Mg2+(0.001M)||Cu2+(0.000.1M)|Cu(s)

Answer:

The nernst equation gives :

ECell=Ecell 0.059nlog[Mg2+][Cu2+]

This gives,

=0.34(2.36)0.0592log0.0010.0001

=2.70.02955

=2.67 V

So the emf of the cell is 2.67 V.

Question 2.5 Write the Nernst equation and emf of the following cells at 298 K:

(ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)|Pt(s)

Answer:

The nernst equation for this gives :

ECell=Ecell 0.0591nlog[Fe+2][H+]2

This gives : =0(0.44)0.05912log0.00112

or =0.440.02955(3)=0.53 V

Thus the emf of the given galvanic cell is 0.53 V.

Question 2.5 Write the Nernst equation and emf of the following cells at 298 K:

(iii) Sn(s)|Sn2+(0.050M)||H+(0.020M)|H2(g)(1bar)Pt(s)

Answer:

The nernst equation for this reaction gives :-

ECell=Ecell 0.0591nlog[Sn+2][H+]2

Now for emf, just put all the values.

ECell=0(0.14)0.05912log0.0500.0202

or =0.140.0295×log125

or =0.140.062=0.078 V

Thus emf of the cell is 0.078 V.

Question 2.5 Write the Nernst equation and emf of the following cells at 298 K:

(iv) Pt(s)|Br(0.010M)|Br2(1)||H+(0.030M)|H2(g)(Ibar)Pt(s)

Answer:

The Nernst equation of the given reaction gives :

ECell=Ecell 0.0591nlog1[Br]2[H+]2

or =(01.09) 0.05912 log1(0.010)2(0.030)2

or =1.09 0.02955× log(1.11×107)

or =1.09 0.208= 1.298 V

So the required emf of the cell is -1.298 V.

Question 2.6 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 ΔrGe and Ee for the reaction.

Answer:

The given reaction is obtained from :-

Zn(s)Zn(aq)2++2e;E=0.76 V

Ag2O(s)+H2O(t)+2e2Ag(s)+2OH(aq);E=0.344 V

So the E o cell can be obtained directly.

Ecell=0.76(0.344)=1.104 V

Now for free energy calculation, we have :-

ΔGr=nFEcell

or =2×96487×1.04

or =213043.29 J

or =213.04 KJ

Question 2.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer:

Conductivity(k) or specific conductance of a solution is defined as the inverse of resistivity.

Mathematically, it can be written as:-

G=κAL

In the above equation is κ the conductivity of a solution. Thus the definition of conductivity becomes as the conductance of a substance which is 1 cm long and has 1 sq. cm of cross-sectional area.

With dilution conductivity of a solution decreases due to an increase in distance between ions.

Molar conductivity: - It is defined as the conductivity of a solution per unit concentration

i.e., ΛM = κC

It is clear from the above mathematical expression of the molar conductivity that, if we dilute the solution or decrease its concentration then molar conductivity increases. This is because, on dilution of a solution, a decrease in is κ more than compensated by the increase in its volume.

Question 2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.

Answer:

We know that the molar conductivity of a solution is defined as:-

ΛM =κC

Putting the value of conductivity and concentration in the above equation:-

ΛM =0.0248×10000.20=124 Scm2 mol1

Question 2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 ohm . What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146×103 Scm1 .

Answer:

We are given with conductivity of cell κ= 0.146×103 Scm1 and resistance R = 1500 Ω .

Also, Cell constant = κ×R

or = 0.146×103×1500

or = 0.219 cm1

Question 2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Calculate Λm for all concentrations and draw a plot between Λm and c1/2. Find the value of Λm0.

