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ICSE • कक्षा XII • Chemistry • अध्याय 2
अनुमानित समय: 90 Mins
प्रगति: अध्ययनरत

विद्युतरसायन (Electrochemistry)

In Class 12 Chemistry, "Electrochemistry" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

⚡ Have You Ever Wondered?

How does a hydrogen fuel cell generate zero-emission electrical energy to propel urban city buses drinking only pure water from its exhaust pipe? Nernst Equation and Kohlrausch's Law govern electrochemical energy conversion.

यह अध्याय क्यों महत्वपूर्ण है

In Class 12 Chemistry, "Electrochemistry" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Redox reactions from Class 11.
  • Electrode potential.
  • Molarity.

इस अध्याय के लक्ष्य

  • Explain Galvanic cells (Daniell cell) and Electrolytic cells; write cell representations.
  • State and apply the Nernst Equation: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q$ at $298\text{ K}$.
  • Relate cell potential to Gibbs Free Energy: $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ and equilibrium constant: $\log K_c = \frac{nE^\circ_{\text{cell}}}{0.059}$.
  • Define Conductivity ($\kappa$), Molar Conductivity ($\Lambda_m = \frac{\kappa \times 1000}{M}$), and state Kohlrausch's Law of Independent Migration of Ions.
  • State Faraday's Laws of Electrolysis and explain commercial Batteries (Lead storage, Dry cell) and Corrosion.

अध्याय रूपरेखा एवं प्रगति

1 1. Nernst Equation & Thermodynamics
2 2. Kohlrausch's Law of Independent...
3 3. Batteries & Corrosion Prevention

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Nernst Equation & Thermodynamics

For cell reaction $aA + bB \to cC + dD$, the cell potential at temperature $T=298\text{ K}$ is: $$\mathbf{E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log_{10}\left(\frac{[C]^c [D]^d}{[A]^a [B]^b}\right)}$$ Maximum work equals electrical free energy: $\mathbf{\Delta G^\circ = -n F E^\circ_{\text{cell}}}$ ($F = 96,487\text{ C/mol}$). Equilibrium constant: $\mathbf{\log K_c = \frac{n E^\circ_{\text{cell}}}{0.0591}}$.

2. Kohlrausch's Law of Independent Migration

Kohlrausch's Law: Limiting molar conductivity of an electrolyte is the sum of the individual contributions of its anions and cations: $$\mathbf{\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ}$$ Crucial application: calculating limiting conductivity of weak electrolytes like acetic acid: $\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl})$!

3. Batteries & Corrosion Prevention

Lead Storage Battery: Spongy lead anode and $\text{PbO}_2$ cathode in $38\%$ $\text{H}_2\text{SO}_4$. In discharging: $\text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \to 2\text{PbSO}_4 + 2\text{H}_2\text{O}$. Recharging reverses the process. Cathodic Protection: Sacrificial zinc blocks (Galvanization) protect steel ship hulls.

महत्वपूर्ण रासायनिक सूत्र, अभिक्रियाएँ एवं समीकरण

Answer architecture
$$Concept \to Evidence \to Application \to Evaluation$$
Use the chapter principle, show the working or evidence, and state the conclusion.
Revision loop
$$Learn \to Practise \to Check \to Correct \to Reattempt$$
Keep an error log and revisit questions that exposed a misconception.

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

उदाहरण 1
Calculate the emf of the Daniell cell at $298\text{ K}$: $\text{Zn}(s) | \text{Zn}^{2+}(0.1\text{M}) || \text{Cu}^{2+}(0.01\text{M}) | \text{Cu}(s)$ given $E^\circ_{\text{cell}} = 1.10\text{ V}$.
विस्तृत समाधान / उत्तर:
Cell reaction: $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$ ($n=2$). $E_{\text{cell}} = 1.10 - \frac{0.0591}{2}\log\left(\frac{0.1}{0.01}\right) = 1.10 - 0.02955\log(10) = 1.10 - 0.02955 = 1.07\text{ Volts}$.
उदाहरण 2
State Kohlrausch's Law of independent migration of ions and state two of its applications.
विस्तृत समाधान / उत्तर:
Kohlrausch's law states that 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. Applications: (1) Determination of limiting molar conductivity ($\Lambda_m^\circ$) for weak electrolytes, (2) Determination of degree of dissociation ($\alpha = \Lambda_m / \Lambda_m^\circ$) and dissociation constant $K_a$.
उदाहरण 3
Calculate $\Delta G^\circ$ for a cell having $E^\circ_{\text{cell}} = 1.1\text{ V}$ and $n = 2$ ($F = 96500\text{ C/mol}$).
विस्तृत समाधान / उत्तर:
$\Delta G^\circ = -n F E^\circ = -2 \times 96500 \times 1.1 = -212,300\text{ J/mol} = -212.3\text{ kJ/mol}$.

