Cell potential tells you, in a single number, whether a redox reaction will run on its own and how hard it pushes electrons through a wire. A positive E°cell and you’ve got a battery. Negative and you need to plug in a power supply. The whole edifice of electrochemistry — galvanic cells, electrolysis, corrosion, biological electron transport — hangs off this measurement, which makes the calculation worth getting right the first time.

The formula that does the bookkeeping for you

E°cell = E°cathode − E°anode

That subtraction is doing real work. It looks innocent, but it’s the reason you should never manually flip the sign of an oxidation half-reaction. Standard reduction potential tables give every half-reaction as a reduction. You leave them as reductions, identify which one acts as cathode (the more positive E°) and which as anode (less positive), and the subtraction handles the sign reversal automatically.

The most common screw-up in undergraduate electrochemistry is flipping signs by hand and then subtracting — you double-negate the anode and end up with the wrong magnitude. Resist the urge. Look up two reductions, plug into E°cathode − E°anode, done.

A worked example: the Daniell cell

Zinc and copper electrodes in their respective sulfate solutions, the canonical galvanic cell that opens every electrochemistry chapter:

Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V
Zn²⁺ + 2e⁻ → Zn, E° = −0.76 V

Copper has the more positive reduction potential, so copper gets reduced (cathode). Zinc gets oxidized (anode). Plug in:

E°cell = (+0.34) − (−0.76) = +1.10 V

Positive, so the reaction runs spontaneously, and you get a 1.10 V battery. This matches every reference for the Daniell cell — the cell potential of a fresh AA battery is in the same neighborhood, which is why this teaching example matters: the math you just did predicts the voltage of a real device.

Predicting whether a reaction will go

Will tin metal reduce Fe³⁺ under standard conditions? Look up:

Fe³⁺ + 3e⁻ → Fe, E° = −0.04 V
Sn²⁺ + 2e⁻ → Sn, E° = −0.14 V

Iron’s reduction potential is more positive (less negative), so iron gets reduced and tin gets oxidized:

E°cell = (−0.04) − (−0.14) = +0.10 V

Small but positive — the reaction is spontaneous, but the driving force is weak, so it’ll be slow and easily reversed by changes in concentration.

Connecting to Gibbs free energy

The whole point of cell potential is that it’s just ΔG° in different units:

ΔG° = −nFE°cell

with F = 96,485 C/mol and n = moles of electrons transferred per mole of reaction. For the Daniell cell (n = 2, E°cell = 1.10 V):

ΔG° = −(2)(96,485)(1.10) = −212,267 J ≈ −212.3 kJ/mol

Strongly negative, which is the same statement as “strongly positive E°cell” — both mean the reaction releases a lot of usable energy.

When concentrations aren’t 1 M: the Nernst equation

Real solutions are rarely at 1.0 M, and a battery’s voltage drifts as it discharges. The Nernst equation corrects for that:

E_cell = E°cell − (RT/nF) × ln(Q)

At 25 °C, the ugly constants collapse to:

E_cell = E°cell − (0.0592/n) × log₁₀(Q)

For a Daniell cell with [Zn²⁺] = 0.10 M and [Cu²⁺] = 2.0 M:

Q = [Zn²⁺] / [Cu²⁺] = 0.050

E_cell = 1.10 − (0.0592/2) × log(0.050) = 1.10 − (0.0296)(−1.301) = 1.14 V

Slightly above standard because low product concentration and high reactant concentration both push the reaction forward. As the battery discharges, [Zn²⁺] climbs and [Cu²⁺] falls, Q rises, and E_cell drops — eventually to zero, which is when the battery is “dead” (system at equilibrium, not literally out of material).

What people get wrong

Three recurring mistakes:

Flipping signs before subtracting. Don’t. Use the reduction potentials exactly as tabulated and let E°cathode − E°anode handle the algebra.

Multiplying E° by stoichiometric coefficients. Cell potential is intensive — it doesn’t scale with how much reaction you write. If you double a half-reaction to balance electrons, you double n in ΔG° = −nFE°, but E° itself stays put.

Mixing up Q in the Nernst equation. Q for the overall cell reaction goes products over reactants, just like in any other equilibrium expression. Don’t try to write Q for individual half-reactions — that’s a different (and much messier) calculation.

Practice

Run these against the Electrochemistry Calculator:

E°cell for Al (E° = −1.66 V) paired with Ag (E° = +0.80 V).
Is 2Ag + Cu²⁺ → 2Ag⁺ + Cu spontaneous under standard conditions?
ΔG° for a cell with E°cell = +0.46 V, n = 2.
Zn | Zn²⁺ (0.010 M) || Cu²⁺ (1.0 M) | Cu — find E_cell at 25 °C.
A cell has E°cell = 0.50 V, n = 2. At what [products]/[reactants] ratio does E_cell hit zero?

How to Calculate Cell Potential