Answer:

Given,

κ=1.237×102 S m1,c=0.001M

Then, κ=1.237×104 S cm1,c1/2=0.0316M1/2

Λm=kc=1.237×104Scm10.001molL1×1000 cm3L=123.7 S cm2 mol1

Given,

κ=11.85×102 S m1,c=0.010M
Then, κ=11.85×104 S cm1,c1/2=0.1M1/2

Λm=kc=11.85×104Scm10.010 mol L1×1000 cm3L=118.5 S cm2 mol1

Given,

κ=23.15×102 S m1,c=0.020M
Then, κ=23.15×104 S cm1,c1/2=0.1414M1/2

Λmkc=23.15×104Scm10.020 mol L1×1000 cm3L=115.8 S cm2 mol1

Given,

κ=55.53×102 S m1,c=0.050M Then, κ=55.53×104 S cm1,c1/2=0.2236M1/2k=kc=55.53×104Scm10.050 mol L1×1000 cm3L=111.11 S cm2 mol1

Given,

κ=106.74×102 S m1,c=0.100M
Then, κ=106.74×104 S cm1,c1/2=0.3162M1/2

Λm=kc=106.74×104 S cm10.100 mol L1×1000 cm3L=106.74 S cm2 mol1
Now, we have the following data:

c1/2/m1/2
0.0316
0.10.1414
0.2236
0.3162
Λm(Scm2 mol1)
123.7
118.5
115.8
111.1
106.74

Since the line interrupts Λm at 124.0 S cm2 mol1,Λm=124.0 S cm2 mol1.

Question 2.11 Conductivity of 0.00241 M acetic acid is 7.896×105Scm1 . Calculate its molar conductivity. If Δm0 for acetic acid is 390.5Scm2mol1 , what is its dissociation constant?

Answer:

Molar conductivity of a solution is given by :-

ΛM=κC

So, =7.896×1050.00241×1000

or =32.76 Scm2 mol1

Also, it is given that Λm=390.5 Scm2 mol1 .

α=ΛmΛm

or α=32.76390.5

α=0.084

For dissociation constant we have,

Kd =cα2(1α)

so, =0.00241×0.0842(10.084)

or =1.86×105 mol L1

Question 2.12 How much charge is required for the following reductions:

(i) 1 mol of Al3+ to Al ?

Answer:

The equation becomes:-

Al+3 +3e = Al

So required charge is 3F.

Q = n*96500

Q = 3*96500 = 289500 C

Question 2.12 How much charge is required for the following reductions:

(ii) 1 mol of Cu2+ to Cu?

Answer:

The equation can be written as:-

Cu2+ + 2e = Cu

Thus charge required is = 2F

=2(96500)=193000 C

Question 2.12 How much charge is required for the following reductions:

(iii) MnO4 to Mn2+ ?

Answer:

The given reaction can be written as:-

Mn+7 + 5e= Mn+2

Thus charge required in above equation = 5F

=5(96500)

=482500 C

Question 2.13 How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl2 ?

Answer:

The equation for the question is given by :-

Ca2+ + 2e = Ca

In this equation, for 1 mol of Ca, 2F charge is required or we can say that for 40 g of Ca charge required is 2F.

So, for 20 g of Ca charge required will be = F = 96500 C.

Question 2.13 How much electricity in terms of Faraday is required to produce

(ii) 40.0 g of AI from molten Al2O3 ?

Answer:

The equation for the given question is :-

Al+3 +3e = Al

Thus for 1 mol of Al, charge required is 3F.

So the required amount of electricity in terms of charge will be :-

= 327×40F=4.44F

Question 2.14 How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H2O to O2 ?

Answer:

According to question the equation of oxidation will be :-

O2  12O2 + 2e

Thus, for oxidation of O 2- , 2F charge is required.

=2×96500 C

=193000 C

Question 2.14 How much electricity is required in coulomb for the oxidation of

(ii) 1 mol of FeO to Fe2O3 ?

Answer:

The oxidation equation for the given reaction will be :-

Fe+2  Fe+3 + e

So for oxidation of 1 mol Fe+2 charge required =1F

=96500 C

Question 2.15 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?

Answer:

We are given:

I = 5A

and t = 20(60) = 1200 sec.

So total charge = 5(1200) = 6000 C.

The equation for nickel deposition will be:-

Ni+2 + 2e  Ni

Thus, from 2F charge 58.7 g of nickel deposition takes place.

i.e., 2(96487) C58.7 g Ni

So for 6000 C charge total nickel deposition will be:-

=58.72×96487×6000

or =1.825 g

Hence 1.825 g Ni will be deposited in the given conditions.

