10.4 Electrolytic Cells & Faraday's Laws of Electrolysis

10.4 Electrolytic Cells & Faraday's Laws of Electrolysis

Quick Summary: Electrolytic cells drive thermodynamically nonspontaneous oxidation-reduction reactions ($\Delta G > 0, E_{\text{cell}} < 0$) using electrical energy supplied by an external direct current (DC) power source. While oxidation occurs at the anode and reduction occurs at the cathode across all electrochemical systems, electrode polarities are inverted in electrolytic cells: the anode is positive and the cathode is negative. In aqueous electrolytes, water can competitively oxidize or reduce alongside solute ions, with product selectivity governed by standard reduction potentials and kinetic overpotentials. Faraday's laws quantify the stoichiometric relationship between electrical current, elapsed time, and deposited mass.

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1. Comparing Galvanic & Electrolytic Cells

Electrochemical cells operate in two fundamental thermodynamic modes:

Universal Rules & Polarity Reversal

Two core rules apply to every electrochemical cell:

1. AN OX: Oxidation occurs at the Anode.
2. RED CAT: Reduction occurs at the Cathode.
3. Electron Flow: Electrons travel from Anode to Cathode through the external circuit.

In an electrolytic cell, an external DC power supply acts as an electron pump. It pulls electrons away from the anode (imparting a positive charge) and forces electrons onto the cathode (imparting a negative charge). Inside the cell, anions migrate to the positive anode, and cations migrate to the negative cathode.

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2. Electrolysis of Molten Salts

In pure molten salts, only the salt's constituent ions are present, eliminating solvent competition.

The Downs Cell (Production of Sodium Metal)

Metallic sodium is produced industrially by electrolyzing molten sodium chloride ($\text{NaCl}$):

• Pure $\text{NaCl}$ melts at $801\ ^\circ\text{C}$; adding $\text{CaCl}_2$ flux depresses the melting point to $\sim 600\ ^\circ\text{C}$.
• Cathode (Reduction): $\text{Na}^+(l) + e^- \to \text{Na}(l) \quad (E^\circ = -2.71\text{ V})$ Molten sodium metal floats to the top and is collected under an inert atmosphere.
• Anode (Oxidation): $2\text{Cl}^-(l) \to \text{Cl}_2(g) + 2e^- \quad (E^\circ = +1.36\text{ V})$ Chlorine gas is captured at the central graphite anode.
• Overall Reaction: $2\text{NaCl}(l) \xrightarrow{\text{electrolysis}} 2\text{Na}(l) + \text{Cl}_2(g) \quad (E^\circ_{\text{cell}} = -4.07\text{ V})$ A cylindrical iron diaphragm separates the compartments to prevent explosive recombination of sodium and chlorine.

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3. Competitive Aqueous Electrolysis & Overpotential

When electrolyzing aqueous salt solutions, water molecules can compete with dissolved ions at both electrodes.

Cathode Competition (Reduction)

• Cation Reduction: $\text{M}^{n+}(aq) + n e^- \to \text{M}(s)$
• Water Reduction: $2\text{H}_2\text{O}(l) + 2e^- \to \text{H}_2(g) + 2\text{OH}^-(aq) \quad (E^\circ = -0.83\text{ V at } 1.0\text{ M }\text{OH}^-; -0.41\text{ V at pH } 7.0)$
• Prediction Rule: Active metals whose cations have reduction potentials more negative than water reduction (Group 1, Group 2, and $\text{Al}^{3+}$) cannot be reduced from aqueous solution; $\text{H}_2$ gas evolves instead. Less active metals ($\text{Cu}^{2+}, \text{Ag}^+, \text{Ni}^{2+}$) reduce directly to solid metal.