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

Reciting a definition without applying it to the question or data.

सही वैज्ञानिक तथ्य

Identify the concept, show the relevant evidence or calculation, and explain the final implication.

सामान्य भ्रम / गलत उत्तर

Skipping conditions, units, domain restrictions, or adjustment effects.

सही वैज्ञानिक तथ्य

State assumptions, preserve units, check boundary cases, and verify the answer against the original problem.

सामान्य भ्रम / गलत उत्तर

Treating a correct intermediate result as proof that the whole solution is correct.

सही वैज्ञानिक तथ्य

Perform an independent reasonableness check and connect the result back to the chapter principle.

Electrochemistry - Key Molecular Architecture & Reaction Mechanism Model

Electrochemistry - Molecular Architecture Electronic & Orbital Mechanisms Stereochemistry, reaction kinetics & pathways Thermodynamic & Coordination Frameworks Crystal field splitting, cell potentials & free energy High-Stakes Examination & Industrial Synthesis CISCE Class 12 Board criteria, JEE/NEET diagnostic applications & conversions

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
Nernst Equation: Voltage scaling dynamically with ion concentration quotient.
मुख्य बिंदु 2
Faraday Constant: $F = 96,487\text{ C/mol}$ of electrons driving electrolysis.
मुख्य बिंदु 3
Kohlrausch Law: Additive ionic migration at infinite dilution determining weak electrolyte conductivity.
मुख्य बिंदु 4
Lead Storage Accumulator: Reversible secondary cell powering automotive ignition systems.
मुख्य बिंदु 5
Corrosion Mechanism: Spontaneous electrochemical rusting cell forming hydrated iron(III) oxide.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
Calculate the emf of the Daniell cell at $298\text{ K}$: $\text{Zn}(s) | \text{Zn}^{2+}(0.1\text{M}) || \text{Cu}^{2+}(0.01\text{M}) | \text{Cu}(s)$ given $E^\circ_{\text{cell}} = 1.10\text{ V}$.
उत्तर एवं व्याख्या देखें
उत्तर: Cell reaction: $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$ ($n=2$). $E_{\text{cell}} = 1.10 - \frac{0.0591}{2}\log\left(\frac{0.1}{0.01}\right) = 1.10 - 0.02955\log(10) = 1.10 - 0.02955 = 1.07\text{ Volts}$.
E_cell = 1.07 V.
2
State Kohlrausch's Law of independent migration of ions and state two of its applications.
उत्तर एवं व्याख्या देखें
उत्तर: Kohlrausch's law states that 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. Applications: (1) Determination of limiting molar conductivity ($\Lambda_m^\circ$) for weak electrolytes, (2) Determination of degree of dissociation ($\alpha = \Lambda_m / \Lambda_m^\circ$) and dissociation constant $K_a$.
Additive ionic conductivities; finds weak electrolyte Λ0 and dissociation degree.
3
Calculate $\Delta G^\circ$ for a cell having $E^\circ_{\text{cell}} = 1.1\text{ V}$ and $n = 2$ ($F = 96500\text{ C/mol}$).
उत्तर एवं व्याख्या देखें
उत्तर: $\Delta G^\circ = -n F E^\circ = -2 \times 96500 \times 1.1 = -212,300\text{ J/mol} = -212.3\text{ kJ/mol}$.
ΔG° = -212.3 kJ/mol.
4
What is the effect of dilution on: (i) Conductivity ($\kappa$), (ii) Molar Conductivity ($\Lambda_m$)?
उत्तर एवं व्याख्या देखें
उत्तर: (i) Conductivity $\kappa$ decreases on dilution because the number of current-carrying ions per unit volume of solution decreases. (ii) Molar conductivity $\Lambda_m$ increases on dilution because the increase in volume ($V$) containing 1 mole of electrolyte vastly outweighs the decrease in $\kappa$.
Conductivity decreases; Molar conductivity increases.
5
Write the chemical reactions taking place at the anode and cathode during the discharging of a lead-storage battery.
उत्तर एवं व्याख्या देखें
उत्तर: Anode: $\text{Pb}(s) + \text{SO}_4^{2-}(aq) \to \text{PbSO}_4(s) + 2e^-$. Cathode: $\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \to \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$.
Discharge forms PbSO4 at both electrodes.
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