# JEE Chapter - Redox Reactions and Electrochemistry

Get interactive courses taught by top teachers

## Introduction to Redox Reactions and Electrochemistry

Both topics, Redox Reactions and Electrochemistry, are very important topics of chemistry. Redox reactions are oxidation and reduction reactions that define all chemical reactions that lower or raise the oxidation number. Electrochemistry, on the other hand, is the study of producing electricity from energy produced during spontaneous chemical reactions. It also deals with the utilisation of energy in non-spontaneous chemical transformations which are some of the primary applications of redox reactions.

This article includes the topics related to the redox reaction such as oxidation and reduction reaction, types of oxidation reaction i.e., combination, decomposition, displacement, and disproportionation reactions. In addition, the electrode and redox couple processes are noted here. Redox reactions also have a wide range of applications in the study of electrode and cell processes. This article also describes several redox reaction laws. It also discusses how to balance a redox reaction and the importance of titration in redox reactions.

The electrochemistry revision notes are also covered in detail in this article. The chapter is particularly significant in terms of both practical and theoretical implications. Most notably, electrochemical processes are used to produce a large number of metals and compounds. For example, batteries and fuel cells are responsible for converting chemical energy into electrical energy.

The reactions that emerge from electrochemical processes are both energy efficient and environment friendly. As a result, the importance of electrochemistry increases as it aids in the development of new revolutionary technologies that are environmentally benign. Also, electrochemistry allows sensory impulses to be transmitted from the brain to the cells and vice versa.

Without a doubt, both redox reactions and electrochemistry are large and interdisciplinary topics, and both are important for JEE and NEET exams.

### Important Topics of Redox Reactions and Electrochemistry

• Electrical Conductors

• Electrolytic Conductance, Molar Conductance And Specific Conductance

• Kohlrausch Law

• Faraday's Laws of Electrolysis

• Electrolysis and Electroplating

• Cell Potential and Nernst Equation

### Electrical Conductor

• Conductors or electrical conductors are materials that allow an electric current to travel through them.

• Non-conductors or insulators, on the other hand, are substances that do not enable electricity to pass through them. Rubber, wood, paper, and all non-metals except carbon are the examples of non-conductors.

• There are two types of conductors: (i) metallic conductors and (ii) electrolytic conductors.

### Electrolytic Conductance

• Free ions are present in molten electrolytes and aqueous electrolyte solutions, which conduct electricity due to ion mobility.

• Ohm's law is applicable on metallic conductors, it also applies to electrolytic conductors.

• According to Ohm’s law, the resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross-section, i.e.,

R ∝ l/a or R = ⍴ x (l/a);
Here, 'l' denotes the length and 'a' denotes the cross-sectional area of the solution column held between the electrodes, and 'R' denotes the solution's resistivity or specific resistance.

### Conductivity of Solutions

• It is easier to think of conductance (G) than resistance when looking for solutions.

• The link between conductance (G) and resistance (R) is as follows:
Conductance = 1 / Resistance or G = (1/R)

• The ohm-1 or mho is unit of conductance (G). siemens (S) is also the unit of conductance.

• As a result, 1 ohm-1 = 1 S.
G = (1/⍴) x (a/l) = 𝜅 x (a/l);
where 𝜅 is called the solution's conductivity or specific conductance.

• This results in
𝜅 = G x (1/a) = (1/R) x (l/a)

• Conductivity (𝜅) =  (l/a)/R = Cell Constant/Resistance

• S m-1 (siemen /metre) is the SI unit for conductivity (𝜅).

• Note that 1 S m-1 = 1 ohm-1 m-1.

### Equivalent Conductivity

• An electrolyte's equivalent conductivity (Λeq) in solution is defined as
“Equivalent conductivity refers to the ability of one equivalent of an electrolyte in a solution to conduct all of the ions it produces.”

• As a result, equivalent conductivity is written as
Equivalent Conductivity (Λeq) = Conductivity (𝜅)/ Concentration in equivalents per unit volume (Ceq).

• ∴ Λeq = 𝜅/Ceq

### Molar Conductivity (ΛM)

• The molar conductivity (ΛM) of a solution can also be used to define its conducting power.

• Molar conductivity (ΛM) is defined as follows:
“The molar conductivity of any solution is defined as the total conducting power of all the ions supplied by one mole of an electrolyte.”

