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NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry provides complete coverage of galvanic cells, Nernst equation, conductance in electrolytic solutions, Kohlrausch's law, Faraday's laws of electrolysis, batteries, fuel cells, and corrosion.

Overview of Topics & Subtopics (Click to Jump)

1. Electrochemical Cells, EMF & Nernst Equation

Short Note:

• Galvanic Cell (Voltaic Cell): Device converting chemical energy stored in a spontaneous redox reaction into electrical energy (e.g., Daniell Cell: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$).

• Standard Cell Potential ($E^\circ_{\text{cell}}$): $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ (measured against Standard Hydrogen Electrode, $E^\circ = 0\text{ V}$).

• Nernst Equation (at 298 K): Relates cell potential to ion concentration:

  $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]}$

• Equilibrium Constant ($K_c$) & Gibbs Free Energy ($\Delta G^\circ$):

  - $\log K_c = \frac{n \cdot E^\circ_{\text{cell}}}{0.0591}$

  - $\Delta G^\circ = -n F E^\circ_{\text{cell}}$ (where $F = 96487\text{ C mol}^{-1} \approx 96500\text{ C mol}^{-1}$)

2. Conductance of Electrolytic Solutions & Kohlrausch's Law

Short Note:

• Conductivity ($\kappa$): Inverse of resistivity ($\kappa = 1/\rho = \frac{1}{R} \cdot \frac{l}{A}$). Unit: $\text{S m}^{-1}$ or $\text{S cm}^{-1}$.

• Molar Conductivity ($\Lambda_m$): Conductance of a solution containing $1\text{ mole}$ of electrolyte:

  $\Lambda_m = \frac{\kappa \times 1000}{M}$ (where $M$ is molarity in $\text{mol L}^{-1}$). Unit: $\text{S cm}^2 \text{mol}^{-1}$.

• Variation with Dilution:

  - $\kappa$ decreases with dilution because number of ions per unit volume decreases.

  - $\Lambda_m$ increases with dilution for both strong and weak electrolytes.

• Kohlrausch's Law of Independent Migration of Ions: At infinite dilution, molar conductivity of an electrolyte is equal to the sum of ionic conductivities of its constituent cations and anions:

  $\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ$

• Degree of Dissociation ($\alpha$) for Weak Electrolyte: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

3. Electrolysis, Batteries, Fuel Cells & Corrosion

Short Note:

• Faraday's Laws of Electrolysis:

  - 1st Law: Mass deposited $m = Z \cdot Q = Z \cdot I \cdot t$ (where $Z$ is electrochemical equivalent).

  - 2nd Law: For same quantity of electricity, masses deposited are proportional to equivalent weights ($m_1 / m_2 = E_1 / E_2$).

• Commercial Batteries:

  - Primary Batteries: Cannot be recharged (e.g., Dry Cell, Mercury Cell).

  - Secondary Batteries: Rechargeable (e.g., Lead storage battery, Ni-Cd cell).

• Fuel Cells: Devices converting energy from combustion of fuels (like $\text{H}_2, \text{CH}_4$) directly into electrical energy (e.g., $\text{H}_2-\text{O}_2$ fuel cell with ~70% efficiency).

• Corrosion: Electrochemical phenomenon where metal gets oxidized. Anodic region: $\text{Fe(s)} \rightarrow \text{Fe}^{2+} + 2e^-$; Cathodic region: $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$. Prevents by sacrificial coating (Galvanization using Zinc).

Intext & Conceptual Questions

Q1: Explain how molar conductivity ($\Lambda_m$) varies with concentration for strong and weak electrolytes.
Answer:
• For Strong Electrolytes: Strong electrolytes are completely ionized even at high concentrations. On dilution, interionic interactions decrease, allowing ions to move faster. Thus, $\Lambda_m$ increases slightly with dilution following the Debye-Hückel-Onsager equation: $\Lambda_m = \Lambda_m^\circ - A \sqrt{C}$. A plot of $\Lambda_m$ vs $\sqrt{C}$ gives a straight line with $y$-intercept $\Lambda_m^\circ$.

