# Lecture 16: Electrolysis and Applications
## General Chemistry II
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## Learning Objectives
By the end of this lecture, students will be able to:
1. Distinguish between galvanic and electrolytic cells
2. Describe the process of electrolysis and predict the products at each electrode
3. Apply Faraday's laws to calculate quantities in electrolysis
4. Explain the electrolysis of molten salts and aqueous solutions
5. Describe industrial applications of electrolysis including electroplating and metal refining
6. Define and calculate overpotential
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## Lecture Content
### I. Electrolytic vs. Galvanic Cells
A galvanic cell harnesses the energy of a spontaneous reaction to produce electrical energy (E_cell > 0, Delta G < 0). An electrolytic cell does the opposite: it uses electrical energy from an external power source to drive a nonspontaneous reaction (E_cell < 0, Delta G > 0). In an electrolytic cell, the anode is connected to the positive terminal of the power source and is therefore positive, while the cathode is connected to the negative terminal and is negative. This is the reverse of the sign convention in galvanic cells. However, the fundamental chemistry remains the same in both types: oxidation always occurs at the anode and reduction always occurs at the cathode. The minimum voltage required to drive an electrolytic cell equals the magnitude of the cell potential for the nonspontaneous reaction, though in practice more voltage is needed due to overpotential.
### II. Electrolysis of Molten Salts
The simplest case of electrolysis involves a molten (liquid) salt, where only two ions are present and there is no water to complicate matters. In the Downs cell, molten NaCl is electrolyzed: Na+ is reduced at the cathode to form liquid sodium metal (Na+ + e- -> Na(l)), while Cl- is oxidized at the anode to form chlorine gas (2Cl- -> Cl2(g) + 2e-). The overall reaction is 2NaCl(l) -> 2Na(l) + Cl2(g). The salt must be heated above its melting point (801 degrees C for NaCl), and this process is the primary industrial method for producing sodium metal and chlorine gas.
### III. Electrolysis of Aqueous Solutions
Electrolysis of aqueous solutions is more complex because water itself can be oxidized or reduced, competing with the dissolved ions. At the cathode, the species with the most positive (least negative) reduction potential is preferentially reduced. If the cation's E^0 is more positive than -0.83 V (the reduction potential of water at pH 7), the metal ion is reduced. If the cation's E^0 is more negative than water's, then water is reduced instead: 2H2O + 2e- -> H2(g) + 2OH-(aq).
At the anode, the species with the most negative (least positive) reduction potential is preferentially oxidized. If the anion's E^0 is less positive than +0.82 V (the oxidation potential of water at pH 7), the anion is oxidized. Otherwise, water is oxidized: 2H2O -> O2(g) + 4H+(aq) + 4e-. However, overpotential can alter these predictions. Because O2 evolution has a particularly high overpotential, anions like Cl- and Br- are often oxidized preferentially even when the standard potentials would predict water oxidation.
A side-by-side comparison of the electrolysis of molten NaCl vs. aqueous NaCl. Left panel (Molten NaCl): An electrolytic cell with a battery, showing Na+ migrating to the cathode where Na(l) is deposited, and Cl- migrating to the anode where Cl2(g) is evolved. Products: Na metal and Cl2 gas. Right panel (Aqueous NaCl): A similar cell but with water present. At the cathode, H2O is reduced (not Na+) because E^0 for Na+ is too negative, producing H2(g) and OH-. At the anode, Cl- is oxidized (due to O2 overpotential), producing Cl2(g). Products: H2 gas, Cl2 gas, and NaOH. Each panel lists the half-reactions and explains why each product forms.
### IV. Faraday's Laws of Electrolysis
Faraday's first law states that the mass of a substance deposited or dissolved at an electrode is directly proportional to the quantity of electric charge passed. The second law states that the mass deposited is proportional to the molar mass divided by the number of electrons transferred per formula unit.