• Thus, molar conductivity is expressed as
Molar conductivity (ΛM) = Conductivity (𝜅)/ Concentration in moles per unit volume (Ceq) = 𝜅/Ceq.

### Relationship Between Molar and Equivalent Conductivities

• The definition says that Λm = 𝜅/Cm, and Λeq = 𝜅/Ceq.

• By combining the two equations,
Cm/Ceq = Λeqm.

• Then, based on equation,
Λeqm = 1/z; where z = 1,2,3,..

### Kohlrausch's Law

• It asserts that the conductivity of an electrolytic solution is equal to the sum of the conductivities of both ions at infinite dilution (which are present in the electrolyte).
λeq = λc + λa

• Here, λeq = Equivalent conductivity at infinite dilution;
λc = cation conductivity, λa = anion conductivity.

### Electrolysis

• Electrolysis is a chemical degradation of the electrolyte that occurs when an electric current is passed across it.

• It takes place in a cell known as an electrolytic cell.

• The electrical energy in this cell is converted into chemical energy.

• The following elements influence the electrolysis product:
The electrolyte's composition.
The electrode's nature.
Ion concentration in a solution.
Flowing current amount.

### Faraday’s Laws of Electrolysis

• The link between the amount of electricity transmitted through an electrolyte and the amount of material freed or deposited at the electrode was established by Faraday.

First Law of Electrolysis

• The amount of any substance deposited or dissolved at a particular electrode is proportional to the quantity of electricity utilised.

• Therefore, from the above definition,  w∝Q or w∝(I x t) or w = ZIt.;
where w is the mass of the substance deposited or liberated in grammes, Q is the quantity of charge used in coulombs, I is the current intensity in amperes, t is the time in seconds that current passes through the cell, and Z is the electrochemical equivalent.

Second Law of Electrolysis

• The quantity of the deposit is directly proportional to its equivalent weight when the same amount of power is transferred through different electrolytes. (Electrolyte equivalent weight).

• Therefore, W/E = constant = F; F= 96500 C per mole = Faraday constant.

### Electrochemical Cell

• The cells in which chemical energy is converted to electrical energy are known as electrochemical cells. This means that chemical processes result in the generation of electric current.

• The Daniel cell is the most basic electrochemical cell to examine.

• The reactions that take place at the two electrodes are as follows:
At anode: Zn(s) → Zn2+(aq) + 2e-
At cathode: Cu2+(aq) + 2e- → Cu(s)

### Electrode Potential

• For charge separation, the equilibrium between the metal and its ions is exploited, which results in a potential being formed between the metallic strip and its solution.

• The potential difference at equilibrium is determined by the net charge separation:
Metal's and one's own nature.
The temperature and ion concentration in the solution.

### EMF of the Cell

• Electromotive force (EMF) or cell voltage refers to the difference in electrode potential between the two electrodes of a cell.

• EMF = Ered(cathode) – Ered(Anode) or simply as EMF = Ecation – Eanion

### Cell Potential and Nernst Equation

• The Nernst equation is used to connect a cell's half-cell potential or EMF to the concentration of the species involved.

• Consider the case of a redox change in an electrochemical cell: xA + yB ⇌ zC + aD;
where A, B, C, and D are the species with varying concentrations, i.e., gases or solution phases.

• Ecell = E°cell = 0.059/n * $log\dfrac{[C]^{z}[D]^{a}}{[A]^{x}[B]^{y}}$.

• ∴ Ecell = E°cell = 0.059/n * log($\left[Product\right]/\left[Reactant\right]$).

• These equations are known as the Nernst equation, and they apply to both half-cell and complete-cell reactions.

### Relation Between Gibbs Free Energy and EMF

• Gibbs free energy can be calculated by multiplying the total charge driven through the cell by the potential difference.

• Thus, -ΔG = Total charge x EMF of the cell

-ΔG = nF x Ecell

• The negative sign denotes a decrease in free energy, implying that as the cell gets more positive, the G will become more negative, resulting in a spontaneous reaction.

• Similarly, -ΔG° = nFE°cell

### Solved Examples from the Chapter

Question 1: In the following, identify the oxidising and reducing agents:
(a) 2H2(g) + O2(g) → 2H2O(g)
(b) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Solution:

• Calculate the oxidation numbers and compare them.

• An increase in the oxidation number represents oxidation.

