Electrochemistry

A galvanic (voltaic) cell converts chemical energy from a spontaneous redox reaction into electrical energy. The cell consists of two half-cells, each containing one half-reaction’s redox couple. Oxidation occurs at the anode (negative terminal) and reduction at the cathode (positive terminal).

Electrons flow through an external wire from anode to cathode, producing usable electric current. A salt bridge connects the two half-cell solutions, allowing ions to migrate and maintain electrical neutrality — without it, charge buildup would quickly halt the reaction.

Cell notation provides a shorthand description, read left to right from anode to cathode. Vertical lines represent phase boundaries: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). The double line (||) represents the salt bridge. When neither member of a half-cell’s redox couple can serve as an electrode, an inert conductor such as platinum is used.

An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction (ΔG > 0). Unlike galvanic cells, the anode is connected to the positive terminal of the power supply and the cathode to the negative terminal.

Key applications of electrolysis include:

• Electroplating — depositing a thin metal coating (e.g., chromium or gold) onto a surface by reducing metal cations at the cathode.
• Metal refining — purifying metals such as copper by dissolving impure metal at the anode and depositing pure metal at the cathode.
• Production of chemicals — electrolysis of brine produces chlorine gas (anode) and sodium hydroxide (cathode); electrolysis of water produces hydrogen and oxygen.

The minimum voltage required to drive electrolysis equals the magnitude of E°_cell for the reverse (spontaneous) reaction, though in practice a higher overpotential is needed due to kinetic barriers at the electrode surfaces.

The tendency of a species to be reduced is quantified by its standard reduction potential (E°), measured relative to the standard hydrogen electrode (SHE), which is assigned exactly 0.00 V. The SHE consists of a platinum electrode in 1 M H⁺ with H₂ gas at 1 bar.

A table of standard reduction potentials ranks half-reactions by their E° values. Species with large positive E° are strong oxidizing agents (easily reduced), while species with large negative E° are strong reducing agents (easily oxidized). For example, F₂ (E° = +2.87 V) is the strongest common oxidizer, while Li (E° = −3.04 V) is the strongest common reducer.

Reduction potentials are intensive properties — they do not change when a half-reaction is multiplied by a coefficient. Only the identity of the species and the conditions (concentration, temperature) affect E°.

The standard cell potential is the potential difference between cathode and anode when each species is in its standard state: dissolved species have unit activity (often approximated as 1 M in introductory calculations), gases have a standard pressure of 1 bar, and pure solids and liquids have unit activity. Standard potentials can be defined at any stated temperature; the course table and calculations here use values evaluated at 25 °C:

E°_cell = E°_cathode − E°_anode

A positive E°_cell indicates a spontaneous reaction; a negative value means the reaction is nonspontaneous under the stated standard-state conditions. To predict spontaneity, compare the two half-reactions in a standard reduction potential table: the half-reaction higher in the table (more positive E°) will proceed as written (reduction), while the lower one will be reversed (oxidation).

E°_cell links directly to thermodynamic quantities. The relationship ΔG° = −nFE°_cell (where n is the moles of electrons transferred and F = 96,485 C/mol is Faraday’s constant) connects cell potential to free energy. Since ΔG° = −RT ln K, a large positive E°_cell corresponds to a large equilibrium constant, meaning the reaction goes nearly to completion.

Under nonstandard conditions, cell potential depends on the concentrations (or pressures) of reactants and products. The Nernst equation accounts for this:

E_cell = E°_cell − (RT / nF) ln Q

At 25 °C, substituting constants and converting to base-10 logarithms gives the convenient form: E_cell = E°_cell − (0.0592 / n) log Q.

As a reaction proceeds, Q increases and E_cell decreases. At equilibrium, E_cell = 0 and Q = K, yielding: E°_cell = (0.0592/n) log K. This relationship connects electrochemistry directly to equilibrium.

A concentration cell uses two half-cells with the same redox couple at different concentrations. Since E°_cell = 0 (identical half-reactions), the potential arises entirely from the concentration difference and drives the system toward equal concentrations in both compartments.

Standard cell potential, standard free energy change, and the equilibrium constant are three views of the same thermodynamic driving force, connected pairwise:

• ΔG° = −nFE°_cell — n is the number of electrons transferred in the balanced overall reaction and F = 96,485 C/mol e⁠^-. A positive E°_cell means a negative ΔG° (spontaneous), and vice versa. The product comes out in J/mol; divide by 1000 for kJ/mol.
• ΔG° = −RT ln K — the free-energy/equilibrium bridge carried over from the thermodynamics topic.
• log K = nFE°_cell/(2.303RT) — the direct electrochemistry-to-equilibrium link; at 298 K it simplifies to log K = nE°_cell/0.0592.

Because the three are interconvertible, you can enter this triangle at any vertex and compute the other two. Worked pattern: for 2 Ag⁺ + Fe → 2 Ag + Fe²⁺ with E°_cell = +1.247 V and n = 2, ΔG° = −(2)(96,485)(1.247) ≈ −240.6 kJ/mol and log K = (2)(1.247)/0.0592 ≈ 42.1, so K ≈ 10⁴² — enormously product-favored.

