JEE Main Study Notes for Redox Reaction and Electrochemistry: Important Concepts, Solved Examples and Tips

    Anam Shams Anam Shams
    Content Curator

    Redox reactions and electrochemistry are the most important and scoring topics of JEE Main Chemistry section. Considered one of the easiest part of Chemistry and there is no doubt in the fact that the topic is scoring too. The topic is very interesting and consists of 2- 3 questions in the paper that totals to 8- 10 marks in the overall marks distribution of the paper. Some of the topics are Conductors, Non- conductors, Specific Conductance, Electrolysis, Various laws of electrolysis, Electrochemical Cells, Electrode representation, etc. These topics are quite fascinating but involve concepts which must be understood properly. ReadJEE Main Preparation Tips

    Some Important Subtopics from Redox Reaction and Electrochemistry are mentioned below: 

    • Specific Conductance
    • Various laws of Electrolysis
    • Electrochemical Cells
    • Electrode Representation
    • Electrolysis
    • Oxidation Number
    • Oxidising Agent and Reducing Agent
    • Batteries
    • Corrosion
    • Types of Redox Reactions
    • Fuel Cells

    Redox Reaction

    Redox Reaction

    Redox Reaction is a reaction mechanism wherein the oxidation state of the metal changes after the reaction. The oxidation state may increase or decrease depending on the type of the Redox Reaction. Formation of rust on the metal surface is one such example wherein iron gets oxidized into iron oxide. Similarly, another example is ghee preparation, wherein the reduction reaction takes place. These reactions- Oxidation and Reduction reaction together are known Redox Reactions.

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    Types of Redox Reactions

    • Combination Reactions

    When two or more substances combine to form a single substance, the reaction is called Combination Reaction. If combination reactions have to be Redox, then one or more substance must be in the elemental form.

    • Decomposition Reactions

    The opposite of Decomposition Reaction is the Combination Reaction. In decomposition reactions, a single substance breaks into two or more simpler substances. At least one out of the formed substances has to be in its elemental form.

    • Displacement Reactions

    When in a chemical reaction, one ion of a compound is replaced by an ion of the other element, the reaction is called a displacement reaction. Displacement Reactions are of two types:

    a) Metal Displacement Reaction

    In the case of the metal displacement reaction, the metal of the compound is replaced by some other metal (in the form of elemental state).

    b) Non - metal displacement reaction

    Non - metal displacement reaction mainly comprises of hydrogen displacement or oxygen replacement reactions.

    • Disproportionation Reactions

    Disproportionation Reactions are the chemical reactions in which the same element is oxidized as well as reduced simultaneously. For element to be involved in disproportionation reaction, the reacting species must have at least three oxidation states. This reaction happens only in the basic medium.


    Example: The reaction of white phosphorous with aqueous NaOH gives phosphine along with another phosphorous containing compound. The reaction type; the oxidation on states of phosphorus in phosphine and the other product are respectively (IIT JEE -2012)

    (A) redox reaction; −3 and −5

    (B) redox reaction; +3 and +5 .

    (C) disproportionation reaction; −3 and +5.

    (D) disproportionation reaction; −3 and +3 .

    Answer: C


    The balanced reaction:

    It is clear from the balanced reaction that it is a disproportionation reaction as P undergo both oxidation as well as reduction in this reaction.

    NaH2PO2 formed in this reaction is less stable and decompose to from two products Na2HPO4 and PH3. Oxidation number of P in Na2HPO4 is +5.

    Hence, the correct option is C.



    Corrosion is a naturally occurring process in which it converts a refined material to a more chemically stable form, such as hydroxide, oxide, sulphide, etc. The process results in the gradual destruction of metals by chemical or electrochemical reaction with their environment. Examples of corrosion are rusting of iron or steel, copper oxide.

    Corrosion can be prevented by the formation of alloy, coating with a layer of less active metal, passivation, by cathodic protection.

    Some causes of Corrosion are:

    • Reactivity of metal-Highly reactive metals corrode faster.
    • Presence of impurity-Presence of salts like NaCl etc acts as catalyst to corrosion.
    • Presence of air, moisture, gases like SO2 and CO2 near metal.
    • Presence of electrolytes.
    • Characteristics of water in contact.(Like alkalinity of it / presence of ions / hardness / pH etc.)
    • Bacteria in surrounding (It’s presence increase O2 concentration near it.)

