NEET Chapter Page - Electrochemistry

Electrochemistry is a discipline of physical chemistry concerned with the relationship between electrical potential as a quantifiable and quantitative phenomena and recognisable chemical change, with electrical potential as an outcome of a specific chemical change or vice versa.

Electrons move between electrodes via an electronically conducting phase (usually, but not always, an external electrical circuit like as in electroless plating), which is separated by an ionically conducting and electrically insulating electrolyte (or ionic species in a solution). Electrochemistry consists of different topics and different solved examples such as redox reaction NEET questions, and different solved examples and previous year 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 can occur on its own and reduces the system's Gibbs energy. 

• The energy is then transformed into electrical energy. Electrochemical cells are used to carry out these interconversions.

• External energy in the form of electrical energy can also be used to cause non-spontaneous processes to occur.

• There are two types of electrochemical cells: galvanic and electrolytic cells. Galvanic cells convert chemical energy into electrical energy, whereas electrolytic cells turn electrical energy into chemical energy.

Galvanic Cell

• To extract cell energy, a spontaneous chemical process or reaction is employed, which is subsequently converted to electric current.

• A Daniell Cell, for example, is a Galvanic Cell that performs the redox process utilising Zinc and Copper.

• The reaction of Daniell Cell is given below:
  Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

• The two halves of the reaction are,
  Oxidation Half: Zn(s) → Zn2+(aq) + 2e-
  Reduction Half: Cu2+(aq) + 2e- → Cu(s)

• These reactions take place separately.

• Zn is the reducing agent, whereas Cu2+ is the oxidising agent.

• Half cells are also known as electrodes. The oxidation half is the anode, while the reduction half is the cathode.

• Electrons travel from the anode to the cathode in the external circuit. The anode is given negative polarity. The cathode is given positive polarity.

• Daniell Cell is a fictional character that Daniell Cell has developed. The cathode is Cu, and the anode is Zn.

Electrode Potential

• The discharge process becomes competitive when more than one cation or anion is present. 

• Any ion that needs to be discharged requires energy, and if numerous ions are present, the one that requires the most energy is discharged first.

• When an element comes into contact with its own ions, it tends to lose or gain electrons, leading it to become positively or negatively charged.

• The electrode potential is referred to as oxidation or reduction potential depending on whether oxidation or reduction has happened.

• Characteristics:

• The oxidation and reduction potentials have the same magnitude and sign.

• The values of E do not add up since it is not a thermodynamic characteristic.

Standard Electrode Potential (Eo)

• It's defined as the electrode potential of a given electrode when compared to a standard hydrogen electrode under standard conditions.

• The standard conditions are as follows:

• Each ion in the solution has a concentration of 1M.

• A temperature of 298 K.

• The pressure of each gas is one bar.

Cell Potential or EMF of a Cell

• The difference between the electrode potentials of two half cells is known as cell potential. 

• Electromotive force occurs when no current is drawn from the cell (EMF).

• Ecell = Ecathode + Eanode. We use the anode's oxidation potential and the cathode's reduction potential in this equation.

• Because the anode is on the left and the cathode is on the right, the following is the result:
  = ER + EL

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

Nernst Equation

• It connects the electrode voltage with the ion concentration. As a result, the reduction potential grows in tandem with the ion concentration. 

• 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

• An electrolyte is a material that dissociates in solution to produce ions and so conducts electricity when dissolved or molten.

• Strong electrolytes, such as HCl, NaOH, KCl, and weak electrolytes, such as CH3COOH, NH4OH, are examples.

• The conductance of electricity by ions in solutions is known as electrolytic or ionic conductance. 

• The following parameters influence the flow of electricity via an electrolyte solution.

• Electrolyte Nature or Interionic Attractions: The greater the freedom of ion mobility and the higher the conductance, the lower the solute-solute interactions.

• Ion Solvation: The extent of solvation increases as the number of solute-solvent interactions grows, and the electrical conductance drops.

• The Nature of the Solvent and its Viscosity: The higher the viscosity and the greater the solvent's resistance to ion flow, and hence the lower the electrical conductance, the larger the solvent-solvent interactions are.

• Temperature: Solute-solute, solute-solvent, and solvent-solvent interactions decrease when the temperature of an electrolytic solution rises, leading electrolytic conductance to rise.

Solved Examples from Chapter

Question 1: Find the charge in coulomb on 1 g-ion of N3-.

Solution: 

• An ion is formed by the gain or loss of an electron. 

• Therefore, Charge of an ion = Charge of an electron
  ∴ Charge of an ion = 1.6 x 10-19 Coulomb (C) .

• Following this, the charge on the given N3- ion is,
  Charge on N3- ion = 3 x 1.6 x 10-19.

• The number of ions in 1gram of N3- is given by the Avogadro’s Number = 6.02 x 1023 ions. 

• Therefore, the total charge in 1 g of N3- is given as
  Charge in 1g of N3- = 3 x 1.6 x 10-19 x 6.02 x 1023 = 2.89 x 105 Coulomb (C) .

Hence, the answer is 2.89 x 105 Coulomb (C).

