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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 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.
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Page 36
Question 2.1 How would you determine the standard electrode potential of the system
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.
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
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:-
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.
By Nernst equation we have :-
So,
or
So the required potential is - 0.591 V.
Question 2.5 Calculate the emf of the cell in which the following reaction takes place:
Answer:
Here we can directly apply the Nernst equation :-
Putting the value in this equation :-
Hence the required potential is 0.914 V.
Answer:
For finding Gibbs free energy we know the relation :-
Now, for equilibrium constant we will use :-
So,
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
Answer:
We know :
If we draw a straight line between
In this way, we can obtain the value of limiting molar conductivity.
Question 2.9 The molar conductivity of
Answer:
We know that :-
For degree of dissociation, we have :-
or
For dissociation constant, we have :-
or
or
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
So toal number of electrons :
or
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:
What is the quantity of electricity in coulombs needed to reduce 1 mol of
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)
Thus 578922 C charge is required for reduction of 1 mol of Cr2O72- .
Page no 58
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:-
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.
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:-
Question 2.2 Given the standard electrode potentials,
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
(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
(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
(iii) Individual reaction at each electrode.
Answer:
The reaction taking place at both cathode and anode are shown below :-
(i) Cathode reaction :-
(ii) Anode reaction :-
Question 2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
Calculate the
Answer:
The galvanic cell of the given reaction is shown below:-
The standard electrode potential of Cr and Cd can be found in table of standard electrode potential.
So, we get :
Now
Putting values :
Now for finding equlilibrium constant we have :
or
or
Question 2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
Calculate the
Answer:
The galvanic cell of the given reaction is shown below :-
We can know about the electrode potential of Fe and Ag with the help of table of standard electrode potential.
We have :
or
or
Now consider :
or
Now for equilibrium constant :
or
or
Thus
Question 2.5 Write the Nernst equation and emf of the following cells at 298 K:
(i)
Answer:
The nernst equation gives :
This gives,
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)
Answer:
The nernst equation for this gives :
This gives :
or
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)
Answer:
The nernst equation for this reaction gives :-
Now for emf, just put all the values.
or
or
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)
Answer:
The Nernst equation of the given reaction gives :
or
or
or
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:
Determine
Answer:
The given reaction is obtained from :-
So the E o cell can be obtained directly.
Now for free energy calculation, we have :-
or
or
or
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:-
In the above equation is
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.,
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
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:-
Putting the value of conductivity and concentration in the above equation:-
Answer:
We are given with conductivity of cell
Also, Cell constant =
or =
or =
Question 2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Calculate
Answer:
Given,
Then,
Given,
Then,
Given,
Then,
Given,
Given,
Then,
Now, we have the following data:
0.0316 | 0.1 | 0.1414 | 0.2236 | 0.3162 | |
123.7 | 118.5 | 115.8 | 111.1 | 106.74 |
Since the line interrupts
Question 2.11 Conductivity of 0.00241 M acetic acid is
Answer:
Molar conductivity of a solution is given by :-
So,
or
Also, it is given that
or
For dissociation constant we have,
so,
or
Question 2.12 How much charge is required for the following reductions:
(i)
Answer:
The equation becomes:-
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)
Answer:
The equation can be written as:-
Thus charge required is
Question 2.12 How much charge is required for the following reductions:
(iii)
Answer:
The given reaction can be written as:-
Thus charge required in above equation
Question 2.13 How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten
Answer:
The equation for the question is given by :-
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
Answer:
The equation for the given question is :-
Thus for 1 mol of Al, charge required is 3F.
So the required amount of electricity in terms of charge will be :-
Question 2.14 How much electricity is required in coulomb for the oxidation of
(i) 1 mol of
Answer:
According to question the equation of oxidation will be :-
Thus, for oxidation of O 2- , 2F charge is required.
Question 2.14 How much electricity is required in coulomb for the oxidation of
(ii) 1 mol of
Answer:
The oxidation equation for the given reaction will be :-
So for oxidation of 1 mol
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:-
Thus, from 2F charge 58.7 g of nickel deposition takes place.
i.e.,
So for 6000 C charge total nickel deposition will be:-
or
Hence 1.825 g Ni will be deposited in the given conditions.
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.
Since for deposition of 108 g silver 96487 C charge is required, thus for 1.45 g deposition of silver charge required will be:-
Now we can find the time taken by 1.5 A current to deposit 1.45 g silver.
For copper:-
Since 2F charge will deposit 63.5 g of Cu, then deposition by 1295.43 C will be:-
Hence 0.426 g of copper will be deposited.
For zinc:-
Since 2F charge will deposit 65.4 g of Zn, then deposition by 1295.43 C will be:-
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)
Answer:
The concept used here will be that a reaction is feasible only if
Anode and cathode reactions will be as follows:-
So
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)
Answer:
A reaction is feasible only if
So, anode and cathode reactions will be as follows :-
and
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)
Answer:
A reaction is feasible only if
So, anode and cathode reactions will be as follows :-
and
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:
Answer:
A reaction is feasible only if
So, anode and cathode reactions will be as follows:-
and
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)
Answer:
A reaction is feasible only if
So, anode and cathode reactions will be as follows :-
and
So this reaction is feasible.
Question 2.18 Predict the products of electrolysis in each of the following:
(i) An aqueous solution of
Answer:
For the given solution :
At cathode :- Reaction with greater E 0 will take place.
At anode :-
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
Answer:
For the given solution :
At cathode :- Reaction with greater E 0 will take place.
At anode :- Self ionisation will take place due to presence of water.
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
Answer:
For the given solution:
At cathode :- Reaction with greater E 0 will take place.
At anode :- Self ionisation of water will take place due to presence of platinum electrode.
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
Answer:
For the given solution :
At cathode :- Reaction with greater E 0 will take place.
At anode :-
Hence, Cu will get deposited at cathode and Cl 2 will be produced from anode.
Question: The standard cell potential
Choose the correct statement:
(1) The standard half cell reduction potential for the reduction of
(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
Here
Standard cell potential:
Given:
Substituting the values,
Solving for
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
Given :
Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be
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
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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,
Reaction | E° (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 |
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:
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).
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:
Before attempting questions, ensure you're clear on:
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
Here's a comparison table highlighting what to study beyond NCERT for JEE:
1. Conductance(G) is the reciprocal of resistance (R) and specific conductance or conductivity(k) is inverse of resistivity
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.
4. Nernst equation
aA+bB
Below are the chapter-wise solutions-
The hyperlinks of NCERT exemplar of class 12 are given below:
The hyperlinks of the NCERT solution of class 12 are given below:
Students can refer to the links given below for the NCERT books and Syllabus:
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.
There are two main types of electrochemical cells:
Galvanic cells convert chemical energy into electrical energy through spontaneous reactions, while electrolytic cells use electrical energy to drive non-spontaneous chemical reactions.
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,
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.
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.
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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.
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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!
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