• A drop in the oxidation number represents reduction.

• (a) 2H2(g) + O2(g) 2H2O(g)
H2 was oxidised (Oxidation Number of H: 0 → +1); H2 is the reducing agent.
O2 was reduced (Oxidation Number of O: 0 → -2); O2 is the oxidising agent.

• (b) Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
Cu has been oxidised (Oxidation Number of Cu: 0 → +2), and it is the reducing agent.
The oxidising agent HNO3 was reduced (Oxidation Number of N: +5 → +4).

Key Points to Remember: Accurate calculation of the changes in the oxidation and reduction number after having identified the oxidising agent and the reducing agent respectively, is the most important concept while solving these types of problems.

Question 2: Balance the following equations using the oxidation number method:
Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g)

Solution:

• Step 1: Give each element an oxidation number.

In Al(s), Al → 0;
In H2SO4(aq), H → +1, S → -6; O → +1;
In Al2(SO4)3(aq), Al → +3, S → -6; O → +1;

In H2, H → 0.

• Step 2: Determine which species are oxidised and which are reduced.
From the above, it is clear that Al is oxidised and Hydrogen gets reduced.

• Step 3: Determine e-gained and e-lost.
Al - 3e-→ Al3+;
2H+ + 2e- → H2.

• Step 4: To make e- lost equal to e- acquired, multiply by factors and apply the factors as coefficients.

• Step 5: Complete the balancing process by inspecting the results.

• Hence, the final answer is: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

Key Points to Remember: Accurate calculation of the changes in the oxidation and reduction number after having identified the oxidising agent and the reducing agent, respectively, is the most important concept while solving these types of problems. It might so happen that for certain chemical species the oxidation state will remain unchanged. Care should be taken while handling such chemical species.

### Solved Examples of Previous Year Question Papers

Question 1: Given : XNa2HAsO3 +YNaBrO3+ZHCl → NaBr + H3AsO4 + NaCl

The values of X, Y and Z in the above redox reaction are respectively :

(1) 2, 1, 3

(2) 3, 1, 6

(3) 2, 1, 2

(4) 3, 1, 4

Solution:

• The equation for a balanced equation is shown below:

3Na2HAsO3 + NaBrO3 + 6HCl → NaBr + 3H3AsO4 + 6NaCl

• X, Y, and Z have the values 3, 1, and 6 correspondingly.

• As a result, option (2) is the correct answer.

Question 2: Consider the reaction

H2SO3(aq) + Sn4+(aq) + H2O(l) → Sn2+(aq) + HSO4–(aq) + 3H+(aq)

Which of the following statements is correct?

(1) H2SO3 is the reducing agent because it undergoes oxidation

(2) H2SO3 is the reducing agent because it undergoes reduction

(3) Sn4+ is the reducing agent because it undergoes oxidation

(4) Sn4+ is the oxidising agent because it undergoes oxidation

Solution:

• The loss of electrons by a molecule during a reaction is referred to as oxidation.

• Because it undergoes oxidation, H2SO3 is the reducing agent in the above equation.

• As a result, option 1 is the correct answer.

Question 3: In which of the following reactions H2O2 acts as a reducing agent ?

(1) H2O2 + 2H+ + 2e → 2H2O

(2) H2O2 - 2e → O2 + 2H+

(3) H2O2 + 2e → 2OH

(4) H2O2 + 2OH - 2e → O2 + 2H2O

(1) (1), (3)

(2) (2), (4)

(3) (1), (2)

(4) (3), (4)

Solution:

• In a redox chemical process, a reducing agent is an element or molecule that loses an electron to an electron recipient.

• H2O2 functions as a reducing agent in (2) and (4).

• As a result, option (2) is the correct answer.

### Practice Questions

Question 1: In each of the following reactions, identify the species that is being oxidised and reduced:
(a) Cr+ + Sn4+ → Cr3+ + Sn2+

(b) 3Hg2+ + 2Fe(s)  → 3Hg2 + 2 Fe3+

(c) 2As(s) + 3Cl2(g) → 2AsCl3

Answer:

• (a) Cr+: oxidised, Sn4+: reduced.

• (b) Hg2+: reduced, Fe: oxidised.

• (c) As: oxidised, Cl2: reduced.

