krit.club logo

Electrochemistry - Electrochemical Cells

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

An Electrochemical Cell (Galvanic or Voltaic cell) converts chemical energy from a spontaneous redox reaction into electrical energy.

•

The cell consists of two half-cells: the Anode where oxidation occurs (negative terminal) and the Cathode where reduction occurs (positive terminal).

•

A Salt Bridge is used to maintain electrical neutrality in the half-cells and to complete the internal circuit by allowing the flow of ions.

•

Electrode Potential is the potential difference between the electrode and the electrolyte. Under standard conditions (1 M1\text{ M} concentration, 298 K298\text{ K}), it is called Standard Electrode Potential (E∘E^\circ).

•

The Standard Hydrogen Electrode (SHE) serves as a reference electrode with an assigned potential of 0.00 V0.00\text{ V}.

•

The Electromotive Force (EMF) of a cell is the potential difference between the two electrodes when no current is drawn through the cell.

•

Gibbs Free Energy (DeltaG\\Delta G) and Cell Potential: A reaction is spontaneous only if ΔG\Delta G is negative, which requires the EcellE_{cell} to be positive.

•

The Nernst Equation describes the effect of ion concentration and temperature on the cell potential.

📐Formulae

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Ecell=Ecell∘−2.303RTnFlog⁡QE_{cell} = E^\circ_{cell} - \frac{2.303 RT}{nF} \log Q

Ecell=Ecell∘−0.0591nlog⁡[Anode Ion][Cathode Ion] (at 298 K)E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[\text{Anode Ion}]}{[\text{Cathode Ion}]} \text{ (at 298 K)}

ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}

log⁡Kc=nEcell∘0.0591\log K_c = \frac{n E^\circ_{cell}}{0.0591}

💡Examples

Problem 1:

Calculate the emf of the following cell at 298 K298\text{ K}: Mg(s)∣Mg2+(0.001 M)∣∣Cu2+(0.0001 M)∣Cu(s)Mg(s) | Mg^{2+}(0.001\text{ M}) || Cu^{2+}(0.0001\text{ M}) | Cu(s). Given EMg2+/Mg∘=−2.37 VE^\circ_{Mg^{2+}/Mg} = -2.37\text{ V} and ECu2+/Cu∘=+0.34 VE^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}.

Solution:

  1. Calculate Ecell∘E^\circ_{cell}: Ecell∘=Ecathode∘−Eanode∘=0.34−(−2.37)=2.71 VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-2.37) = 2.71\text{ V}.
  2. Identify nn: For Mg→Mg2++2e−Mg \rightarrow Mg^{2+} + 2e^- and Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu, n=2n = 2.
  3. Apply Nernst Equation: Ecell=2.71−0.05912log⁡[Mg2+][Cu2+]E_{cell} = 2.71 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}.
  4. Substitute values: Ecell=2.71−0.02955log⁡10−310−4=2.71−0.02955log⁡(10)E_{cell} = 2.71 - 0.02955 \log \frac{10^{-3}}{10^{-4}} = 2.71 - 0.02955 \log(10).
  5. Ecell=2.71−0.02955(1)=2.68045 VE_{cell} = 2.71 - 0.02955(1) = 2.68045\text{ V}.

Explanation:

The standard cell potential is calculated first. Since the concentrations are not 1 M1\text{ M}, the Nernst equation is used to find the actual cell potential. The reaction quotient QQ is the ratio of product ion concentration to reactant ion concentration.

Problem 2:

Calculate the standard Gibbs energy ΔG∘\Delta G^\circ for the reaction: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s). Given Ecell∘=1.1 VE^\circ_{cell} = 1.1\text{ V} and F=96500 C/molF = 96500\text{ C/mol}.

Solution:

  1. Identify nn: In this redox reaction, 22 electrons are transferred, so n=2n = 2.
  2. Use the formula: ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}.
  3. ΔG∘=−(2)×(96500 C/mol)×(1.1 V)\Delta G^\circ = -(2) \times (96500\text{ C/mol}) \times (1.1\text{ V}).
  4. ΔG∘=−212300 J/mol=−212.3 kJ/mol\Delta G^\circ = -212300\text{ J/mol} = -212.3\text{ kJ/mol}.

Explanation:

Standard Gibbs energy is directly proportional to the standard cell potential. The negative sign indicates that the reaction is thermodynamically spontaneous under standard conditions.

Electrochemical Cells - Revision Notes & Key Formulas | CBSE Class 12 Chemistry