Key Points to Remember: The charge of an electron and the amount of constituent particles in a given unit amount of sample i.e. the Avogadro’s number are the two important concepts in use in the solution.

Question 2: Estimate the Eo from the half-reaction M+(aq) + e- → M(s) based on the following observations:

(i) M interacts with H2SO4(aq) but not with HI(aq); M displaces Au+(aq) but not Fe3+(aq).

(ii) The metal M reacts with HI(aq) to produce H2(g), but neither Al3+(aq) nor Na+(aq) are displaced.

Solution: 

• (i) When a metal dissolves in H2SO4, it has a lower reduction potential than ESO42-(aq)/SO2(g) = 0.17 V. 

• It has a reduction potential greater than EH+(aq)/H2(g) = 0 V if it does not dissolve in HI. 

• It has a reduction potential smaller than EAu+(aq)/Au(s) = 1.68 V if it displaces Au+(aq) from solution.

• However, if it does not remove Fe3+(aq) from solution, its reduction potential is greater than,
  EoFe3+(aq)/Fe2+(s) = 0.769 V.

• As a result,  0 V < Eo < 0.17 V .

• (ii) When a metal dissolves in HI(aq), it has a lower reduction potential than EH+(aq)/H2(g) = 0 V. 

• Its reduction potential is greater than EAl3+(aq)/Al(s) = 1.676 V if it does not displace Al3+(aq) from solution.

• Its reduction potential is greater than ENa+(aq)/Na(s) = 2.7144 V if it does not displace Na+(aq) from solution. As a result, -1.7676 < E0 < 0 V.

Key Points to remember: Electrode potential is different for different sets of ions and whether they are undergoing an oxidation or a reduction reaction. 

Solved Examples from Previous Year Questions

Question 1: Which of the following relationships for the values of ΔGo and Keq is valid if Eocell for a particular reaction is negative?

(a) ΔGo < 0; Keq > 1

(b) ΔGo < 0; Keq < 1

(c) ΔGo > 0; Keq < 1

(d) ΔGo > 0; Keq > 1

Solution: 

• The general relationship between free energy, ΔGo and Eocell is given as,
  ΔGo = -nFEocell.

• Also, the relationship between ΔGo and Keq at equilibrium is given as;

• ΔGo > 0 then Keq < 1 and at equilibrium, reactants are preferred above products.

• ΔGo = 0 then Keq = 1 and products and reactants are equally preferred in an equilibrium state.

• ΔGo < 0 then Keq > 1 and at equilibrium, products are preferred over reactants.

• Since, the given value of Eocell is negative, the resulting value of ΔGo is positive.
  ∴ ΔGo > 0

• Since, ΔGo > 0 then Keq < 1.

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

Question 2: For the cell reaction

2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq).

E-cell = 0.24 V at 298 K. The standard Gibbs Energy (ΔrG-) of the cell reaction is:

(Given that Faraday Constant is F = 96,500 C mol-1)

(a) - 46.32 kJ/mol

(b) - 23.6 kJ/mol

(c) 46.32 kJ/mol

(d) 23.16 kJ/mol.

Solution: 

• Using the formula for free energy,
  ΔrG- = -nFEocell

• ΔrG- = - 2 x 96,500 x 0.24 J mol-1.

• ΔrG- = - 46,320 J mol-1

• ΔrG- = - 46.32 Jk mol-1 

Hence, the final answer is (d) ΔrG- = - 46.32 kJ mol-1 

Question 3: Without losing its concentration ZnCl2 solution cannot be kept in contact with: 

(a) Au

(b) Al

(c) Pb

(d) Ag

Solution: 

• Al is situated above Zn in the electrochemical series, while all other elements are found below Zn. 

• As a result, zinc is displaced from the ZnCl2 solution by aluminium. As a result, it is unable to communicate with Al.

• And the reaction is 2Al + 3ZnCl2 → 2AlCl3 + 3Zn.

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

Practice Questions

Question 1: When a copper wire is submerged in an AgNO3 solution, the solution turns blue because copper:

(a) With AgNO3, creates a soluble compound.

(b) Is converted to Cu2+ by oxidation

(c) Is converted to Cu2+ by reduction

(d) Splits and dissolves into atomic form

Answer: (b) Is converted to Cu2+ by oxidation

Question 2: The half-cell reactions are listed below:

Mn2+ + 2e- → Mn; Eo = -1.18 V

2Mn3+ + 2e- → 2Mn2+; Eo = +1.51 V.

Therefore, 3Mn2+ → 2Mn3+ + Mn will be,

(a) -2.69 V; The reaction will not take place.

(b) -2.69 V; The reaction will take place.

(c) -0.33 V; The reaction will not take place.

(d) -0.33 V; The reaction will take place.

Answer: (a) -2.69 V; The reaction will not take place.

Conclusion of Electrochemistry NEET

The study of chemical reactions that cause electrons to move is known as electrochemistry. This flow of electrons is known as electricity, and it can be generated by electrons moving from one element to another in an oxidation-reduction reaction. The electrochemistry NEET 2022 notes thus are valuable to your preparation.