Reverse the half-reaction with the more negative standard reduction potential (it becomes the oxidation/anode) so E°cell comes out positive.

E°cell = E°(cathode) − E°(anode), both as reduction potentials. Multiplying a half-reaction by a coefficient does NOT change its potential.

anode | anode ion || cathode ion | cathode. Oxidation on the left; no coefficients. Use an inert Pt electrode when neither side is a conducting solid.

E = E° − (0.0592/n) log Q at 25 °C (n = electrons transferred, Q = products/reactants). Use it for any non-standard concentration or pressure.

Put the measured E into the Nernst equation, solve for [H⁺], then pH = −log[H⁺].

Charge (C) = current(A) × time(s) (convert h/min to s). Moles of e⁻ = charge / F, with F = 96,485 C/mol e⁻.

ΔG° = −nFE°cell. E°cell > 0 ↔ ΔG° < 0 (spontaneous galvanic); E°cell < 0 ↔ ΔG° > 0 (electrolytic if driven). n = e⁻ per balanced reaction.

ln K = nFE°cell/(RT), or log K = nFE°cell/(2.303RT). E°cell > 0 → K > 1. Together with ΔG° = −RT ln K this closes the E°cell ↔ ΔG° ↔ K triangle.

Same chemistry, different [ion], so E° = 0 and E = −(0.0592/n) log Q. Write Q from the balanced overall reaction (Mⁿ⁺: Q = [anode]/[cathode]). At equal concentrations Q = 1, E = 0.

The low-[ion] half-cell is the anode (its concentration rises); the high-[ion] half-cell is the cathode (falls). Electrons flow anode → cathode until concentrations equalize.