    Corrosion of iron:




    Corrosion is basically a:

    A) Altered reaction in presence of H2O

    B) Electrochemical phenomenon

    C) Interaction

    D) Union between light metal and heavy metal

    Answer: B

    Eelectrochemical Cells

    Electrochemical Cells

    Electrical energy is converted into chemical energy.Chemical energy is converted into electrical energy.
    Anode positive electrode. Cathode negative electrode.Anode negative electrode. Cathode positive electrode.
    Both the electrodes can be fitted in same compartment.The electrodes are fitted in different compartments.
    Ions are discharged on both the electrodes.Ions are discharged on the cathode.
    If the electrodes are inert, concentration of the electrolyte decreases when the electric current is circulated.Concentration of the anodic half-cell increases while that of cathodic half-cell decreases when the two electrodes are joined by a wire.

    Standard electrode potential: The potential difference developed between metal electrode and the solution of its ions of unit molarity (1M) at 25°C (298 K)

    IUPAC Cell Representation: Anode (Molarity of electrolyte at anode) || Cathode (Molarity of electrolyte at cathode)

    E0cell = E0cathode - E0anode


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    NERST Equation

    NERST Equation

    For a general reaction such as

    1A + m2B ..... n1X + n2Y + .... .......(i)

    Occurring in the cell, the Gibbs free energy change is given by the equation


    'a' represents the activities of reactants and products under a given set of conditions and

    ∆Go refers to free energy change for the reaction when the various reactants and products are present at standard conditions.

    The free energy change of a cell reaction is related to the electrical work that can be obtained from the cell, i.e.,

    ∆Go = -nFEcell and ?Go = -nFEo.

    On substituting these values in Eq. (ii) we get

    This equation is known as Nearnst equation.


    The solubility product (Ksp; moldm–9) of MX2at 298 K based on the information available for the given concentration cell is (take 2.303 ×R ×298/F = 0.059 V). (IIT JEE -2012)

    1) 1 ×10–15

    2) 4 ×10–15

    3) 1 ×10–12

    4) 4 ×10–12

    Answer: C


    Hence, the correct option is C.

    Oxidation and Reduction

    Oxidation and Reduction

    Oxidation is defined as the loss of electrons by a chemical species, i.e., atom, ion or molecule while Reducing is considered as the gain of electrons by those chemical species only.

    An Oxidising agent is an electron acceptor that takes electrons while reducing agent is the chemical species that gives electrons and thus acts as an electron donor.


    Based on the data given above, strongest oxidising agent will be. (IIT JEE-2013)

    1) Cr3+

    2) Mn2+

    3) MnO4-

    4) Cl-

    Answer: (3)


    More positive value of standard reduction potential indicated the poor tendency of the species to

    undergo reduction. In the given data, MnO4- has the highest positive value of standard

    reduction potential this means that it has least tendency to undergo reduction and thus highest

    tendency to undergo oxidation among the given species, thus it is the strongest oxidising agent.

    Hence, the correct option is 3.

    Also Check

    Faraday Law of Electrolysis

    Faraday Law of Electrolysis

    First Law

    The mass of an atom or ion oxidized or reduced at either electrode during the process of electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

    Second Law

    When the same quantity of electricity is passed through different electrolytes, the mass of substances deposited is proportional to their respective chemical equivalent or equivalent weight.


    Read How to Effectively Prepare for the Coming JEE Main Session from Home?


    Conductance is the property of a conductor which facilitates the flow of electricity through it.

    Conductivity or Specific Conductance: Conductance of a solution of definite dilution enclosed in a cell having two electrodes of unit area separated by 1 cm.

    Equivalent Conductance: Conductance of all the ions produced by 1 gram equivalent of an electrolyte in a given solution.

    Molar Conductance: Conductance of all the ions produced by ionization of 1 gram mole of an electrolyte present in V ml of a solution.


    Battery: A battery consists of two or more voltaic cells connected in series.

    Primary batteries

    In primary batteries, the reaction occurs only once and cannot be reused again.


    Leclanché cell: A zinc container acts as the anode, and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon.



    Mercury cell: It consists of zinc and mercury amalgam as the anode and a paste of HgO and carbon as the cathode.

    Secondary batteries

    Secondary batteries are portable voltaic cells which are rechargeable.

    The most important secondary cell is the lead storage battery commonly used in automobiles and invertors.