Anode Competition (Oxidation)

• Anion Oxidation: $2\text{X}^-(aq) \to \text{X}_2 + 2e^-$
• Water Oxidation: $2\text{H}_2\text{O}(l) \to \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \quad (E^\circ = +1.23\text{ V at } 1.0\text{ M }\text{H}^+; +0.82\text{ V at pH } 7.0)$
• Prediction Rule: Highly oxidized polyatomic anions ($\text{SO}_4^{2-}, \text{NO}_3^-, \text{ClO}_4^-, \text{PO}_4^{3-}$) and fluoride ($\text{F}^-$) resist oxidation. Water is oxidized instead, producing $\text{O}_2$ gas.

Overpotential Phenomena

Thermodynamic potentials do not account for reaction kinetics. The evolution of gaseous products (like $\text{O}_2$) requires significant multi-step activation energy. The additional voltage required to drive an electrode reaction at a practical rate is called overpotential ($0.4\text{--}0.6\text{ V}$ for $\text{O}_2$ on graphite). Because of this large overpotential, the electrolysis of aqueous brine ($\text{NaCl}$) oxidizes chloride to $\text{Cl}_2(g)$ rather than water to $\text{O}_2$: $2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{electrolysis}} \text{Cl}_2(g) + \text{H}_2(g) + 2\text{NaOH}(aq)$

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4. Major Industrial Applications

• Electroplating: Coats objects with protective or decorative metals ($\text{Cr}, \text{Au}, \text{Ag}$). The object to be plated is the cathode ($-$), immersed in a bath containing metal cations, while a sacrificial metal anode ($+$) dissolves to replenish ions.
• Electrorefining of Copper: Impure blister copper ($\sim 99\%$ pure) is made the anode in an acidic $\text{CuSO}_4$ bath. Copper dissolves and selectively plates onto a pure copper cathode ($>99.99\%$ purity). Noble impurities ($\text{Ag}, \text{Au}, \text{Pt}$) resist oxidation and settle as valuable anode sludge.
• Hall-Héroult Process: Extracts aluminum metal by dissolving alumina ($\text{Al}_2\text{O}_3$) in molten cryolite ($\text{Na}_3\text{AlF}_6$) at $950\ ^\circ\text{C}$. Molten $\text{Al}$ forms at the carbon cathode, while carbon anodes oxidize to $\text{CO}_2(g)$, requiring routine replacement.

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5. Faraday's Laws & Quantitative Stoichiometry

Michael Faraday established that the mass of substance altered at an electrode is directly proportional to electrical charge passed:

1. Total charge: $Q = I \times t$ (where $I$ is current in amperes, $t$ is time in seconds).
2. Moles of electrons: $n_{e^-} = \frac{Q}{F} = \frac{I \times t}{96,485\text{ C/mol }e^-}$.
3. Electrodeposited mass: $\text{Mass} = \frac{I \times t \times \text{Molar Mass}}{n \times F}$ where $n$ represents moles of electrons per mole of substance.

Worked Problem

Problem: Calculate the mass of gold deposited from an aqueous $\text{AuCl}_3$ solution by a current of $8.00\text{ A}$ running for $45.0\text{ minutes}$. (Molar mass of $\text{Au} = 196.97\text{ g/mol}$).

1. Convert time to seconds: $t = 45.0\text{ min} \times 60\text{ s/min} = 2,700\text{ s}$
2. Calculate charge: $Q = 8.00\text{ A} \times 2,700\text{ s} = 21,600\text{ C}$
3. Calculate moles of electrons: $n_{e^-} = \frac{21,600\text{ C}}{96,485\text{ C/mol }e^-} = 0.2239\text{ mol }e^-$
4. Convert to moles of gold ($\text{Au}^{3+} + 3e^- \to \text{Au}$, so $n = 3$): $n_{\text{Au}} = \frac{0.2239\text{ mol }e^-}{3} = 0.07463\text{ mol Au}$
5. Calculate mass: $\text{Mass} = 0.07463\text{ mol} \times 196.97\text{ g/mol} = 14.7\text{ g Au}$ The cell deposits $14.7\text{ g}$ of gold onto the cathode.