Two cautions: away from 298 K use the general form log K = nFE°_cell/(2.303RT) with the actual temperature (the 0.0592 shortcut is 298-K-specific); and when K is astronomically large or small, tiny rounding in E° or in the constant shifts K's leading digit — quote K to a couple of significant figures and don't be surprised when different constant precisions disagree slightly.

A concentration cell has identical half-cells — same electrode, same redox couple — differing only in ion concentration. Because both half-cells share the same standard reduction potential, E°_cell = 0 V: the driving force is purely the concentration imbalance, and the cell runs until the two concentrations equalize.

E_cell = 0 − (0.0592/n)·log Q at 298 K. Write Q from the balanced overall cell reaction. For an Mⁿ⁺/M cell the overall reaction is Mⁿ⁺(cathode) → Mⁿ⁺(anode) at 1:1 stoichiometry, so Q = [Mⁿ⁺]_anode/[Mⁿ⁺]_cathode with exponent 1 regardless of n — the n belongs only in the 0.0592/n denominator. For an H⁺/H₂ cell the balanced reaction has 2 H⁺ on each side, so Q carries squared concentrations. Never apply a blanket “raise the ratio to n” shortcut: the exponent always comes from the balanced reaction, and the blanket shortcut silently doubles the computed voltage for M²⁺/M cells.

The half-cell with the lower ion concentration is the anode (its metal oxidizes, raising that concentration); the higher-concentration half-cell is the cathode (its ions plate out, lowering that concentration). Electrons flow anode → cathode through the external circuit until [ion]_anode = [ion]_cathode, at which point Q = 1 and E_cell = 0 (equilibrium).

Example: Zn | Zn²⁺(0.10 M) || Zn²⁺(0.50 M) | Zn gives Q = 0.10/0.50 = 0.20 and E_cell = −(0.0592/2)·log 0.20 ≈ +0.021 V — small, because a 5× imbalance is a modest driving force. A pH-driven H⁺/H₂ cell across pH 4 vs pH 1 gives +0.178 V, and coupling one half-cell to a sparingly soluble salt (an AgCl-saturated solution) produces large potentials from the enormous concentration ratio.

Faraday’s law relates the quantity of substance produced or consumed during electrolysis to the total electric charge passed through the cell:

moles of substance = (I × t) / (n × F)

• I = current in amperes (C/s)
• t = time in seconds
• n = electrons transferred per formula unit (e.g., 2 for Cu²⁺ + 2e⁻ → Cu)
• F = 96,485 C/mol (Faraday’s constant, the charge of one mole of electrons)

Once you know the moles produced, convert to mass using the molar mass, or to gas volume using the ideal gas law. For example, electroplating a spoon with 0.50 g of silver from AgNO₃ solution requires passing (0.50/107.87) × 1 × 96,485 = 447 C of charge. At a current of 1.5 A, this takes about 298 seconds (roughly 5 minutes).

A battery is one or more galvanic cells packaged for practical use. Primary cells (e.g., alkaline batteries) cannot be recharged because their reactions are not easily reversed. Secondary cells (e.g., lithium-ion, lead-acid) are rechargeable because applying an external voltage reverses the cell reaction, restoring the original reactants.

The common alkaline battery uses a zinc anode and a manganese dioxide cathode in potassium hydroxide electrolyte, producing about 1.5 V. Lead-acid batteries in cars stack six cells in series for a 12 V output, using Pb and PbO₂ electrodes in sulfuric acid.

Fuel cells differ from batteries in that reactants are continuously supplied. A hydrogen fuel cell combines H₂ and O₂ to produce water and electricity with high efficiency and no combustion byproducts. Fuel cells are used in spacecraft, vehicles, and stationary power generation. Corrosion is an unwanted electrochemical process where metals are oxidized by environmental exposure — it can be prevented by protective coatings, sacrificial anodes (cathodic protection), or alloying.

Use this workflow to navigate electrochemistry problems:

1. Identify cell type: galvanic (spontaneous, E°_cell > 0) or electrolytic (driven by external voltage, E°_cell < 0).
2. Write the two half-reactions. Oxidation occurs at the anode; reduction occurs at the cathode.
3. Calculate E°_cell = E°_cathode − E°_anode (using standard reduction potentials for both).
4. For nonstandard conditions, use the Nernst equation: E = E° − (RT/nF) lnQ.
5. For mass/charge calculations, use Faraday’s law: moles = (current × time)/(n × F).

Common mistakes: reversing the sign of a reduction potential when writing an oxidation half-reaction (never change the sign in the table—the subtraction formula handles it), multiplying E° values by stoichiometric coefficients (cell potentials are intensive, not extensive), confusing anode and cathode between galvanic and electrolytic cells, and using seconds for time in Faraday’s law when minutes or hours were given.