Electrochemistry Class 12 NCERT: Concepts, Formulas & Solved Examples

Galvanic Cells and Electrode Potential

A Galvanic (or Voltaic) cell is an electrochemical cell that converts chemical energy from a spontaneous redox reaction into electrical energy. The most classic example is the Daniell cell, which operates on the reaction between Zinc and Copper. In this setup, oxidation occurs at the anode (Zinc electrode: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$), which holds a negative polarity. Reduction takes place at the cathode (Copper electrode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$), which is positive. The two half-cells are connected externally by a metallic wire and internally by a salt bridge filled with an inert electrolyte like $KCl$ or $KNO_3$ in agar-agar gel. The salt bridge completes the circuit and maintains electrical neutrality. The cell potential, measured in Volts, is the difference between the reduction potentials of the cathode and anode: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$.

How to Calculate Cell Potential Using the Nernst Equation

1. Identify the Cell Reaction and n-value — Write down the balanced overall redox equation. Count the total number of moles of electrons ($n$) exchanged between the oxidizing and reducing agents.
2. Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) — Obtain standard reduction potentials from the electrochemical series. Apply the formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$.
3. Formulate the Reaction Quotient (Q) — Express $Q$ as the ratio of product concentration to reactant concentration: $Q = \frac{[\text{Products}]^p}{[\text{Reactants}]^r}$. Do not include pure solids or liquids in this expression.
4. Substitute Values into the Nernst Equation — At $298\text{ K}$, apply the simplified equation: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log_{10} Q$. Solve for the unknown potential.

Worked Calculations on Gibbs Energy and Cell EMF

• Example 1: Calculate the standard Gibbs free energy change ($\Delta G^\circ$) for the reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$ given that $E^\circ_{\text{cell}} = 1.1\text{ V}$ and $F = 96487\text{ C mol}^{-1}$. Solution: Step 1: Identify the number of electrons transferred, $n = 2$. Step 2: Use the formula $\Delta G^\circ = -nFE^\circ_{\text{cell}}$. Step 3: Calculate $\Delta G^\circ = -2 \times 96487 \times 1.1 = -212271.4\text{ J mol}^{-1} = -212.27\text{ kJ mol}^{-1}$.
• Example 2: Find the equilibrium constant ($K_c$) for the Daniell cell reaction at 298 K with $E^\circ_{\text{cell}} = 1.1\text{ V}$. Solution: Step 1: Use the relation $\log K_c = \frac{n E^\circ_{\text{cell}}}{0.0591}$. Step 2: Substitute values: $\log K_c = \frac{2 \times 1.1}{0.0591} = 37.22$. Step 3: Take antilog: $K_c = 10^{37.22} \approx 1.66 \times 10^{37}$.

CBSE Exam Traps & Key Conceptual Tips

• Do not multiply standard electrode potentials ($E^\circ$) by stoichiometric coefficients when balancing cell reactions. Electrode potential is an intensive property and does not scale with chemical quantities.
• Gibbs free energy change ($\Delta G$) is an extensive property. Its value scales proportionally with the stoichiometric coefficients used in the cell reaction.
• A highly negative standard reduction potential implies that the metal is a powerful reducing agent (e.g., Lithium). Conversely, a highly positive reduction potential means the species is a strong oxidizing agent (e.g., Fluorine).
• Remember that molar conductivity ($\Lambda_m$) increases with dilution for both strong and weak electrolytes, but weak electrolytes experience a sharp, dramatic curve upward as concentration approaches zero.

Practice Questions with Solutions

• Q: Calculate the electrode potential of a copper electrode in contact with a $0.1\text{ M } Cu^{2+}$ solution. Given $E^\circ(Cu^{2+}/Cu) = 0.34\text{ V}$. A: Step 1: Write the reduction half-reaction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. Here, $n = 2$. Step 2: Use Nernst equation for a single electrode: $E = E^\circ - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$. Step 3: Substitute the known values: $E = 0.34 - \frac{0.0591}{2} \log \frac{1}{0.1}$. Step 4: Solve the logarithm: $\log(10) = 1$. Therefore, $E = 0.34 - 0.02955 \times 1 = 0.31\text{ V}$. Final answer: The electrode potential of the copper electrode is 0.31 V.
• Q: What is the limiting molar conductivity ($\Lambda^\circ_m$) of $Al_2(SO_4)_3$ if limiting molar conductivities of $Al^{3+}$ and $SO_4^{2-}$ are $189\text{ S cm}^2\text{ mol}^{-1}$ and $160\text{ S cm}^2\text{ mol}^{-1}$ respectively? A: Step 1: According to Kohlrausch's law, the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its anions and cations. Step 2: Write the dissociation reaction: $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$. Step 3: Express the law mathematically: $\Lambda^\circ_m[Al_2(SO_4)_3] = 2\lambda^\circ(Al^{3+}) + 3\lambda^\circ(SO_4^{2-})$. Step 4: Calculate the final sum: $\Lambda^\circ_m = 2(189) + 3(160) = 378 + 480 = 858\text{ S cm}^2\text{ mol}^{-1}$. Final answer: The limiting molar conductivity is 858 S cm² mol⁻¹.
• Q: How many hours does it take to deposit $5.6\text{ g}$ of iron from a solution of $FeCl_3$ using a constant current of $2.0\text{ A}$? (Atomic mass of $Fe = 56\text{ g mol}^{-1}$, $1\text{ F} = 96500\text{ C}$) A: Step 1: Write the cathodic reduction reaction: $Fe^{3+} + 3e^- \rightarrow Fe(s)$. Here, $3\text{ moles}$ of electrons are needed to deposit $1\text{ mole}$ ($56\text{ g}$) of iron. Step 2: Calculate the charge ($Q$) required for $5.6\text{ g}$ of Fe. $Q = \frac{w \times n \times F}{\text{Molar Mass}} = \frac{5.6 \times 3 \times 96500}{56} = 28950\text{ C}$. Step 3: Relate charge to current and time: $Q = I \times t \Rightarrow t = \frac{Q}{I} = \frac{28950}{2.0} = 14475\text{ seconds}$. Step 4: Convert seconds to hours: $t = \frac{14475}{3600} \approx 4.02\text{ hours}$. Final answer: It takes approximately 4.02 hours to deposit the iron.
• Q: Why does conductivity of a solution decrease with dilution, while molar conductivity increases? A: Step 1: Conductivity ($\kappa$) is defined as the conductance of a unit volume (e.g., $1\text{ cm}^3$) of solution. Upon dilution, the number of current-carrying ions per unit volume decreases, leading to a drop in conductivity. Step 2: Molar conductivity ($\Lambda_m = \frac{\kappa \times 1000}{C}$) is the conducting power of all the ions produced by dissolving one mole of electrolyte. As dilution increases, concentration ($C$) drops much faster than $\kappa$, or the total volume containing one mole increases significantly. Step 3: Therefore, the total conductance of the entire solution containing one mole of solute increases. Final answer: Conductivity decreases because ion concentration per unit volume decreases, while molar conductivity increases due to a larger volume of solution containing 1 mole of electrolyte.

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