Question 2.16 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 1.45g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Answer:

Since the cells are connected in series so the current passing through each cell will be equal.(1.5 A)

Now we are given that 1.45 g of silver is deposited. So firstly we will consider the cell containing silver.

Ag+ + e Ag

Since for deposition of 108 g silver 96487 C charge is required, thus for 1.45 g deposition of silver charge required will be:-

=96487108×1.45 =1295.43 C

Now we can find the time taken by 1.5 A current to deposit 1.45 g silver.

Time taken=1295.431.5864 sec.

For copper:-

Cu+2 +2e = Cu

Since 2F charge will deposit 63.5 g of Cu, then deposition by 1295.43 C will be:-

=63.52×96487×1295.43 =0.426 g

Hence 0.426 g of copper will be deposited.

For zinc:-

Zn+2 + 2e  Zn

Since 2F charge will deposit 65.4 g of Zn, then deposition by 1295.43 C will be:-

=65.42×96487×1295.43 =0.439 g

Hence 0.439 g of zinc will be deposited.

Question 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Feaq3+ and Iaq

Answer:

The concept used here will be that a reaction is feasible only if Ecell is positive.

Anode and cathode reactions will be as follows:-

Fe3+ + e = Fe2+ E =0.77 V

2I = I2 + 2e E =0.54 V

So Ecell=0.770.54=0.23 V

So this reaction is feasible.

Question 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(ii) Agaq+ and Cu(s)

Answer:

A reaction is feasible only if Ecell is positive.

So, anode and cathode reactions will be as follows :-

(Ag+ + e = Ag)×2 E =0.80 V

Cu = Cu+2 + 2e E =0.34 V

and Ecell=0.800.34=0.46 V

So this reaction is feasible.

Question 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(iii) Feaq3+and Braq

Answer:

A reaction is feasible only if Ecell is positive.

So, anode and cathode reactions will be as follows :-

(Fe+3 + e = Fe+2)×2 E =0.77 V

2Br = Br2 + 2e E =1.09 V

and Ecell=0.771.09=0.32 V

So this reaction is not feasible.

Question 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(iv) Ags and Feaq3+

Answer:

A reaction is feasible only if Ecell is positive.

So, anode and cathode reactions will be as follows:-

Ag = Ag+ + e E =0.80 V

Fe+3 + e = Fe+2 E =0.77 V

and Ecell=0.80+0.77=0.03 V

So this reaction is not feasible.

Question 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(v) Br2(aq) and Feaq2+

Answer:

A reaction is feasible only if Ecell is positive.

So, anode and cathode reactions will be as follows :-

Br2 + 2e = 2Br E =1.09 V

Fe2+ = Fe3+ + e E =0.77 V

and Ecell=1.090.77=0.32 V

So this reaction is feasible.

Question 2.18 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes.

Answer:

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

Ag+ + e = Ag(s)

At anode :-

Ag + NO3 = AgNO3 + e

Hence, silver will get deposited at the cathode and it will be getting dissolved at anode.

Question 2.18 Predict the products of electrolysis in each of the following:

(ii) An aqueous solution of AgNO3 with platinum electrodes.

Answer:

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

Ag+ + e = Ag(s)

At anode :- Self ionisation will take place due to presence of water.

H2O  2H+ +12O2 + 2e

Hence, silver will get deposited at the cathode and O 2 will be produced from anode.

Question 2.18 Predict the products of electrolysis in each of the following:

(iii) A dilute solution of H2SO4 with platinum electrodes.

Answer:

For the given solution:

At cathode :- Reaction with greater E 0 will take place.

H++e1/2H2

At anode :- Self ionisation of water will take place due to presence of platinum electrode.

H2O  2H+ +12O2 + 2e

Hence, H 2 gas will be generated at cathode and O 2 will be produced from anode.

Question 2.18 Predict the products of electrolysis in each of the following:

(iv) An aqueous solution of CuCI2 with platinum electrodes.

