Electrochemistry

1. Electrochemical Cells

Galvanic Cells: Spontaneous chemical energy $\to$ electrical energy ($\Delta G < 0$). Electrolytic Cells: Electrical energy $\to$ non-spontaneous chemical change.

Nernst Equation

Calculates cell potential under any conditions:

Interactive diagram: open this page in a browser to use it.

2. Conductance and Molar Conductivity

Kohlrausch's Law: At infinite dilution, each ion migrates independently. $\Lambda_m^\infty = \nu_+ \lambda_+^\infty + \nu_- \lambda_-^\infty$. This allows calculation of $\Lambda_m^\infty$ for weak electrolytes.

3. Cell Potential and Free Energy

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ using standard reduction potentials. The link to thermodynamics:

A positive $E^\circ_{cell}$ means a spontaneous cell reaction. In the electrochemical series, a lower (more negative) reduction potential means a stronger reducing agent; Li is the strongest reducing agent in aqueous solution.

4. Conductivity Terms

• Conductivity $\kappa = \frac{1}{\rho} = \frac{G^*}{R}$, where $G^* = l/A$ is the cell constant.
• Molar conductivity $\Lambda_m = \frac{\kappa \times 1000}{C}$ ($S\,cm^2mol^{-1}$, with $C$ in mol/L).
• On dilution, $\kappa$ decreases (fewer ions per cm³) but $\Lambda_m$ increases.
• Strong electrolytes follow $\Lambda_m = \Lambda_m^\circ - A\sqrt{C}$ (Debye–Hückel–Onsager). Weak electrolytes show a steep rise at low concentration; their degree of dissociation is $\alpha = \Lambda_m / \Lambda_m^\circ$.

5. Faraday's Laws of Electrolysis

First law: $m = ZIt = \frac{M}{nF}It$, with $F = 96485\ C\,mol^{-1}$. Second law: for the same charge, masses deposited are in the ratio of their equivalent weights.

6. Products of Electrolysis

• Aqueous NaCl: $H_2$ at the cathode, $Cl_2$ at the anode (overpotential makes $Cl_2$ form instead of $O_2$), leaving NaOH.
• Molten NaCl: Na at the cathode, $Cl_2$ at the anode.
• Dilute $H_2SO_4$: $H_2$ at the cathode, $O_2$ at the anode.

7. Batteries, Fuel Cells and Corrosion

The lead storage battery uses Pb (anode), $PbO_2$ (cathode) and 38% $H_2SO_4$. The $H_2$–$O_2$ fuel cell converts chemical energy to electricity at about 70% efficiency. Rusting is an electrochemical process; galvanising with zinc protects iron because zinc is oxidised preferentially (sacrificial protection).

Practice questions

1. On dilution of an electrolyte solution:

   1. Both and ₘ increase
   2. Both decrease
   3. decreases, ₘ increases
   4. increases, ₘ decreases
   Answer

   C. Fewer ions per unit volume lowers , but each mole of electrolyte conducts better, so ₘ rises.

2. The charge needed to deposit 1 mol of Al from Al³⁺ is:

   1. 1 F
   2. 2 F
   3. 3 F
   4. 6 F
   Answer

   C. Al³⁺ + 3e⁻ → Al needs 3 mol of electrons, i.e. 3 F.

3. For a cell with E°cₑₗₗ > 0:

   1. Δ G° > 0, K < 1
   2. Δ G° < 0, K > 1
   3. Δ G° = 0, K = 1
   4. Δ G° < 0, K < 1
   Answer

   B. Δ G° = -nFE° is negative, and Δ G° = -RT K then gives K > 1.

4. During electrolysis of aqueous NaCl, the gas evolved at the anode is:

   1. H₂
   2. O₂
   3. Cl₂
   4. Na vapour
   Answer

   C. Because of the overpotential of oxygen, Cl⁻ is oxidised to Cl₂ at the anode.

5. Zinc protects iron from rusting because:

   1. Zinc is less reactive than iron
   2. Zinc has a higher reduction potential
   3. Zinc is oxidised in preference to iron
   4. Zinc forms an alloy with iron
   Answer

   C. Zinc has a more negative reduction potential, so it acts as a sacrificial anode.