NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry
NCERT Solutions for Class 12 Chemistry Chapter 3 Solutions contains all the questions with detailed solutions. Students are advised to practice these questions for better understanding of the concepts given in the chapter.
Krati Saraswat24 Feb, 2024
Share
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 3 Electrochemistry while going before solving the NCERT questions. Our Physics Wallah team Prepared Other Subjects NCERT Solutions for class 12.NCERT Solutions for Class 12 Chemistry Chapter 1
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Overview
NCERT Solutions for Class 12 Chemistry Chapter 3 cover several important topics. It is highly recommended for students to review each topic thoroughly in order to gain a comprehensive understanding of the concepts taught in the chapter and make optimal use of the provided solutions. These solutions are the result of dedicated efforts by the Physics Wallah teachers aimed at assisting students in grasping the concepts covered in this chapter. By going through and practicing these solutions, the objective is for students to achieve excellent results in their exams effortlessly.CBSE Class 12th Toppers List 2024
NCERT Solutions for Class 12 Chemistry Chapter 3
Answer the following Questions of NCERT Solutions for Class 12 Chemistry Chapter 3
Question 1. How would you determine the standard electrode potential of the system Mg 2+ | Mg? Solution : 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 ) (1 M). A cell, consisting of Mg | MgSO 4 (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
E
ø
= E
ø
R - E
ø
L
Here,E
ø
R for the standard hydrogen electrode is zero.
∴E
ø
= 0 - E
ø
L
= - E
ø
L
NCERT Solutions for Class 12 Chemistry Chapter 2
Question 2. Can you store copper sulphate solutions in a zinc pot? Solution : Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution. Zn + CuSO 4 → ZnSO 4 + Cu Hence, copper sulphate solution cannot be stored in a zinc pot. Question 3. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. Solution : Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions. Fe 2+ → Fe 3+ + e -1 ; E ø = −0.77 V This implies that the substances having higher reduction potentials than +0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are F 2 , Cl 2 , and O 2 .NCERT Solutions for Class 12 Chemistry Chapter 4
Question 4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. Solution : For hydrogen electrode,
, it is given that pH = 10
∴[H
+
] = 10
−10
M
Now, using Nernst equation:
= −0.0591 log 10
10
= −0.591 V
NCERT Solutions for Class 12 Chemistry Chapter 5
Question 5. Calculate the emf of the cell in which the following reaction takes place: Ni (s) + 2Ag + (0.002 M) → Ni 2+ (0.160 M) + 2Ag(s) Given that E ø cell = 1.05 V Solution : Applying Nernst equation we have:
= 1.05 − 0.02955 log 4 × 10
4
= 1.05 − 0.02955 (log 10000 + log 4)
= 1.05 − 0.02955 (4 + 0.6021)
= 0.914 V
Question
6. The cell in which the following reactions occurs:
has E
ø
cell = 0.236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution :
Here, n = 2, E
ø
cell = 0.236 T = 298 K
We know that:
Question
7. Why does the conductivity of a solution decrease with dilution?
Solution :
The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.
Question 8. Suggest a way to determine the Λ
∘
m value of water.
Solution :
Applying Kohlrausch’s law of independent migration of ions, the value of water can be determined as follows:
Λ°m(H
2
O) = Λ°m(HCl) + Λ°m(NaOH) – Λ°m(NaCl)
Hence, by knowing the Λ°m values of HCl, NaOH, and NaCl, the Λ°m value of water can be determined.
Question
9. The molar conductivity of 0.025 mol L
−1
methanoic acid is
46.1 S cm2 mol
−1.
Calculate its degree of dissociation and dissociation constant. Given λ
°(H+)
= 349.6 S cm
2
mol
−1
and λ °(HCOO
−
) = 54.6 S cm
2
mol
Solution :
C = 0.025 mol L
−1
Now, degree of dissociation:
Thus, dissociation constant:
Question
10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Solution :
I = 0.5 A
t = 2 hours = 2 × 60 × 60 s = 7200 s
Thus, Q = It
= 0.5 A × 7200 s
= 3600 C
We know that 96487C = 6.023 X 10
23
number of electrons.
Then,
Hence, 2.25 X 10
22
number of electrons will flow through the wire.
Question 11. Suggest a list of metals that are extracted electrolytically.
