Faraday Efficiency

Introduction

Passing charge through an electrolyzer does not guarantee every electron makes the desired product. Water splitting, metal deposition or another side reaction may share the current. Faraday efficiency compares the product actually formed with the amount expected if all charge followed the chosen electrode reaction.

Core explanation

Charge passed during constant current is Q=It, where I is amperes and t seconds; for varying current, Q=∫I dt. Faraday constant F≈96,485 C mol⁻¹ of electrons converts charge to moles of electrons: n e=Q/F. If making one mole of a product requires z moles of electrons by the balanced half-reaction, theoretical product amount is n theory=Q/(zF). For a deposited metal, theoretical mass is m theory=QM/(zF), where M is its molar mass.

Faradaic efficiency for a chosen product is FE=(actual moles of product × zF/Q)×100%, equivalently actual amount divided by theoretical amount under the same z. It can be less than 100% if current makes other products, reverses through product oxidation, or drives unintended reactions. A measured value above 100% usually signals inaccurate product or charge measurement, wrong electron stoichiometry, a non-electrochemical product source, or an unaccounted process; it is not an ordinary charge-allocation efficiency greater than unity.

For copper deposition, Cu²⁺+2e⁻→Cu(s), so z=2. Hydrogen evolution, 2H⁺+2e⁻→H₂ under acidic conditions, can compete at a cathode. The exact side reactions depend on electrode potential, pH, surface and electrolyte. A high current does not ensure high copper FE; at excessive current, Cu²⁺ depletion can increase the share of hydrogen or other reductions.

Faradaic efficiency differs from energy efficiency. FE counts charge allocation, whereas energy efficiency compares useful product energy or reversible work with electrical energy supplied. A process can have nearly 100% FE but require a large extra voltage due to resistance and overpotential, wasting energy as heat. Conversely, an energy metric requires defining the system boundary and desired output, not just measuring product moles.

Gas products require careful collection because dissolution, leaks or crossover can make apparent FE lower than true electrode selectivity. Deposited solids may flake off or include impurities, affecting mass-based inference. A rigorous charge balance may account for all products on both electrodes, not only one selected product.

In a rechargeable battery, coulombic efficiency compares discharge and charge amounts over a cycle, but its definition and interpretation differ from FE for one electrolytic product. Side reactions consume some charge and degrade capacity. Context should therefore accompany the term “efficiency.”

Step-by-step reasoning

1. Balance the desired electrode half-reaction and find z. 2. Integrate current over time to obtain Q. 3. Calculate theoretical product Q/(zF). 4. Measure actual product with corrections for loss or impurity. 5. Compute FE and separately assess energy efficiency from voltage and energy input.

Visual explanation

Draw a current arrow splitting at a cathode into a thick desired-product path and a thinner side-reaction path. Label total charge Q and desired branch charge zFn actual. A separate bar shows electrical energy input Q×operating voltage, emphasizing that voltage losses are a different accounting question.

Real-world analogy

A factory may use 100 delivered parts, with 80 going into the intended product and 20 into another line. That split resembles Faradaic efficiency. How much electricity the factory spends per intended part is a separate energy-efficiency question.

Real-world example

An electroplating bath can deposit less copper than a Faraday-law calculation predicts if hydrogen evolves at the same cathode. Comparing carefully dried deposited mass with charge passed estimates plating FE. Uneven adhesion or trapped bath liquid must be excluded from the mass measurement.

Why?

Why can a metal-deposition efficiency fall at high current density? Metal ions may not reach the electrode fast enough, lowering their surface concentration. Competing reactions such as hydrogen evolution can then consume more of the electron flow while target deposition becomes transport-limited.

Common misconception

“100% Faradaic efficiency means the process wastes no electricity.” It means the counted electrons form the selected product according to stoichiometry. Ohmic resistance and overpotential may still require substantial extra energy per mole.

Worked example

A 2.00 A current passes for 1,000 s through a Cu²⁺ plating cell. Q=2,000 C. Theoretical copper mass is 2000×63.55/(2×96,485)=0.659 g. If the cleaned deposit has 0.593 g Cu, FE=0.593/0.659×100≈90.0%. About 10% of charge was not accounted for by collected copper under this simple analysis.

Quick check

1. How many electrons are needed per Cu atom deposited from Cu²⁺? Answer: Two. 2. Does FE include overpotential energy loss directly? Answer: No; it is a charge-allocation measure.

Exam focus

Use Q=It with seconds, divide by zF and convert moles to mass only afterward. Distinguish anode and cathode products and identify possible side reactions. Keep Faradaic/coulombic efficiency conceptually separate from voltage or energy efficiency.

Advanced insight

Selective electrosynthesis may have different electron counts for competing products. A full electron balance sums zᵢF nᵢ for all quantified products and compares with Q. Missing charge can indicate unmeasured products, corrosion, crossover or measurement error, not one uniquely identifiable side reaction.

Summary

Faraday's law turns charge into theoretical product through electron stoichiometry. Faradaic efficiency measures the fraction of charge yielding a selected product, while energy efficiency also accounts for voltage. Reliable analysis needs accurate charge, product and side-reaction measurements.

Practice questions

1. What theoretical moles of a two-electron product result from Q=19,297 C? Answer: 19,297/(2×96,485)=0.100 mol. 2. If actual product is 0.080 mol in that run, what is FE? Answer: 0.080/0.100×100=80%. 3. Why might gas collection underestimate true FE? Answer: Some gas may dissolve, leak or cross to another compartment before measurement.