Question 2: Write balanced equations for the following redox reactions:

(a) NaBr + Cl2 → NaCl + Br2

(b) Fe2O3 + CO → Fe + CO2 in acidic solution

(c) CO + I2O5 → CO2 + I2 in basic solution

Answer:

• (a) 2NaBr + Cl2 → 2NaCl + Br2

• (b) Fe2O3 + 3CO  → 2Fe + 3CO2 in acidic solution.

• (c) 5CO + I2O5  → 5 CO2 + I2 in basic solution.

### Conclusion

Redox reactions are oxidation-reduction chemical reactions in which the oxidation states of the reactants change. The term 'redox' refers to the reduction-oxidation process. All redox reactions can be divided into two types of reactions: reduction and oxidation.

In a redox reaction or Oxidation-Reduction process, the oxidation and reduction reactions always happen at the same time. The oxidising agent is the substance that is being reduced in a chemical process, while the reducing agent is the substance that is being oxidised. The same is exhibited in the redox reaction examples.

The discipline of chemistry that deals with the changes in matter induced by passing an electric current and converting chemical energy to electrical energy and vice versa is known as electrochemistry. Chemical energy is converted into electrical energy by devices called cells and batteries and the parameters are defined in the electrochemistry formulas.

See More

## JEE Main Important Dates

View all JEE Main Exam Dates
JEE Main 2022 June and July Session exam dates and revised schedule have been announced by the NTA. JEE Main 2022 June and July Session will now be conducted on 20-June-2022, and the exam registration closes on 5-Apr-2022. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2022 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.
See More
View all JEE Main Exam Dates

## JEE Main Information

Application Form
Eligibility Criteria
Reservation Policy
Admit Card
NTA has announced the JEE Main 2022 June session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2022 June and July session Application Form is available on the official website for online registration. Besides JEE Main 2022 June and July session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. July Session's details will be updated soon by NTA.

## JEE Main Syllabus

View JEE Main Syllabus in Detail
It is crucial for the the engineering aspirants to know and download the JEE Main 2022 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2022 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2022 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.
See More
Download full syllabus
Download full syllabus
View JEE Main Syllabus in Detail

## JEE Main 2022 Study Material

View all study material for JEE Main
JEE Main 2022 Study Materials: Strengthen your fundamentals with exhaustive JEE Main Study Materials. It covers the entire JEE Main syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Main study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Main exam 2022 with Vedantu right on time.
See More
All
Mathematics
Physics
Chemistry
See All

## JEE Main Question Papers

see all
Download JEE Main Question Papers & ​Answer Keys of 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.
See More

## JEE Main 2022 Book Solutions and PDF Download

View all JEE Main Important Books
In order to prepare for JEE Main 2022, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2022 exam so that they can grab the top rank in the all India entrance exam.
See More
Maths
NCERT Book for Class 12 Maths
Physics
NCERT Book for Class 12 Physics
Chemistry
NCERT Book for Class 12 Chemistry
Physics
H. C. Verma Solutions
Maths
R. D. Sharma Solutions
Maths
R.S. Aggarwal Solutions
See All

## JEE Main Mock Tests

View all mock tests
JEE Main 2022 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2022 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.
See More

## JEE Main 2022 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
See More

## JEE Main 2022 Results

JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.
See More
Rank List
Counselling
Cutoff
JEE Main 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in NITs and CFTIs colleges. JEE Main 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2022 state rank list on the official website or on our site.

## JEE Top Colleges

View all JEE Main 2022 Top Colleges
Want to know which Engineering colleges in India accept the JEE Main 2022 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.
See More

## FAQs on JEE Chapter - Redox Reactions and Electrochemistry

FAQ

1. How do redox reactions work in electrochemical cells?

Electrons are transported from one species to another in redox reactions. When a reaction occurs spontaneously, energy is liberated, which can then be put to good use. To extract this energy, the reaction must be divided into two half-processes: oxidation and reduction reactions.

2. Why is an electrochemical reaction defined as a redox reaction?

An electrochemical reaction is defined as a redox reaction because they are complementary in nature, as both include the oxidation and reduction processes. The processes include oxidising agent and reducing agent. The oxidising agent is the reagent that causes the oxidation, while the reducing agent is the reagent that is reduced.

3. What is a redox reaction?

An oxidation-reduction (redox) reaction is a type of chemical reaction in which two species exchange electrons. Any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron is known as an oxidation-reduction reaction.