Solution :
Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.
Question
12. What is the quantity of electricity in coulombs needed to reduce 1 mol of
cr
2
O
7
2
? Consider the reaction:
Cr
2
O
7
2-
+ 14H+ + 6e
-
→ Cr
3
+ + 8H
2
O
Solution :
The given reaction is as follows:
Cr
2
O
7
2-
+ 14H+ + 6e
-
→ Cr
3
+ + 8H
2
O
Therefore, to reduce 1 mole of
cr
2
O
7
2
, the required quantity of electricity will be:
=6 F
= 6 × 96487 C
= 578922 C
Question
13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Solution :
A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (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
2
-4(aq) → PbSO
4
(s) + 2e-
At cathode:
The overall cell reaction is given by,
When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, present at the anode and cathode is converted into and respectively.
Question
14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Solution :
Methane and methanol can be used as fuels in fuel cells.
Question
15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Solution :
In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,
Fe
(s)
→ Fe
2+
(aq) + 2e-
Electrons released at the anodic spot move through the metallic object and go to another spot of the object.
There, in the presence of H
+
ions, the electrons reduce oxygen. This spot behaves as the cathode. These H
+
ions come either from H
2
CO
3,
which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.
The reaction corresponding at the cathode is given by,
The overall reaction is:
Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide(Fe
2
O
3
, x)H
2
Oi.e., rust.
Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.
Question
16. 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
Solution :
The following is the order in which the given metals displace each other from the solution of their salts.
Mg, Al, Zn, Fe, Cu
Question
17. Given the standard electrode potentials,
K
+/K
= −2.93V, Ag
+
/Ag = 0.80V,
Hg
2+
/Hg = 0.79V
Mg
2+
/Mg = −2.37 V, Cr
3+
/Cr = − 0.74V
Arrange these metals in their increasing order of reducing power.
Solution :
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 increases in the following order:
Ag < Hg < Cr < Mg < K
Question
18. Depict the galvanic cell in which the reaction Zn(s) + 2Ag
+
(aq)
→
Zn
2+
(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.
Solution :
The galvanic cell in which the given reaction takes place is depicted as:
(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
2+
(aq) + 2e-
The reaction taking place at the cathode is given by,
Ag
+
(aq) + e
-
→ Ag(s)
Question
19. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd
2+
(aq)
→
2Cr
3+
(aq) + 3Cd
(ii) Fe
2+(aq) + Ag+
(aq)
→
Fe
3+
(aq) + Ag(s)
Calculate the Δ
r
G
θ
and equilibrium constant of the reactions.
Solution :
(i) E
ø
Cr
3+
/ Cr = - 0.74 V
Eø Cd
2+
/ Cd = - 0.40 V
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
E
ø
= E
ø
R - E
ø
L
= -0.40 - (-0.74)
= + 0.34 V
ΔrG
ø
= -nFE
ø
cell
In the given equation,
n = 6
F = 96487 C mol
−1
E
ø
cell = +0.34 V
Then, ΔrG
ø
= −6 × 96487 C mol
−1
× 0.34 V
= −196833.48 CV mol
−1
= −196833.48 J mol
−1
= −196.83 kJ mol
−1
Again,
ΔrG
ø
= −RT ln K
= 34.496
⇒ K = antilog (34.496)
= 3.13 × 10
34
(ii) E
ø
Fe3+
/ Fe
2+
= - 0.77 V
E
ø
Ag
+
/ Ag = - 0.80 V
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
E
ø
= E
ø
R - E
ø
L
= 0.80 - 0.77
= 0.03 L
Here, n = 1.
Then, ΔrG
ø
= -nFE
ø
= −1 × 96487 C mol
−1
× 0.03 V
= −2894.61 J mol
−1
= −2.89 kJ mol
−1
Again, ΔrG
ø = - 2.303 RT In K
= 0.5073
∴ K = antilog (0.5073)
= 3.2 (approximately)
Question
20. Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s) | Mg
2+
(0.001M) || Cu
2
+(0.0001 M) | Cu(s)
(ii) Fe(s) | Fe
2+
(0.001M) || H+(1M)|H
2
(g)(1bar) | Pt(s)
(iii) Sn(s) | Sn
2+
(0.050 M) ||
H+
(0.020 M) | H
2
(g) (1 bar) | Pt(s)
(iv) Pt(s) | Br
2
(l) | Br
−
(0.010 M) ||
H+
(0.030 M) | H
2
(g) (1 bar) | Pt(s).
