Electrochemistry & redox
The idea
Electrochemistry turns redox chemistry into measurable voltage. In a galvanic cell, oxidation at the anode and reduction at the cathode are separated so the transferred electrons must travel through an external wire. Each half-reaction carries a standard reduction potential E°; the half-cell with the higher E° runs as the reduction, and E°cell = E°cathode − E°anode. A positive cell potential signals a spontaneous reaction, and the bridge to thermodynamics is exact: ΔG° = −nFE°, with F = 96,485 C/mol, while log K = nE°/0.0592 at 25 °C links potential to the equilibrium constant.
Your balancing skills feed straight in: split the overall reaction into half-reactions, balance atoms and charge with electrons, and scale until electron counts match. But here is the subtlety that catches almost everyone — multiplying a half-reaction by a coefficient does not multiply its E°. Potential is an intensive property, energy per unit charge, like a height difference; doubling the amount of substance doubles the energy and the charge together, leaving the voltage unchanged.
Keep the sign conventions straight by anchoring to physical meaning: electrons always flow from anode to cathode through the wire, and a table of reduction potentials simply ranks how strongly each species pulls electrons toward itself.
Worked example
A galvanic cell pairs the half-reactions Zn²⁺ + 2e⁻ → Zn (E° = −0.76 V) and Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V). Identify the anode and cathode, then calculate E°cell, ΔG°, and the order of magnitude of K at 25 °C.
- Compare reduction potentials: copper's +0.34 V beats zinc's −0.76 V, so Cu²⁺ is reduced at the cathode and Zn metal is oxidized at the anode. The overall reaction is Zn + Cu²⁺ → Zn²⁺ + Cu with n = 2 electrons transferred.
- Compute the cell potential: E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V. The positive sign confirms the reaction runs spontaneously as written.
- Convert to free energy: ΔG° = −nFE° = −(2)(96,485 C/mol)(1.10 V) = −2.12 × 10⁵ J/mol = −212 kJ/mol, comfortably spontaneous.
- Estimate the equilibrium constant: log K = nE°/0.0592 = (2 × 1.10)/0.0592 ≈ 37.2, so K ≈ 10³⁷ — the reaction goes essentially to completion.
- Sanity-check the coherence: positive E°, large negative ΔG°, and an astronomically large K are three statements of one fact, and the physical setup (zinc dissolving while copper plates out) matches what is observed in the classic Daniell cell.
Answer. Zn is the anode and Cu the cathode; E°cell = +1.10 V, ΔG° = −212 kJ/mol, and K ≈ 10³⁷, so the reaction is overwhelmingly product-favored.
Check your understanding
- Why does multiplying a half-reaction by two leave its E° unchanged while ΔG° doubles, and what property of voltage explains this?
- How would the cell potential change under nonstandard concentrations, and which direction does the Nernst equation push E as products accumulate?
- What distinguishes a galvanic cell from an electrolytic cell in terms of the sign of ΔG and the role of the power supply?
- How would you use a table of reduction potentials to predict whether a given metal will dissolve in 1 M acid?
Build the foundations first
Electrochemistry & redox builds on these concepts. If any feel shaky, start there.