Answer:

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

Cu2++2eCu(s)

At anode :-

2Cl = Cl2 + 2e

Hence, Cu will get deposited at cathode and Cl 2 will be produced from anode.

Class 12 Chemistry NCERT Chapter 2: Higher Order Thinking Skill (HOTS) Questions

Question: The standard cell potential (Ecell ) of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V . The standard half cell reduction potential for O2(EO2/H2O0) is 1.229 V.

Choose the correct statement:

(1) The standard half cell reduction potential for the reduction of CO2(ECO2/CH3OH0) is 19 mV

(2) Oxygen is formed at the anode.

(3) Reactants are fed at one go to each electrode.

(4) Reduction of methanol takes place at the cathode.

Answer:

Fuel cell reaction

CH3OH(I)+32O2( g)CO2( g)+2H2O(I)
Here O2 reduces to H2O and CH3OH oxidises to CO2.

Standard cell potential:

Ecell =Ecathode Eanode 

Given:

Ecell =1.21 VE02/H2O=1.229 V

Substituting the values,

1.21=1.229Eanode 

Solving for Eanode  :

Eanode =1.2291.21=0.019 V

0.019 V convert into mV multiply by 1000 = 0.019 * 1000 = 19mV

Hence, the correct answer is option (1).

Question:

Given below are two statements :
1 M aqueous solution of each of Cu(NO3)2, AgNO3,Hg2(NO3)2;Mg(NO3)2 are electrolysed using inert electrodes,
Given : EAg/Ag0=0.80 V,EHg22+/Hg0=0.79 V, ECu2+/Cu0=0.24 V and EMg2+/Mg0=2.37 V

Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag,Hg and Cu

Statement (II) : Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.

In the light of the above statement, choose the most appropriate answer from the options given below

(1) Both statement I and statement II are incorrect

(2) Statement I is correct but statement II is incorrect

(3) Both statement I and statement II are correct

(4) Statement I is incorrect but statement II is correct

Answer:

We are electrolyzing 1 M aqueous solutions of the following nitrates using inert electrodes:

  • Cu(NO₃)₂

  • AgNO₃

  • Hg₂(NO₃)₂

  • Mg(NO₃)₂

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

At the cathode, reduction occurs. So we must look at the standard reduction potentials (E°) for the metal cations.

From the electrochemical series, we can see that,

ReactionE° (V)
Ag⁺ + e⁻ → Ag+0.80 V
Hg₂²⁺ + 2e⁻ → 2Hg+0.79 V
Cu²⁺ + 2e⁻ → Cu+0.24 V
Mg²⁺ + 2e⁻ → Mg–2.37 V
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook
  • In electrolysis, the ion with the highest E° gets reduced (i.e., deposited) first at the cathode.

  • So as the voltage increases, the cations get deposited in order of decreasing E°.

Thus, the expected order of deposition at the cathode is:

Ag>Hg>Cu

So, Statement I: "With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg, and Cu" is correct.

Let’s analyze the possibility of Mg²⁺ getting reduced:

  • E° for Mg²⁺/Mg = –2.37 V (very negative).

  • Water can also be reduced at the cathode:

2H2O+2e>H2+2OHE=0.83 V

  • Since water has a less negative E° than Mg²⁺, water gets reduced before Mg²⁺.

So, Mg²⁺ is not deposited — instead, hydrogen gas is evolved at the cathode.

But Statement II says:
"Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode", is incorrect because oxygen is evolved at the anode, not at the cathode.

At the cathode, hydrogen gas evolves if Mg is not deposited.

Hence, the correct answer is option (2).

