Overpotential and Polarization

Introduction

The standard potentials in data tables predict that water should split at 1.23 V, yet a practical water electrolyser typically runs at 1.8 to 2.0 V. A fresh battery rated at 1.5 V may read 1.5 V on an open circuit but deliver noticeably less when it powers a motor. The gap between the thermodynamic voltage and the working voltage is not an error in the tables. It is the price of making charge flow at a useful rate, and electrochemists call it overpotential or polarization .

Core explanation

Equilibrium versus operation. The Nernst equation gives the equilibrium potential, E eq, of an electrode when no net current flows. At that point the forward and reverse electron-transfer rates are equal and no energy is being dissipated. As soon as a net current is drawn, the electrode potential moves away from E eq. The shift is the overpotential:

η = E − E eq

By convention, η is positive for an anode (oxidation needs the potential pushed more positive) and negative for a cathode (reduction needs it pushed more negative).

Three sources of polarization.

1. Activation overpotential, η act. Electron transfer across the electrode–solution interface has an activation barrier. Shifting the potential lowers the barrier in one direction and raises it in the other, so a larger η drives a larger net current. Sluggish reactions, such as oxygen evolution or oxygen reduction, need hundreds of millivolts of activation overpotential even on good catalysts.

2. Concentration (mass-transport) overpotential, η conc. Current consumes reactant at the surface. If diffusion, migration and convection cannot replace it quickly enough, the surface concentration falls below the bulk value. The Nernst equation, applied to surface concentrations, then gives a shifted potential. Near the limiting current this term grows very steeply.

3. Ohmic drop, IR. The electrolyte, separator and electrical contacts have resistance R. Driving current I through them costs a voltage IR, dissipated as heat. Strictly this is not an electrode overpotential, but it adds to the cell's polarization and grows linearly with current.

Effect on a whole cell. For an electrolytic cell the applied voltage must exceed the reversible voltage:

V applied = E rev + η a + η c + IR

For a galvanic cell the delivered voltage falls below the reversible value:

V delivered = E rev − η a − η c − IR

In both cases the losses grow with current, so running faster always costs efficiency. Plotting cell voltage against current density gives a polarization curve : a steep initial drop (activation), a roughly linear middle region (ohmic) and a sharp fall at high current (mass transport).

Energy consequence. The electrical work lost per mole of electrons is F multiplied by the total overpotential, and it appears as heat. The voltage efficiency of an electrolyser is E rev ÷ V applied; for a galvanic cell it is V delivered ÷ E rev.

Formulae

η = E − E eq. Electrolyser: V = E rev + Σ η + IR. Galvanic cell: V = E rev − Σ η − IR. Voltage efficiency (electrolyser) = E rev ÷ V. Heat released from losses per mole of electrons ≈ F × (Σ η + IR).

Step-by-step reasoning

To estimate the voltage needed to run an electrolysis cell at a chosen current:

1. Find E rev from standard potentials and the Nernst equation for actual conditions. 2. Estimate the anodic and cathodic activation overpotentials at that current density from kinetic data. 3. Check whether the current approaches the limiting current; if so, add a concentration term. 4. Multiply current by total cell resistance to obtain the IR drop. 5. Add all terms to E rev to obtain the operating voltage.

Visual explanation

Picture a polarization curve for a fuel cell: voltage on the vertical axis, current density on the horizontal. It starts near the reversible voltage at zero current, drops sharply over the first small currents (activation loss), slopes gently downwards in a long straight section (ohmic loss), then plunges near the right-hand edge (mass-transport loss). The shaded area between this curve and a horizontal line at E rev represents wasted energy.

Real-world analogy

Think of pushing water uphill through a pipe. The height of the hill is the reversible voltage, fixed by thermodynamics. Friction in the pipe, a stiff valve at the entrance and a narrow section that starves the flow all need extra pressure. The faster you want the water to move, the more extra pressure these losses demand.

Real-world example

Industrial chlor-alkali cells produce chlorine and sodium hydroxide from brine. The reversible voltage is about 2.2 V, but cells operate at roughly 3 V or more. Engineers use dimensionally stable anodes coated with ruthenium and iridium oxides to reduce the chlorine overpotential, narrow electrode gaps to cut the IR drop, and circulation to limit concentration polarization.

Why?

Why must there be a loss at all? At equilibrium the forward and reverse processes cancel exactly, so the net rate is zero. To obtain a finite net rate the system must be driven away from equilibrium, and any process driven at a finite rate is thermodynamically irreversible, generating entropy. Overpotential is the electrical measure of that driving force.

Common misconception

"Overpotential means the tabulated standard potentials are wrong." Standard potentials describe equilibrium and remain correct. Overpotential is an additional, current-dependent quantity describing kinetics and transport, which thermodynamic tables do not include.

Worked example

Question: A water electrolyser has E rev = 1.23 V. At 1.0 A cm⁻² the anode overpotential is 0.35 V, the cathode overpotential is −0.10 V and the cell resistance is 0.15 Ω cm². Find the cell voltage and the voltage efficiency.

Reasoning: IR drop = 1.0 A cm⁻² × 0.15 Ω cm² = 0.15 V. V = 1.23 + 0.35 + 0.10 + 0.15 = 1.83 V. Efficiency = 1.23 ÷ 1.83 = 0.67.

Answer: About 1.83 V, giving a voltage efficiency of about 67%.

Quick check

1. Which contribution to polarization increases linearly with current and is caused by electrolyte resistance? Answer: The ohmic or IR drop, equal to current multiplied by the cell resistance.

Exam focus

Be able to define η = E − E eq, name the three sources of polarization and write the voltage balance for both galvanic and electrolytic cells with correct signs. Examiners often ask you to sketch a polarization curve and label its three regions.

Advanced insight

The ohmic term can be measured separately by current interruption: when current is switched off, the IR drop vanishes within microseconds, whereas activation and concentration overpotentials decay more slowly as the double layer discharges and concentrations relax. Impedance spectroscopy separates the contributions by their different characteristic frequencies. This is why careful kinetic studies report "IR-corrected" potentials.

Summary

Overpotential is the departure of an electrode potential from its equilibrium value when current flows. Activation, concentration and ohmic losses all grow with current. They raise the voltage an electrolyser needs and lower the voltage a galvanic cell supplies, and the lost energy appears as heat. Good catalysts, efficient mass transport and low-resistance cell designs reduce polarization.

Practice questions

1. Define overpotential and state its sign convention for anodes and cathodes. Answer: η = E − E eq; it is positive for an anode and negative for a cathode. 2. A galvanic cell with E rev = 1.10 V has total electrode overpotentials of 0.12 V and an IR drop of 0.08 V. What voltage does it deliver? Answer: 1.10 − 0.12 − 0.08 = 0.90 V. 3. Which loss dominates at very high current density, and why? Answer: Concentration overpotential, because reactant cannot be supplied to the surface fast enough and the surface concentration approaches zero. 4. Suggest two design changes that reduce the IR drop in an electrolysis cell. Answer: Reduce the electrode gap and use a more conductive electrolyte or thinner, more conductive separator.