JEE Chapter - 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.

JEE Main Chemistry Chapter-wise Solutions 2022-23 

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

Important Concepts of Redox Reactions and Electrochemistry

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 = Λeq/Λm.

• Then, based on equation,
  Λeq/Λm = 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).
  λ∞eq = λ∞c + λ∞a  

• Here, λ∞eq = Equivalent conductivity at infinite dilution;
  λ∞c = cation conductivity, λ∞a = 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.

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