Revision Notes for CBSE Class 12 Chemistry Chapter 3: Electrochemistry

Welcome to the ultimate CBSE Class 12 Electrochemistry Revision Notes. Electrochemistry is a high-yield, formula-heavy chapter in the CBSE Class 12 Chemistry syllabus, carrying significant weightage (typically 9 to 11 marks) in the board exams. This chapter bridges chemical energy and electrical energy, covering critical concepts like the Nernst Equation, Kohlrausch's Law, Faraday's Laws of Electrolysis, and Conductance in electrolytic solutions. To excel, you must be comfortable with both qualitative conceptual derivations and rigorous quantitative numericals. Use these notes as your high-density, formula-packed revision sheet right before your exam. To optimize your revision, leverage YoLearn AI Tools: study with our AI Flashcards for instant recall of reaction formulas, generate interactive Mind Maps to visualize electrolytic cells, and run automated Quizzes to master numerical accuracy on the fly.

Essential Glossary of Terms

Electrochemical Cell
A device that converts the chemical energy released during a spontaneous redox reaction into electrical energy.
Electrolytic Cell
A device that uses electrical energy from an external source to force a non-spontaneous chemical reaction to occur.
Standard Electrode Potential ($E^0$)
The potential difference developed between a metal electrode and its solution containing metal ions at unit concentration ($1\text{ M}$) at a temperature of $298\text{ K}$ and $1\text{ bar}$ pressure.
Conductivity (Specific Conductance, $\kappa$)
The reciprocal of resistivity, representing the conductance of a solution of $1\text{ cm}$ length and $1\text{ cm}^2$ area of cross-section. Measured in $\text{S cm}^{-1}$ or $\text{S m}^{-1}$.
Molar Conductivity ($\Lambda_m$)
The conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution. Mathematically, $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Limiting Molar Conductivity ($\Lambda_m^0$)
The molar conductivity of an electrolyte when the concentration of the solute approaches zero (infinite dilution).
Faraday's Constant ($F$)
The total electrical charge carried by one mole of electrons, approximately equal to $96487\text{ C mol}^{-1}$ (commonly rounded to $96500\text{ C}$).

Understanding the Nernst Equation & Cell Thermodynamics

The potential of an electrochemical cell is fundamentally dependent on the concentration of the reacting ions and the system's temperature. The Nernst Equation mathematically represents this relationship. For a general reduction reaction occurring at an electrode: $M^{n+}(aq) + ne^- \rightarrow M(s)$, the electrode potential is given by:

$E_{(M^{n+}/M)} = E^0_{(M^{n+}/M)} - \frac{RT}{nF} \ln \frac{1}{[M^{n+}]}$

When we generalize this for a complete cell reaction at standard temperature ($298\text{ K}$), converting the natural logarithm to base 10, the equation simplifies to:

$E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]}$

Gibbs Free Energy and Cell Potential:
Thermodynamically, the maximum electrical work done by an electrochemical cell is equal to the decrease in Gibbs Free Energy ($\Delta_r G$). The mathematical relationship is expressed as:

$\Delta_r G = -nFE_{\text{cell}}$

Under standard state conditions, this becomes:

$\Delta_r G^0 = -nFE^0_{\text{cell}}$

If the cell reaction reaches chemical equilibrium, the cell potential ($E_{\text{cell}}$) drops to zero, and the reaction quotient ($Q$) becomes equal to the equilibrium constant ($K_c$). Rearranging the Nernst equation at $298\text{ K}$ yields: $E^0_{\text{cell}} = \frac{0.0591}{n} \log K_c$.

Galvanic Cells vs. Electrolytic Cells

AspectDetails

Step-by-Step Molar Conductivity & Kohlrausch's Law Calculations

Step-by-Step Solved Numericals

  • {"title":"Example 1: Calculating Standard Gibbs Energy and Equilibrium Constant","bodyMarkdown":"Question: Calculate the standard Gibbs energy ($\\Delta G^0$) and equilibrium constant ($K_c$) for the cell reaction: $\\text{Zn}(s) + \\text{Cu}^{2+}(aq) \\rightarrow \\text{Zn}^{2+}(aq) + \\text{Cu}(s)$ given $E^0_{\\text{cell}} = 1.10\\text{ V}$.\n\nSolution:\n1. Identify the number of transferred electrons ($n$) = $2$.\n2. Compute $\\Delta G^0$ using $\\Delta G^0 = -nFE^0_{\\text{cell}}$:\n $\\Delta G^0 = -2 \\times 96500 \\times 1.10 = -212,300\\text{ J mol}^{-1} = -212.3\\text{ kJ mol}^{-1}$\n3. Calculate $K_c$ using $\\log K_c = \\frac{n E^0_{\\text{cell}}}{0.0591}$:\n $\\log K_c = \\frac{2 \\times 1.10}{0.0591} = 37.22 \\implies K_c = 1.66 \\times 10^{37}$"}
  • {"title":"Example 2: Applying Faraday's Laws of Electrolysis","bodyMarkdown":"Question: A solution of $\\text{CuSO}_4$ is electrolyzed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? (Atomic mass of $\\text{Cu} = 63.5\\text{ g mol}^{-1}$)\n\nSolution:\n1. Convert time to seconds: $t = 10 \\times 60 = 600\\text{ s}$.\n2. Calculate total charge passed ($Q = I \\times t$):\n $Q = 1.5\\text{ A} \\times 600\\text{ s} = 900\\text{ C}$\n3. The cathode reaction is: $\\text{Cu}^{2+} + 2e^- \\rightarrow \\text{Cu}(s)$.\n This shows $2$ moles of electrons ($2 \\times 96500\\text{ C}$) deposit $1$ mole of $\\text{Cu}$ ($63.5\\text{ g}$).\n4. Use proportion to find mass ($w$):\n $w = \\frac{63.5}{2 \\times 96500} \\times 900 = 0.296\\text{ g}$"}