The key quantitative relationships are straightforward. Charge equals current times time: q = I * t, where q is in coulombs (C), I is in amperes (A), and t is in seconds (s). One Faraday of charge (F = 96,485 C) corresponds to the charge carried by one mole of electrons. The number of moles of electrons transferred is therefore mol e- = q / F = (I * t) / 96,485. From there, the half-reaction stoichiometry converts moles of electrons to moles of substance, and multiplication by the molar mass gives the mass deposited.
As an example, consider how many grams of copper are deposited by passing 3.00 A of current for 2.00 hours. The total charge is q = 3.00 A x 7200 s = 21,600 C. The moles of electrons transferred are 21,600 / 96,485 = 0.224 mol. Since Cu^2+ + 2e- -> Cu, the moles of copper deposited are 0.224 / 2 = 0.112 mol, corresponding to a mass of 0.112 x 63.55 = 7.12 g.
### V. Overpotential
Overpotential, also called overvoltage, is the extra voltage beyond the thermodynamic minimum that must be applied to drive electrolysis at a measurable rate. It arises from several sources. Activation overpotential reflects the kinetic energy barrier for the electrode reaction. Concentration overpotential results from depletion of the reactant at the electrode surface. Resistance overpotential comes from the ohmic resistance of the solution and the electrodes themselves.
The magnitude of overpotential depends on the electrode material and surface condition, the current density, and the nature of the reaction. Gas evolution reactions have particularly high overpotentials: O2 evolution typically requires an extra 0.4-0.8 V, while H2 evolution varies widely depending on the electrode material. The practical consequence is that overpotential can change which reaction actually occurs at an electrode, sometimes overriding the predictions based solely on standard reduction potentials.
### VI. Industrial Applications of Electrolysis
Electroplating uses electrolysis to deposit a thin, uniform layer of metal onto an object. The object to be plated serves as the cathode, and the plating metal or its salt solution provides the cations. Applications include chrome plating, silver plating, gold plating, and corrosion protection.
Electrorefining of copper produces ultra-pure copper (99.99%) from impure copper (about 99% pure). The impure copper serves as the anode and dissolves into solution, while pure copper deposits on the cathode from the CuSO4 electrolyte. Impurities that do not dissolve collect at the bottom of the cell as "anode sludge," which may contain valuable metals such as gold, silver, and platinum.
The Hall-Heroult process produces aluminum by electrolyzing Al2O3 dissolved in molten cryolite (Na3AlF6), which lowers the melting point. Aluminum ions are reduced at the cathode to form liquid aluminum, while the carbon anodes are oxidized to CO2. This process is enormously energy-intensive, which is the primary reason aluminum recycling is so environmentally and economically important.
The chlor-alkali process electrolyzes concentrated brine (NaCl solution) to produce three valuable industrial chemicals: chlorine gas, hydrogen gas, and sodium hydroxide.
A diagram of the electrorefining of copper. A large electrolytic cell contains CuSO4 solution (blue). The left electrode (anode) is labeled "Impure Cu (99% pure)" and is shown dissolving: Cu -> Cu^2+ + 2e-. The right electrode (cathode) is labeled "Pure Cu (99.99% pure)" and is shown growing as Cu^2+ + 2e- -> Cu deposits. At the bottom of the cell, "anode sludge" is shown collecting, with a note listing its contents: "Ag, Au, Pt, Se, Te -- valuable byproducts." A battery symbol provides the external voltage. Arrows show Cu^2+ ions migrating from anode to cathode through the solution.
### VII. Electrolysis and Stoichiometry Summary
The quantitative framework for electrolysis problems follows a consistent chain: charge -> moles of electrons -> moles of substance -> grams. Common problem types include finding the mass deposited given current and time, finding the time required to deposit a given mass at a specified current, finding the current needed to deposit a given mass in a given time, and determining the oxidation state of a metal from electrolysis data. In every case, the essential first step is to write the balanced half-reaction to establish the ratio of electrons to substance.
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