• For Weak Electrolytes: Weak electrolytes dissociate partially at higher concentrations. Dilution significantly increases the degree of dissociation ($\alpha$). Near infinite dilution, $\alpha \rightarrow 1$, causing a steep, exponential rise in $\Lambda_m$. Hence, $\Lambda_m^\circ$ cannot be obtained by direct extrapolation and must be determined using Kohlrausch's Law.
Q2: Distinguish between Primary and Secondary Batteries with examples.
Answer:
Feature Primary Battery Secondary Battery
Rechargeability Non-rechargeable. Reaction occurs only once. Rechargeable by passing current in opposite direction.
Lifespan Becomes dead over time after reactants are consumed. Can undergo several charge-discharge cycles.
Examples Leclanché Dry Cell, Mercury Cell (used in hearing aids). Lead Storage Battery (automobiles), Nickel-Cadmium Cell.

Textbook Chapter End Exercise Questions

Q1: Calculate the standard Gibbs energy ($\Delta G^\circ$) and equilibrium constant ($K_c$) for the cell reaction: 2 Fe³⁺(aq) + 2 I⁻(aq) → 2 Fe²⁺(aq) + I₂(s) (Given: E°_cell = 0.236 V at 298 K).
Answer:
Given:
• $E^\circ_{\text{cell}} = 0.236\text{ V}$
• Reaction transfers $n = 2$ moles of electrons ($2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+}$)
• Faraday constant $F = 96487\text{ C mol}^{-1}$

Step 1: Calculate Standard Gibbs Free Energy ($\Delta G^\circ$)
$\Delta G^\circ = -n F E^\circ_{\text{cell}}$
$\Delta G^\circ = -2 \times 96487 \times 0.236\text{ J mol}^{-1}$
$\Delta G^\circ = -45541.86\text{ J mol}^{-1} \approx -45.54\text{ kJ mol}^{-1}$

Step 2: Calculate Equilibrium Constant ($K_c$)
$\log K_c = \frac{n \cdot E^\circ_{\text{cell}}}{0.0591} = \frac{2 \times 0.236}{0.0591} = 7.986$
$K_c = 10^{7.986} \approx 9.68 \times 10^7$

• $\Delta G^\circ = \mathbf{-45.54\text{ kJ mol}^{-1}}$
• $K_c = \mathbf{9.68 \times 10^7}$
Q2: The molar conductivity of 0.025 mol L⁻¹ methanoic acid (HCOOH) is 46.1 S cm² mol⁻¹. Calculate its degree of dissociation (α) and dissociation constant (K_a). Given: λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.
Answer:
Given:
• Concentration $C = 0.025\text{ mol L}^{-1}$
• Molar conductivity $\Lambda_m = 46.1\text{ S cm}^2 \text{mol}^{-1}$
• $\lambda^\circ(\text{H}^+) = 349.6\text{ S cm}^2 \text{mol}^{-1}$
• $\lambda^\circ(\text{HCOO}^-) = 54.6\text{ S cm}^2 \text{mol}^{-1}$

Step 1: Calculate $\Lambda_m^\circ$ using Kohlrausch's Law
$\Lambda_m^\circ = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{HCOO}^-)$
$\Lambda_m^\circ = 349.6 + 54.6 = 404.2\text{ S cm}^2 \text{mol}^{-1}$

Step 2: Calculate Degree of Dissociation ($\alpha$)
$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{46.1}{404.2} \approx 0.114$ (or 11.4%)

Step 3: Calculate Dissociation Constant ($K_a$)
$K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.025 \times (0.114)^2}{1 - 0.114}$
$K_a = \frac{0.025 \times 0.0130}{0.886} \approx 3.67 \times 10^{-4}\text{ mol L}^{-1}$

• Degree of dissociation ($\alpha$) = 0.114
• Dissociation constant ($K_a$) = 3.67 × 10⁻⁴ mol L⁻¹
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Class 12 › Chemistry › Electrochemistry
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Topics in this chapter: Electrochemical Cells, EMF & Nernst EquationConductance of Electrolytic Solutions & Kohlrausch's LawElectrolysis, Batteries, Fuel Cells & Corrosion