Must-Remember Revision Points

  • Oxidation always occurs at the anode, and reduction always occurs at the cathode, regardless of whether it is a Galvanic or Electrolytic cell.
  • Specific conductivity ($\kappa$) always decreases with a decrease in concentration (dilution) for both strong and weak electrolytes because the number of current-carrying ions per unit volume decreases.
  • Molar conductivity ($\Lambda_m$) increases with dilution. Strong electrolytes experience a gradual, linear increase due to decreased inter-ionic attractions (described by Debye-Hückel-Onsager equation).
  • Weak electrolytes display a steep increase in molar conductivity ($\Lambda_m$) at high dilution due to a significant increase in the degree of dissociation ($\alpha$).
  • In the electrochemical series, elements with lower standard reduction potentials act as stronger reducing agents, while those with higher standard reduction potentials act as stronger oxidizing agents.
  • Primary cells (like Dry Cell, Mercury Cell) cannot be recharged, whereas Secondary cells (like Lead Storage Battery, Ni-Cd Cell) can be recharged by reversing the current flow.
  • Corrosion of iron is an electrochemical phenomenon where rust is formed at anodic spots on the metal surface: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$, combined with cathodic reduction of oxygen in an acidic medium.
  • In a dry cell, the manganese dioxide ($MnO_2$) acts as a depolarizer to oxidize hydrogen gas into water, preventing gas buildup.

CBSE Board Exam Traps & Marking Guidelines

  • Unit Conversions: The biggest trap in Electrochemistry numericals is units! If conductivity ($\kappa$) is given in $\text{S m}^{-1}$ instead of $\text{S cm}^{-1}$, you must use the formula $\Lambda_m = \frac{\kappa}{1000 \times M}$ where $M$ is in $\text{mol L}^{-1}$. Always verify whether you are working in CGS or SI units before starting calculations.
  • Writing Cell Representations: Always write physical states such as $(s)$, $(aq)$, and $(g)$ alongside the concentration of electrolytes in brackets. Follow the standard LOAN acronym (Left, Oxidation, Anode, Negative) to construct cell diagrams.
  • Showing Formula Steps: CBSE marking schemes allocate step-wise marks. Write down the blank formula (e.g., $E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log Q$) before substituting numerical values to secure partial marks even if you make a calculation error.

Practice Questions with Solutions

  • Why does conductivity of a solution decrease with dilution? Conductivity is defined as the conductance of ions present in a unit volume of solution. Dilution increases the overall volume of the solution, which reduces the number of current-carrying ions per unit volume, leading to a decrease in conductivity.
  • Can we store copper sulfate solution in an iron vessel? Why? No. The standard reduction potential of iron is lower than that of copper ($E^0_{\text{Fe}^{2+}/\text{Fe}} = -0.44\text{ V}$, $E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$). This makes iron more reactive than copper, meaning it will displace copper from the solution, corroding the iron vessel.
  • State Kohlrausch's law of independent migration of ions. Kohlrausch's Law states that at infinite dilution, where dissociation of the electrolyte is complete, each ion makes a definite individual contribution towards the limiting molar conductivity of the electrolyte, regardless of the nature of the other ion present.
  • What is the role of a salt bridge in a Galvanic cell? A salt bridge completes the electrical circuit by allowing migration of ions between half-cells, maintaining electrical neutrality in both compartments and preventing the accumulation of charges.

Frequently Asked Questions

What is the difference between EMF and Cell Potential?

Electromotive Force (EMF) is the potential difference between two electrodes when no current is drawn through the circuit (open circuit). Cell Potential is the potential difference measured when current is flowing through the external circuit.

How does the molar conductivity of strong electrolytes change with concentration?

For strong electrolytes, molar conductivity increases slowly and linearly with dilution. It is mathematically governed by the Debye-Hückel-Onsager equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$, where $A$ is a constant and $C$ is the concentration.

What reaction occurs at the anode and cathode during charging of a lead storage battery?

During charging, the discharging reactions are reversed. At the anode, $\text{PbSO}_4(s)$ is reduced back to lead: $\text{PbSO}_4(s) + 2e^- \rightarrow \text{Pb}(s) + \text{SO}_4^{2-}(aq)$. At the cathode, $\text{PbSO}_4(s)$ is oxidized to lead dioxide: $\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^-$.

Why is it impossible to measure the absolute potential of a single electrode?

An oxidation or reduction half-reaction cannot occur in isolation. To measure a potential difference, a closed loop with another reference half-cell (such as the Standard Hydrogen Electrode, SHE) must be established, meaning only relative potentials can be measured.

What is a fuel cell? Name the most common fuel cell used.

A fuel cell is an electrochemical device that directly converts the chemical energy of combustion fuels (like hydrogen, methane) into electrical energy. The most successful and widely used fuel cell is the Hydrogen-Oxygen ($H_2-O_2$) fuel cell.