Electrode potentials look like ordinary numbers. That makes it tempting to manipulate them like ordinary numbers—adding, subtracting, or reversing them whenever a reaction is combined.
That intuition is wrong. The key idea is simple: Electrode potentials are not additive. Gibbs free energies are. Once you understand why, a whole class of electrochemistry problems becomes much easier.
The Problem
Consider these two oxidation half-reactions:
\(A\rightarrow A^{+}+e^{-},\quad E^{\circ }=y\)
\(A\rightarrow A^{2+}+2e^{-},\quad E^{\circ }=x\)
Question: Find the standard electrode potential for:
\(A^{+}\rightarrow A^{2+}+e^{-}\)
At first glance, you might try:
\(E^{\circ }=x-y\quad \text{or}\quad E^{\circ }=y-x\)
Both approaches miss the key issue. So what should we do instead?
What Is Actually Additive?
The standard electrode potential (\(E^{\circ }\)) is not the quantity that follows Hess’s law. The thermodynamic quantity that is additive is the standard Gibbs free-energy change: \(\Delta G^\circ\).
The relationship between Gibbs free energy and electrode potential is given by \(\Delta G^\circ = -nFE^\circ\). Because \(\Delta G^\circ\) follows Hess’s law and is additive, calculations must combine free energies rather than potentials directly.
By converting half-reactions into their corresponding \(\Delta G^\circ\) expressions—accounting for the number of electrons \(n\) transferred in each step—you can correctly derive unknown potentials using the standard pathway: Reaction \(\rightarrow \) Gibbs energy \(\rightarrow \) combine \(\rightarrow \) electrode potential.
For the example above, converting the processes yields \(\Delta G_1^\circ = -Fy\) and \(\Delta G_2^\circ = -2Fx\). Combining the equations properly leads to the derived potential \(E_{\text{desired}}^\circ = 2x – y\), demonstrating why simple subtraction fails. This same step-by-step logic resolves complex problems, such as deriving unknown reduction potentials for manganese species from standard values.
Ultimately, the core principle remains: Combine \(\Delta G^\circ\), not \(E^{\circ }\). The energy is additive, but the voltage is not.