Electrolysis Voltage: Standard & Non-Standard Conditions

Introduction

In our last lesson, we mastered the Nernst equation, which allows us to calculate cell potentials under non-standard conditions. We specifically saw how the potentials for the Hydrogen Evolution Reaction (HER) and Oxygen Evolution Reaction (OER) shift with pH, but that the overall potential difference of 1.23 V remains thermodynamically constant.

Today, we will synthesize all the concepts we've covered—Gibbs free energy, standard cell potentials, and the Nernst equation—to address a crucial question for your work: What is the absolute minimum voltage required to make electrolysis happen?

The learning outcome for this lesson is to calculate the thermodynamic minimum voltage required for water electrolysis under standard and non-standard conditions. This value, often called the reversible potential, represents the theoretical energy barrier you must overcome before any electrolysis can begin. It's the baseline upon which all other energy losses, like the overpotentials you work to minimize with your catalysts, are added.

1. The Minimum Voltage: Standard Conditions

The decomposition of water is a non-spontaneous process, meaning it requires energy input. The minimum electrical energy required is equal to the Gibbs free energy change ($\Delta G$) of the reaction.

Under standard conditions (25 °C, 1 atm, pH 0 for the standard hydrogen electrode scale), the reaction is:
$H_{2}O(l)\to H_2(g)+\frac{1}{2}{O}_2(g)\Delta G°=+237.2\text{ kJ/mol}$

This positive $\Delta G°$ confirms the reaction is non-spontaneous. We can convert this energy value into a voltage using the familiar equation $\Delta G°=-nFE{°}_{\text{cell}}$.

Let's watch a short video that clearly illustrates how the standard half-cell potentials sum up to this minimum voltage.

As the video showed, for water electrolysis, $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ is negative. For example, in an acidic solution:

• Anode (OER): $2H_{2}O\to O_2+4H^{+}+4e^{-}$ , $E{°}_{\text{anode}}=+1.23$ V
• Cathode (HER): $2H^{+}+2e^{-}\to H_2$ , $E{°}_{\text{cathode}}=0.00$ V
• Overall $E{°}_{\text{cell}}$: $0.00\text{ V}-1.23\text{ V}=-1.23\text{ V}$

The negative sign indicates a non-spontaneous reaction (an electrolytic cell). The minimum applied voltage must be equal to or greater than the magnitude of this value, so at least 1.23 V.

Reversible vs. Thermoneutral Voltage

It's important to distinguish between two key thermodynamic voltages. The following reading will clarify this.

To summarize:

• Reversible Voltage (1.23 V): The absolute minimum voltage needed to start splitting water, corresponding to the electrical energy input ($\Delta G°$). The remaining energy ($T\Delta S°$) must be supplied as heat from the surroundings, causing the cell to cool down.
• Thermoneutral Voltage (1.48 V): The voltage at which the electricity supplied ($\Delta H°$) covers both the Gibbs energy and the heat requirement. At this voltage, the cell operates isothermally without heat exchange with the surroundings.

In practice, all electrolyzers operate above 1.48 V due to kinetic overpotentials, and thus they generate waste heat that must be removed. However, 1.23 V remains the fundamental thermodynamic floor.

2. Minimum Voltage under Non-Standard Conditions

Now, let's see how this minimum voltage changes when we move away from standard conditions of temperature, pressure, and pH.

Effect of pH

As we touched upon in the last lesson, the thermodynamic driving force for the overall water splitting reaction is independent of pH. While the individual electrode potentials shift, the difference between them remains constant.

To reinforce this, the video below provides a quick summary.

This confirms that whether you are performing electrolysis in acid (PEM), neutral, or alkaline (AWE) conditions like in your research, the thermodynamic starting line is the same: 1.23 V. The choice of pH is therefore dictated by kinetics and catalyst stability, not overall thermodynamics.

Effect of Temperature and Pressure

Temperature and pressure, however, do change the minimum required voltage. Given your background in chemical engineering, you'll appreciate that these effects can be precisely calculated from fundamental thermodynamic principles.

The following reading explains these dependencies in detail.

Key findings from the reading:

• Temperature: Increasing the operating temperature makes water splitting easier thermodynamically. The required voltage drops from 1.23 V at 25 °C to about 1.18 V at 100 °C. This is a key motivation for developing high-temperature electrolyzers.
• Pressure: Increasing the operating pressure makes water splitting harder thermodynamically, as you are compressing the gaseous products. The required voltage increases logarithmically with pressure.

3. Calculation for Non-Standard Conditions

Let's apply these principles to a practical problem relevant to industrial electrolysis.

Problem: Calculate the minimum thermodynamic voltage required for water electrolysis operating at 80 °C and a pressure of 30 bar.

We can solve this in two steps using the equations from the resource you just read.

Step 1: Calculate the reversible voltage at 80 °C (353.15 K) and 1 bar.
We use Equation 2.29 from the text:
$E(T)=1.5184-(1.5421\times 10^{-3})T+(9.523\times 10^{-5})T\ln (T)+(9.84\times 10^{-8})T^2$
Plugging in $T=353.15$ K:
$E(353.15)=1.5184-0.5446+(0.0336\times 5.867)+0.0123$
$E(353.15)\approx 1.183\text{ V}$
So, just by increasing the temperature to 80 °C, the thermodynamic barrier has already dropped from 1.23 V to 1.18 V.

Step 2: Add the voltage increase due to pressure.
We use Equation 2.37 from the text, which is derived directly from the Nernst equation:
$E(T,P)=E(T,1)+\frac{3RT}{4F}\ln (P)$
The second term is the pressure correction:
$\Delta E_P=\frac{3\times (8.314\text{ J/mol⋅K})\times (353.15\text{ K})}{4\times (96485\text{ C/mol})}\ln (30)$
$\Delta E_P=(0.0228\text{ V})\times (3.401)\approx 0.078\text{ V}$
This is the extra voltage required to compress the H₂ and O₂ products to 30 bar.

Step 3: Combine the effects.
$E_{\text{req}}(80°C,30\text{ bar})=E(80°C,1\text{ bar})+\Delta E_P$
$E_{\text{req}}=1.183\text{ V}+0.078\text{ V}=1.261\text{ V}$

So, under these more realistic operating conditions, the minimum required voltage is approximately 1.26 V.

Conclusion

In this lesson, we consolidated our understanding of electrochemical thermodynamics to define and calculate the minimum voltage for water electrolysis. This theoretical value is the foundation upon which all practical performance is built.

Key Takeaways:

• The minimum voltage required for electrolysis is determined by the Gibbs free energy change ($\Delta G$) of the reaction.
• Under standard conditions, this reversible voltage is 1.23 V.
• The reversible voltage is independent of pH, though the individual electrode potentials shift.
• The reversible voltage decreases with increasing temperature, making electrolysis thermodynamically more favorable.
• The reversible voltage increases with increasing pressure, reflecting the energy cost of compressing the gas products.
• By combining these effects, we can calculate the thermodynamic minimum voltage for any set of operating conditions.

Preview of the Next Lesson:

We've now seen the Pourbaix diagram for water and how it defines the window of electrochemical stability. In our next and final lesson of this module, we will expand on this concept to interpret Pourbaix diagrams for other materials. You will learn how to use these diagrams to predict the stability of metals and their oxides—such as the nickel and iron in your catalysts—at different potentials and pH values. This is a critical tool for understanding and predicting catalyst stability and corrosion during electrolysis.