Solution :
(i) For the given reaction, the Nernst equation can be given as:
= 2.7 − 0.02955
= 2.67 V (approximately)
(ii) For the given reaction, the Nernst equation can be given as:
= 0.52865 V
= 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:
= 0.14 − 0.0295 × log125
= 0.14 − 0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:
Question
21. 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) + 2Ag(s) + 2OH−(aq)
Determine and for the reaction.
Solution :
∵ E
ø
= 1.104 V
We know that,
ΔrG
ø
= -nFE
ø
= −2 × 96487 × 1.04
= −213043.296 J
= −213.04 kJ
Question
22. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Solution :
Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbolκ. If ρ is resistivity, then we can write:
k = 1 / p
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.,
(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.
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
A
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 A
m
with
for strong and weak electrolytes is shown in the following plot:
Question
23. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm
−1
. Calculate its molar conductivity.
Solution :
Given,
κ = 0.0248 S cm
−1
c = 0.20 M
⇒ Molar conductivity, A
m
= (k × 1000) / c
= 0.0248 × 1000 /0.20
= 124 Scm
2
mol
−1
Question
24. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10
−3
S cm
−1.
Solution :
Given,
Conductivity, κ = 0.146 × 10
−3
S cm
−1
Resistance, R = 1500 Ω
⇒ Cell constant = κ × R
= 0.146 × 10
−3 × 1500
= 0.219 cm
−1
Question
25. 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
10
2
× κ/S m
−1
1.237 11.85 23.15 55.53 106.74
Calculate λ
m
for all concentrations and draw a plot between
λ
m
and c½. Find the value of λº
m
.
Solution :
Given,
κ = 1.237 × 10
−2 S m−1,
c = 0.001 M
Then, κ = 1.237 × 10
−4
S cm
−1,
c
½
= 0.0316 M
1/2
= 123.7 S cm
2
mol
−1
Given,
κ = 11.85 × 10
−2
S m
−1
, c = 0.010M
Then, κ = 11.85 × 10
−4
S cm
−1,
c
½
= 0.1 M
1/2
= 118.5 S cm
2
mol
−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
= 115.8 S cm
2
mol
−1
Given,
κ = 55.53 × 10
−2
S m
−1,
c = 0.050 M
Then, κ = 55.53 × 10
−4
S cm
−1
, c
1/2
= 0.2236 M
1/2
= 111.1 1 S cm
2
mol−
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
= 106.74 S cm
2
mol
−1
Now, we have the following data:
| C 1/2 / M 1/2 | 0.0316 | 0.1 | 0.1414 | 0.2236 | 0.3162 |
| λ m (Scm 2 mol- 1 ) | 123.7 | 118.5 | 115.8 | 111.1 | 106.74 |
Since the line interrupts λm at 124.0 S cm
2
mol
−1,
λº
m
= 124.0 S cm
2
mol
−1
.
Question
26. Conductivity of 0.00241 M acetic acid is 7.896 × 10
−5 S
cm
−1
. Calculate its molar conductivity and if A
m
º for acetic acid is 390.5 S cm
2
mol
−1
, what is its dissociation constant?
Solution :
Given, κ = 7.896 × 10
−5
S m
−1
c = 0.00241 mol L
−1
Then, molar conductivity,
A
m = k / c
= 32.76S cm
2 mol−1
Again,
A
m
º
= 390.5 S cm
2 mol−1
Now,
= 0.084
⇒ Dissociation constant,
⇒ (0.00241 mol L
-1
)(0.084)
2
/ (1-0.084)
= 1.86 × 10
−5
mol L
−1
Question
27. 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+.
Solution :
(i)
Al
3+
+ 3e
-
→ Al
⇒ Required charge = 3 F
= 3 × 96487 C
= 289461 C
(ii) Cu
2+
+ 2e
-
→ Cu
⇒ Required charge = 2 F
= 2 × 96487 C
= 192974 C
(iii)
MnO
4
-
→ Mn
2+
i.e.,
Mn
7+
+ 5e
-
→ Mn
2+
⇒ Required charge = 5 F
= 5 × 96487 C
= 482435 C
Question
28. 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
.
