Unit 2: Electrochemistry

2.1 Redox Reactions and Galvanic Cells

Electrochemistry is the study of the production of electricity from the energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.

A Galvanic Cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. A typical example is the Daniell Cell:

• Anode (Oxidation): \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)
• Cathode (Reduction): \( Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \)
• Overall Reaction: \( Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s) \)

2.2 Standard Electrode Potential (\( E^\circ \))

Standard electrode potential is the potential difference developed between the metal electrode and the solution of its ions of unit molarity at 298 K and 1 atm pressure. \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}} \]

2.3 Nernst Equation

The Nernst equation relates the cell potential at non-standard conditions to the standard cell potential, temperature, and concentrations of the reacting species.

For the reaction: \( aA + bB \rightarrow cC + dD \) \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \] where:

• \( R \) is the gas constant (8.314 J/(K·mol))
• \( T \) is the temperature in Kelvin
• \( n \) is the number of moles of electrons transferred
• \( F \) is the Faraday constant (~96487 C/mol)
• \( Q \) is the reaction quotient, \( Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

At 298 K, using base 10 logarithm, the Nernst equation simplifies to: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059V}{n} \log Q \]

Equilibrium Constant from Nernst Equation

At equilibrium, \( E_{\text{cell}} = 0 \) and \( Q = K_c \): \[ E^\circ_{\text{cell}} = \frac{0.059V}{n} \log K_c \]

2.4 Gibbs Energy Change

The reversible work done by a galvanic cell is equal to its decrease in Gibbs energy: \[ \Delta_r G = -nFE_{\text{cell}} \] Under standard conditions: \[ \Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -RT \ln K_c \]

2.5 Conductance of Electrolytic Solutions

• Resistance (\(R\)): \( R = \rho \frac{l}{A} \)
• Conductivity (\(\kappa\)): \( \kappa = \frac{1}{\rho} = \frac{1}{R} \left( \frac{l}{A} \right) \), where \( l/A \) is the cell constant (\( G^* \)).
• Molar Conductivity (\(\Lambda_m\)): The conducting power of all the ions produced by dissolving one mole of electrolyte in solution. \[ \Lambda_m = \frac{\kappa \times 1000}{C} \quad (\text{if } \kappa \text{ is in S cm}^{-1} \text{ and } C \text{ is in mol L}^{-1}) \]

2.6 Kohlrausch’s Law of Independent Migration of Ions

According to this law, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. \[ \Lambda^\circ_m = \nu_+ \lambda^\circ_+ + \nu_- \lambda^\circ_- \] where \( \nu_+ \) and \( \nu_- \) are the number of cations and anions respectively, and \( \lambda^\circ_+, \lambda^\circ_- \) are the limiting molar conductivities of the cation and anion.

2.7 Batteries and Corrosion

Primary Batteries: Non-rechargeable (e.g., Dry cell, Button cell). Secondary Batteries: Rechargeable (e.g., Lead storage battery, Nickel-cadmium battery). Fuel Cells: Produce electricity directly from the energy of combustion of fuels like \(H_2\), \(CO\), \(CH_4\). Corrosion: The slow coating of metal surfaces with oxides or other salts. In rusting of iron, a miniature electrochemical cell is formed at the surface of the iron.

Competency-Based Questions (CBQs)

Q1. (CBSE 2022) The standard reduction potentials for \( Zn^{2+}/Zn \), \( Ni^{2+}/Ni \), and \( Fe^{2+}/Fe \) are -0.76 V, -0.23 V, and -0.44 V, respectively. The reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) will be spontaneous when: (A) X = Ni, Y = Fe (B) X = Ni, Y = Zn (C) X = Fe, Y = Zn (D) X = Zn, Y = Ni