    Overall reaction:

    Previous Year Questions

    Previous Year Questions on Redox Reaction and Electrochemistry

    Question: The compound which could not act both as oxidising as well as reducing agent is

    1. SO2

    2. MnO2

    3. Al2O3

    4. CrO

    Answer: (c)

    Al2O3 could not act as a oxidising and reducing agent.

    Question: When MnO2 is fused with KOH, a coloured compound is formed, the product and its colour is

    1. K2MnO4 and purple green

    2. KMnO4 and purple

    3. Mn2O3 and brown

    4. Mn3O4 and black

    Answer: (a)

    Solution: 2MnO2+4KOH+O2−→Δ2K2MnO4purplegreen+2H2O

    Question: The conductivity of a saturated solution of BaSO4 is 3.06×10−6ohm−1cm−1 and its equivalent conductance is 1.53ohm−1cm−1equivalent−1. The Ksp of the BaSO4 will be:

    1. 4×10−12

    2. 2.5×10−9

    3. 2.5×10−13

    4. 4×10−6

    Answer: (D)

    Λm =1000K/S = 1000×3.06×10-6/ S = 1.53

    S=2×10 -3 mol/litre

    Ksp(BaSO4) = S2 = (2×10 -3)2 = 4×10-6

    Question: Consider the following cell reaction:

    2Fe(s) + O2(g) + 4H+(aq) ® 2Fe2+(aq) + 2H2O(l); E° = 1.67 V

    At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25°C is

    1. 1.47 V

    2. 1.77 V

    3. 1.87 V

    4. 1.57 V

    Answer: (d)

    Question: In a cell that utilizes the reaction

    Zn(s) + 2H+ (aq) ® Zn2+ (aq) + H2(g) addition of H2SO4 to cathode compartment, will:

    1. Lower the E and shift equilibrium to the left

    2. Lower the E and shift equilibrium to the right

    3. Increase the E and shift equilibrium to the right

    4. Increase the E and shift equilibrium to the left

    Answer: 3

    Question: The rusting of iron takes place as follows

    2H+ + 2e- + ® H2O(l); E° = +1.23 V

    Fe2+ + 2e- ® Fe(s); E° = -0.44 V

    Calculate ΔG° for the net process.

    a. -322 kJ mol-1

    b. -161 kJ mol-1

    c. -152 kJ mol-1

    d. -76 kJ mol-1

    Answer: (a)

    Question: Given the data at 25°C,

    Ag + I− → AgI + e− ; Eo = 0.152 V

    Ag → Ag+ + e−; Eo = −0.800 V

    What is the value of log Ksp for AgI?

    1. -8.12

    2. +8.612

    3. -37.83

    4. -16.13

    Answer: (d)

    Question: Zn | Zn2+ (a = 0.1M) || Fe2+ (a = 0.01M)|Fe. The emf of the above cell is 0.2905 V. Equilibrium constant for the cell reaction is


    Answer: (b)

    Question: The charge possessed by 1 mole of electrons:

    a. 1 F

    b. 1 C

    c. 96500 C

    d. 96500 F

    Answer: (a)

    Preparation Tips for Electrochemistry

    Tips and Tricks for Redox Reaction and Electrochemistry

    • Go through the detailed syllabus and divide each topic according to the time left to prepare for the examination.
    • The chapter is a part of the physical chemistry. It is one of the most important and scoring chapters of Chemistry. Its concepts, laws, numerical and graphs are all important both for the basic foundation of chemistry and for scoring good marks in JEE Main 2020.
    • Students must gather complete knowledge of the mole concept before reading the chapter.
    • You must learn carefully how to balance and various methods for balancing the chemical equations, Nernst equation, the conductance of electrolytic solutions.
    • Also, Be regular and consistent with the numerical part. Practice it regularly daily.
    • Class Notes are the perfect example of an initiating step that helps you begin from scratch and build your concepts strong slowly. Pen down all the important information you receive in the class.
    • Candidates can purchase the online test series by some of the best and top coaching institutes in online mode only.
    • Online tutorials are good source of preparation as one can find unlimited video lectures on each subject. The content offered through these channels is reliable, given by field experts. Also, students can rely on these video lectures in case they’ve missed a topic or had doubts even after attending the lecture.
    • Candidates are advised to attempt at least 10- 15 previous year sample papers before appearing for the actual examination to understand the paper pattern, marking scheme and types of questions asked in the examination. This will give you an overview of the paper and the questions asked in the examination..



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