Electrochemistry Formulae

Introduction

Electrochemistry formulae connect electron-transfer thermodynamics with measured voltage and passed charge. A cell potential describes driving force per charge, while current integrated over time gives total charge delivered. The two should not be confused: a high voltage alone does not state how much product is made, and a large charge does not prove that all electrons formed the desired product.

Core explanation

With tabulated reduction potentials, E cell = E cathode − E anode for the reaction as written. The cathode is the reduction site; the anode is the oxidation site. Reversing a half reaction to write oxidation does not require multiplying its potential by the stoichiometric coefficient. Potentials are intensive, while Gibbs-energy changes scale with reaction amount. If a half reaction is doubled for electron balance, its listed E° remains the same.

For a reaction transferring n moles of electrons per mole of reaction, Δ rG = −nFE cell under reversible thermodynamic conditions. Units check: F is C mol⁻¹ and volt is J C⁻¹, giving J mol⁻¹. A positive E cell corresponds to negative Δ rG for the forward galvanic reaction. At equilibrium, Δ rG = 0 and the cell potential for that reaction is zero, even if individual half-cell potentials relative to a reference are not zero.

The Nernst equation is E = E° − (RT/nF)ln Q, where Q is dimensionless and formed from activities for the balanced redox reaction. At 298 K, E ≈ E° − (0.05916 V/n)log₁₀Q. This decimal coefficient changes with temperature. Increasing product activity relative to reactants raises Q and lowers forward cell potential under this convention.

Combining Δ rG° = −nFE° with Δ rG° = −RT ln K gives E° = (RT/nF)ln K. A positive standard cell potential implies K > 1 for the balanced reaction as written, but a particular nonstandard mixture can still have E < 0 if Q is sufficiently large. The reaction equation and n must match the K and E° being used.

Charge is Q electric = ∫I dt, or Q = It for constant current. The symbol Q is also used for reaction quotient; label it to avoid confusion. Ideal product amount for a z-electron product is n product = Q electric/(zF). If only fraction η F of current makes the desired product, n product = η F Q electric/(zF). A current efficiency claim therefore needs product measurement.

In electrolysis, an external power supply can force a nonspontaneous reaction. Applied voltage exceeds the equilibrium potential because of overpotential, resistance and concentration gradients. The Nernst equation describes reversible thermodynamic potential, not necessarily the full voltage required at practical current. Comparing measured operating voltage directly to E° without these losses is misleading.

Half-cell potentials require a reference electrode and a specified medium. Solution activities, pH and complexation shift them. A tabulated standard potential is not a universal constant for a bare element under every condition. Write physical states and species carefully when building Q.

Step-by-step reasoning

Balance oxidation and reduction half reactions to find n. Compute E° from reduction potentials with cathode minus anode. Build dimensionless Q and use Nernst for actual conditions. For product calculations, integrate current to charge, divide by zF and apply measured Faradaic efficiency. Check energy and charge units separately.

Visual explanation

Draw an electron arrow through a wire from anode to cathode. Label E cell as the potential difference, current I as charge flow per time, and product formation at each electrode. A second panel shows E decreasing as ln Q increases.

Real-world analogy

Voltage resembles pressure pushing water through a pipe, while current resembles flow rate and total charge resembles volume delivered. A pipe can have pressure without much flow; likewise a cell can have potential without producing much material. The analogy does not replace electron stoichiometry or electrochemical overpotentials.

Real-world example

Copper plating follows Cu²⁺ + 2e⁻ → Cu(s). Measuring current and duration estimates theoretical deposited copper, while weighing the electrode tests whether charge went into copper rather than hydrogen evolution or other reactions. A surface potential measurement alone cannot yield the deposited mass.

Why?

Electrochemical formulae support batteries, corrosion, electrolysis and analytical sensors. Keeping voltage, charge and product stoichiometry distinct makes energy and material balances trustworthy. It also clarifies why a measured current does not guarantee the desired product.

Common misconception

“Multiply electrode potential by two when the half reaction is doubled” is wrong because potential is intensive. Another error assumes 100% Faradaic efficiency without checking side products or gas evolution.

Worked example

A constant 2.00 A current flows for 965 s through a Cu²⁺ plating cell. Q electric = It = 1930 C. Theoretical copper amount is 1930/(2 × 96,485) ≈ 0.0100 mol, or about 0.635 g using 63.55 g mol⁻¹. If measured deposit is 0.572 g, Faradaic efficiency is approximately 0.572/0.635 = 90.1%, assuming the deposit is pure copper.

Quick check

1. What is the unit of F, and what does it convert? Answer: C mol⁻¹; it converts moles of electrons to charge magnitude.

Exam focus

Use reduction-potential convention consistently and do not scale E by coefficients. Distinguish reaction quotient Q from electrical charge. State n or z explicitly and account for Faradaic efficiency when predicting actual product.

Advanced insight

For a concentration cell, identical electrode materials can produce voltage solely from activity differences. Nernst predicts its reversible value. As the cell operates, composition gradients shrink and the potential approaches zero unless maintained externally.

Summary

Cell potential is cathode minus anode reduction potential; Gibbs energy is −nFE; Nernst corrects potential for composition. Current integrated over time gives charge, which yields an ideal product amount only after electron stoichiometry and efficiency are included.

Practice questions

1. What is E cell if E cathode = 0.80 V and E anode = 0.34 V as reduction potentials? Answer: 0.46 V. 2. Does doubling a balanced half reaction double its potential? Answer: No. Potential is intensive; the associated reaction Gibbs energy scales. 3. What charge passes at 0.50 A for 120 s? Answer: 60 C. 4. Why can measured electrode product be below Q/(zF)? Answer: Some charge may drive side reactions, giving Faradaic efficiency below one.

Sources

- OpenStax Chemistry 2e: Potential, Free Energy and Equilibrium. - NIST 2022 CODATA constants.