Answer: Correct Option: (D) Reasoning and Calculation: For the reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) to be spontaneous, the standard cell potential \( E^\circ_{\text{cell}} \) must be positive. \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cathode (reduction)}} - E^\circ_{\text{Anode (oxidation)}} \] \[ E^\circ_{\text{cell}} = E^\circ_{Y^{2+}/Y} - E^\circ_{X^{2+}/X} > 0 \] Which means \( E^\circ_{Y^{2+}/Y} > E^\circ_{X^{2+}/X} \). In simpler terms, X must be a stronger reducing agent than Y (X must have a more negative standard reduction potential than Y). Looking at the potentials: \( E^\circ_{Zn} = -0.76 \) V \( E^\circ_{Fe} = -0.44 \) V \( E^\circ_{Ni} = -0.23 \) V

For option (D), X = Zn (-0.76 V) and Y = Ni (-0.23 V). Since -0.23 V > -0.76 V, the reaction is spontaneous.

Q2. (CBSE 2019) Calculate the emf of the following cell at 298 K: \[ Cu(s) | Cu^{2+} (0.130 M) || Ag^+ (1.0 \times 10^{-4} M) | Ag(s) \] Given: \( E^\circ_{Cu^{2+}/Cu} = +0.34 \) V and \( E^\circ_{Ag^+/Ag} = +0.80 \) V.

Answer: Step 1: Write the cell reactions and overall reaction.

• Anode (Oxidation): \( Cu(s) \rightarrow Cu^{2+} + 2e^- \)
• Cathode (Reduction): \( [Ag^+ + e^- \rightarrow Ag(s)] \times 2 \)
• Overall cell reaction: \( Cu(s) + 2Ag^+ \rightarrow Cu^{2+} + 2Ag(s) \)

The number of moles of electrons transferred, \( n = 2 \).

Step 2: Calculate \( E^\circ_{\text{cell}} \). \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}} \] \[ E^\circ_{\text{cell}} = 0.80 , \text{V} - 0.34 , \text{V} = 0.46 , \text{V} \]

Step 3: Apply the Nernst equation. \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[Cu^{2+}]}{[Ag^+]^2} \] \[ E_{\text{cell}} = 0.46 - \frac{0.059}{2} \log \frac{0.130}{(1.0 \times 10^{-4})^2} \] \[ E_{\text{cell}} = 0.46 - 0.0295 \log \left(\frac{0.130}{1.0 \times 10^{-8}}\right) \] \[ E_{\text{cell}} = 0.46 - 0.0295 \log (1.3 \times 10^7) \] \[ \log (1.3 \times 10^7) = \log(1.3) + 7 = 0.1139 + 7 = 7.1139 \] \[ E_{\text{cell}} = 0.46 - 0.0295 \times 7.1139 \] \[ E_{\text{cell}} = 0.46 - 0.2098 = 0.25 , \text{V} \]

Q3. (Sample Paper 2024) The molar conductivity of 0.025 mol L\(^{-1}\) methanoic acid is 46.1 S cm\(^2\) mol\(^{-1}\). Calculate its degree of dissociation and dissociation constant. Given \(\lambda^\circ (H^+) = 349.6\) S cm\(^2\) mol\(^{-1}\) and \(\lambda^\circ (HCOO^-) = 54.6\) S cm\(^2\) mol\(^{-1}\).

Answer: Step 1: Find the limiting molar conductivity (\( \Lambda^\circ_m \)). According to Kohlrausch’s law: \[ \Lambda^\circ_m (HCOOH) = \lambda^\circ (H^+) + \lambda^\circ (HCOO^-) \] \[ \Lambda^\circ_m = 349.6 + 54.6 = 404.2 , \text{S cm}^2 \text{ mol}^{-1} \]

Step 2: Calculate the degree of dissociation (\( \alpha \)). \[ \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{46.1}{404.2} = 0.114 \]

Step 3: Calculate the dissociation constant (\( K_a \)). For methanoic acid, \( HCOOH \rightleftharpoons HCOO^- + H^+ \) \[ K_a = \frac{C \alpha^2}{1 - \alpha} \] \[ K_a = \frac{0.025 \times (0.114)^2}{1 - 0.114} \] \[ K_a = \frac{0.025 \times 0.013}{0.886} = 3.67 \times 10^{-4} \]