Solution :
(i) According to the question,
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium(2 × 20) /40
= 1 F
(ii) According to the question,
Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al = 3F
= 4.44 F
Question
29. 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
.
Solution :
(i) According to the question,
H
2
O → H
2
+ ½O
2
.
Now, we can write:
O
2-
→
½O
2
+ 2e
-
Electricity required for the oxidation of 1 mol of H
2
O to O
2
= 2 F
= 2 × 96487 C
= 192974 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
= 96487 C
Question
30. 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?
Solution :
Given,
Current = 5A
Time = 20 × 60 = 1200 s
∵ Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C = (58.71 X 6000) / (2 X 96487) g
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
Question
31. 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?
Solution :
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by = 96487 X 1.45 / 108 C
= 1295.43 C
Given,
Current = 1.5 A
⇒ Time = 1295.43 /1.5s
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit = (63.5x1295.43) / (2x96487) g
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit
= (65.4x1295.43) / (2x96487) g
= 0.439 g of Zn
Question
32. Using the standard electrode potentials given in Table 3.1, 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).
Solution :
Since Eº for the overall reaction is positive, the reaction between Fe
3+
(
aq) and I
−
(aq) is feasible.
Since Eº for the overall reaction is positive, the reaction between Ag
+
(aq
)
and Cu
(
s
)
is feasible.
Since Eº for the overall reaction is negative, the reaction between Fe
3+
(
aq
)
and Br
−
(
aq
)
is not feasible.
Since Eº for the overall reaction is negative, the reaction between Ag
(
s
)
and Fe
3+
(aq)
is not feasible.
Since Eº for the overall reaction is positive, the reaction between Br
2(
aq
)
and Fe
2+
(
aq
)
is feasible.
Question
33. Predict the products of electrolysis in each of 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.
Solution :
(i) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of Eº takes place at the cathode. 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 form Ag
+.
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of Eº takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
Since Pt electrodes are inert, the anode is not attacked by NO
3
-
ions. Therefore, OH
−
or NO
3
-
ions can be oxidized at the anode. But OH
−
ions having a lower discharge potential and get preference and decompose to liberate O
2
.
OH
-
→ OH + e
-
4OH
-
→ 2H
2
O +O
2
(iii) At the cathode, the following reduction reaction occurs to produce H
2
gas.
H+ (aq) + e- → ½ H
2
(g)
At the anode, the following processes are possible.
For dilute sulphuric acid, reaction (i) is preferred to produce O
2
gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of Eº takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
At anode:
The following oxidation reactions are possible at the anode.
At the anode, the reaction with a lower value of Eº is preferred. But due to the over-potential of oxygen, Cl
−
gets oxidized at the anode to produce Cl
2
gas.
NCERT Solutions for Class 12 Chemistry Chapter 3 Solutions FAQs
Why is electrochemistry important in chemistry?
Electrochemistry plays a crucial role in various fields such as energy storage (batteries), corrosion prevention, electroplating, and biochemical processes. It helps understand the principles of electron transfer in chemical reactions.
How can NCERT Solutions for Class 12 Chemistry Chapter 3 help students?
These solutions offer clear explanations and step-by-step solutions to the problems, helping students grasp the concepts of electrochemistry more effectively.
Are the solutions provided in NCERT Solutions for Class 12 Chemistry Chapter 3 accurate?
Yes, the solutions are prepared by experienced subject matter experts and are based on the content provided in the NCERT textbook. They are accurate and aim to provide clear explanations to students.
What are the applications of electrochemical cells?
Electrochemical cells find applications in various devices such as batteries, fuel cells, and sensors. They are also used in electroplating processes to deposit a layer of metal on a surface.
🔥 Trending Blogs
Talk to a counsellorHave doubts? Our support team will be happy to assist you!
Join 15 Million students on the app today!
Live & recorded classes available at easeDashboard for progress trackingMillions of practice questions at your fingertips
Free Learning Resources
PW Books
Notes (Class 10-12)
PW Study Materials
Notes (Class 6-9)
Ncert Solutions
Govt Exams
Class 6th to 12th Online Courses
Govt Job Exams Courses
UPSC Coaching
Defence Exam Coaching
Gate Exam Coaching
Other Exams
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Copyright © 2025 Physicswallah Limited All rights reserved.
