JEE Important Chapter - Electrochemistry

Electrochemistry is a branch of physical chemistry that studies the relationship between electrical potential as a quantifiable and quantitative phenomenon and observable chemical change, as well as the relationship between electrical potential as a result of a specific chemical change and vice versa.

Electrons travel between electrodes through an electronically conducting phase (typically, but not always, an external electrical circuit, as in electroless plating) separated by an ionically conducting and electrically insulating electrolyte (or ionic species in a solution). These electrochemistry notes on electrochemistry cover a variety of themes and solved examples, including electrochemistry IIT JEE questions, as well as other solved examples and previous year’s questions.

Important Topics of Electrochemistry

• Electrochemical Cell

• Electrode Potential

• Cell Potential or EMF of Cell

• Nernst Equation

• Kohlrausch’s Law

• Electrolytic Conductance

Important Definitions of Electrochemistry

Types of Electrochemical Cell

• A spontaneous chemical reaction is one that occurs on its own and lowers the Gibbs energy of the system.

• After that, the energy is converted into electrical energy. These interconversions are carried out using electrochemical cells.

• Non-spontaneous processes can also be induced using external energy in the form of electrical energy.

• Galvanic and electrolytic cells are the two types of electrochemical cells. Electrolytic cells convert electrical energy into chemical energy, whereas galvanic cells convert chemical energy into electrical energy.

Galvanic Cell

• A spontaneous chemical process or reaction is used to extract cell energy, which is then transformed to an electric current.

• A Daniell Cell, for example, is a Galvanic Cell that uses Zinc and Copper to perform the redox process.

• Daniell Cell's reaction is as follows:
  Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

• The reaction's two halves are as follows:
  Oxidation Half: Zn(s) → Zn2+(aq) + 2e-
  Reduction Half: Cu2+(aq) + 2e- → Cu(s)

• These reactions take place at different times.

• The reducing agent is Zn, while the oxidising agent is Cu2+.

• Electrodes are another name for half cells. The anode is the oxidation half, whereas the cathode is the reduction half.

• In the external circuit, electrons pass from the anode to the cathode. The negative polarity is applied to the anode. Positive polarity is applied to the cathode.

• Daniell Cell is a fictitious character created by Daniell Cell. Cu is used as the cathode, and Zn is used as the anode.

Electrode Potential

• When more than one cation or anion is present, the discharge process becomes competitive.

• Any ion that must be discharged requires energy, and if there are multiple ions, the one that requires the greatest energy is discharged first.

• When an element interacts with its own ions, it tends to lose or gain electrons, resulting in a positively or negatively charged state.

• Depending on whether oxidation or reduction has occurred, the electrode potential is referred to as oxidation or reduction potential.

• Characteristics:
  - The size and sign of the oxidation and reduction potentials are the same.
  - Because Eo is not a thermodynamic feature, the values do not add up.

Standard Electrode Potential (Eo)

• It's the difference between a specific electrode's electrode potential and that of a standard hydrogen electrode under standard conditions.

• The following are the usual conditions:
  - Each ion in the solution has a concentration of 1M.
  - A 298 K temperature.
  - Each gas has a pressure of one bar.

Cell Potential or EMF of a Cell

• Cell potential is the difference between the electrode potentials of two half cells.

• When no current is taken from the cell, electromotive force happens (EMF).

• Ecell = Ecathode + Eanode. In this equation, the oxidation potential of the anode and the reduction potential of the cathode are used.

• The result is:
  = ER + EL,
  since the anode is on the left and the cathode is on the right.

• As a result, for a Daniel Cell:
  Eocell = EoCu2+/Cu - EoZn/Zn2+ = 0.34 + (0.76) = 1.10 V.

Nernst Equation

• It establishes a link between the electrode voltage and the ion concentration. As a result, as the ion concentration rises, so does the reduction potential.

• For a sort of electrochemical reaction in general,
  aA + bB →cC + dD.

• The Nernst equation is as follows:
  Ecell = Eocell - $\frac{2.303}{nF}.RT.log\frac{\left [ C \right ]^{c}\left [ D \right ]^{d}}{\left [ A \right ]^{a}\left [ B \right ]^{b}}$

Factors Affecting Electrolytic Conductance

• When dissolved or molten, an electrolyte dissociates in solution to form ions and so transmits electricity.

• Examples include strong electrolytes like HCl, NaOH, and KCl, as well as weak electrolytes like CH3COOH and NH4OH.

• Electrolytic or ionic conductance is the electrical conductivity of ions in solutions.