Approach to Solve Questions of Chapter 2

Electrochemistry is the study of electron movement in an oxidation or reduction reaction at a polarized electrode surface. Here is a more detailed approach on how students can approach solving questions of the NCERT class 12 Electrochemistry:

1. Understand the Key Concepts First

Before attempting questions, ensure you're clear on:

  • Redox reactions and balancing
  • Galvanic and Electrolytic cells

2. Pay special attention on topics like:

  • Electrode potential, standard electrode potential
  • Cell notation and EMF (E°cell)
  • Nernst equation
  • Gibbs free energy and cell potential
  • Conductance and molar conductivity
  • Kohlrausch’s Law
  • Electrolysis and Faraday’s laws

3. Resources to Use

  • NCERT Textbook & NCERT Exemplar
  • Coaching material or reference books like OP Tandon, Pradeep
  • Previous Year Papers

4. Tips to solve questions

  • Always write units in numericals
  • For EMF, ensure you use the sign of the standard potentials carefully.
  • Practice step-by-step problem solving rather than shortcuts.
  • Make a sheet of formulas for quick revision

Topics of NCERT Class 12 Chemistry Chapter 2

2.1 Electrochemical Cells

2.2 Galvanic Cells

2.2.1 Measurement of Electrode Potential

2.3 Nernst Equation

2.3.1 Equilibrium Constant from Nernst Equation

2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction

2.4 Conductance of Electrolytic Solutions

2.4.1 Measurement of the Conductivity of Ionic Solutions

2.4.2 Variation of Conductivity and Molar Conductivity with Concentration

2.5 Electrolytic Cells and Electrolysis

2.5.1 Products of Electrolysis

2.6 Batteries

2.6.1 Primary Batteries

2.6.2 Secondary Batteries

2.7 Fuel Cells

2.8 Corrosion

What Extra Should Students Study Beyond NCERT for JEE?

Here's a comparison table highlighting what to study beyond NCERT for JEE:

Formulas Of NCERT Class 12 Chemistry Electrochemistry

1. Conductance(G) is the reciprocal of resistance (R) and specific conductance or conductivity(k) is inverse of resistivity (ρ)

G=1R=1ρ(al)k=G(la)

2. l/a is called the cell constant of conductivity cell.

3. Equivalent Conductivity is defined as the conductance of a solution containing 1g of an electrolyte.

Λeq=K×V

4. Nernst equation

aA+bB cC+dD

Ecell=Ecello0.0591nlog[C]c[D]d[A]a[B]b

NCERT Solutions for Class 12 Chemistry

Below are the chapter-wise solutions-

NCERT Exemplar Class 12 Solutions Subject-wise

The hyperlinks of NCERT exemplar of class 12 are given below:

NCERT Solutions for Class 12 Subject-wise

The hyperlinks of the NCERT solution of class 12 are given below:


NCERT Books and NCERT Syllabus here:

Students can refer to the links given below for the NCERT books and Syllabus:



Frequently Asked Questions (FAQs)

1. What is electrochemistry?

Electrochemistry is the branch of chemistry that deals with the interconversion of chemical energy and electrical energy. It involves the study of chemical reactions that occur in electrochemical cells, where oxidation and reduction processes take place. This field finds applications in batteries, fuel cells, and corrosion science.

2. What are the different types of electrochemical cells?

There are two main types of electrochemical cells: 

  • Galvanic cells
  • Electrolytic cells

Galvanic cells convert chemical energy into electrical energy through spontaneous reactions, while electrolytic cells use electrical energy to drive non-spontaneous chemical reactions.

3. How do you calculate the EMF of a galvanic cell?

The EMF of a galvanic cell can be calculated using the Nernst equation, which relates the cell potential to the concentration of reactants and products. 

The general form of the equation is:

E = E° - (RT/nF) ln Q

were,

  • E° is the standard cell potential
  • R is the universal gas constant
  • T is the temperature in Kelvin
  • n is the number of moles of electrons exchanged
  • F is Faraday's constant.

4. What is the significance of the standard electrode potential?

The standard electrode potential is a measure of the inherent tendency of a half-cell to be reduced under standard conditions: 1 M concentration, 1 atm pressure, and 25°C. It allows us to predict the direction of spontaneous reactions, determine cell voltages, and compare the reactivity of different electrodes.

5. What is the concept of conductance and its importance in electrochemistry?

Conductance refers to the ability of a solution to conduct electric current. It is influenced by the concentration of ions present in the solution. In electrochemistry, conductance plays a critical role in determining the efficiency of electrochemical reactions, as higher ionic concentration typically leads to better conductivity and faster reaction rates.

Articles

Explore Top Universities Across Globe

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top