• The flow of electricity through an electrolyte solution is influenced by the following variables.
  - Electrolyte Nature or Interionic Attractions: The fewer the solute-solute interactions, the larger the flexibility of ion mobility and the higher the conductance.
  - Ion Solvation: As the number of solute-solvent interactions grows, the extent of solvation expands, and the electrical conductance decreases.
  - The Nature of the Solvent and Its Viscosity: The larger the solvent-solvent interactions are, the higher the viscosity and the greater the solvent's resistance to ion flow, and hence the lower the electrical conductance.
  - Temperature: When the temperature of an electrolytic solution rises, solute-solute, solute-solvent, and solvent-solvent interactions diminish, causing electrolytic conductance to rise.

Solved Examples from the Chapter

Question 1: The molar conductance of a 1.5 M electrolyte solution was measured to be 138.9 S cm2. This solution's specific conductance is

(a) 0.208 S cm−1

(b) 0.102 S cm−1

(c) 0.320 S cm−1

(d) 0.152 S cm−1

Solution: 

• The equation of Molar Conductance is given by,
  Molar Conductance = (K x 1000)/M;
  where K = specific conductance of the solution, and
  M = molarity of the solution.

• 138.9 S cm-1 = (K x 1000) / 1.5
  K = 0.208 S cm-1

• Hence, the answer is (a) 0.208 S cm−1.
  
  

Key Points to Remember: The relation between Molar Conductance and Specific Conductance of the solution. 

Question 2: The galvanic cell is made up of

(a) Thermal energy (heating) to electrical energy conversion

(b) Electrical energy (heating) to chemical energy conversion

(c) Chemical energy to thermal energy conversion

(d) Chemical energy to electrical energy conversion

Solution: 

• A galvanic cell is an electrochemical cell that generates emf by reduction-oxidation.

• Hence, the correct conversion is the conversion of chemical into electrical energy. 

Key Points to remember: A galvanic cell is one in which cell energy is extracted via a spontaneous chemical process or reaction and then converted to electric current.

Solved Examples from Previous Year Questions

Question 1: A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many moles of Ni will be deposited at the cathode?

(a) 0.10

(b) 0.15

(c) 0.20

(d) 0.05

Solution: 

• The change in the oxidation of Ni in the given condition is:
  Ni2+ + 2e- → Ni.

• From the above oxidation reaction, it is clear that 2 F of electricity generates 1 mole of Ni. 

• ∴ 0.1 F of electricity will generate = (0.1/2) moles of Ni. = 0.05 moles of Ni.

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

Trick: Always note down the oxidation or reduction reaction for the metal for which the moles are to be found. That will give the relation with the amount of electricity used for one mole of the metal. Using this correlation, the answer can then be easily determined.

Question 2: For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be

(a) 1 × 10–10

(b) 29.5 × 10–2

(c) 10

(d) 1 × 1010

Solution: 

• The given value of EMF of the cell = Eo = 0.295V.

• The relation between the EMF and the equilibrium constant is given as:
  Eo = $\dfrac{0.0591}{n}\cdot log\left ( K_c \right )$

• The value of n from the question is given as n = 2

∴ 0.295 = $\dfrac{0.0591}{2}\cdot log\left ( K_c \right )$

• Hence, log(Kc) = 0.295/0.0295 = 10

Kc = antilog(10) = 1010

• Hence, the final answer is (d) 1 × 1010

Question 3: The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is [Molar mass of PbSO4 = 303 g mol–1]

(a) 15.2

(b) 11.4

(c) 7.6

(d) 22.8

Solution: 

• The reaction for a lead-acid battery given in the question is as given below:
  PbSO4(s) + 2OH- → PbO2 + H2SO4 + 2e– 

• Following this, the reaction between the electricity and lead sulphate i.e. electrolysis of PbSO4 is,
  ​​PbSO4 + 2e– + 2H+ → Pb(s) + H2SO4

• Hence, the total number of moles of lead generated by 2 F electricity is 1. 

Total number of moles of Lead produced by 0.05 F = 0.05/2 = 0.025.

• Mass of PbSO4 electrolysed is 0.025x303 = 7.575 gm ≊ 7.6 gm.

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

Practice Questions

Question 1: When a current of 2.0 A is fed through a molten metal salt for 5 hours, it deposits 22.2 g of atomic weight 177 metal. The metal's oxidation state in the metal salt is

(a) + 1

(b) + 2

(c) + 3

(d) + 4

Answer: (c) + 3

Question 2: Which of the following is the correct statement:

(a) Conductivity & molar conductivity of solution increase on dilution.

(b) The voltage of the button cell remains unchanged throughout its life time.

(c) On electrolysis of aqueous CuSO4 using a platinum electrode, its concentration [Cu2+] remains constant.

(d) On electrolysis of aqueous NaCl, the solution becomes acidic.

Answer: (b) Voltage of the button cell remains unchanged throughout its life time.

Conclusion 

Electrochemistry is the study of chemical reactions that cause electrons to move. Electricity is the flow of electrons that occurs when electrons move from one element to another in an oxidation-reduction reaction. As a result, the electrochemistry IIT JEE notes are beneficial to your preparation.