Galvanic (Voltaic) Cells & Cell Potential

You’ll be able to

Balance redox reactions by combining oxidation and reduction half-reactions
Identify the anode, cathode, and role of the salt bridge in a galvanic cell
Compute E°cell from standard reduction potentials and link its sign to ΔG°

Redox is electron bookkeeping

Every redox reaction is two half-reactions happening at once. Oxidation is loss of electrons (oxidation number goes up); reduction is gain of electrons (oxidation number goes down). The mnemonic OIL RIG — Oxidation Is Loss, Reduction Is Gain — keeps them straight. To balance, split the reaction, balance mass and charge in each half, scale so the electrons lost equal the electrons gained, and add them back together.

A galvanic cell puts the electrons to work

A galvanic (voltaic) cell separates the two half-reactions into two compartments so the electrons must travel through an external wire — that flow is usable current. Oxidation happens at the anode; reduction happens at the cathode. (Anode/oxidation and cathode/reduction both pair vowel-with-vowel and consonant-with-consonant.) Electrons always flow from anode to cathode through the wire. A galvanic cell runs a spontaneous reaction, so it produces a positive voltage.

The salt bridge and cell notation

As oxidation pumps positive ions into the anode solution and reduction removes them from the cathode solution, charge would build up and stop the reaction. The salt bridge lets spectator ions migrate to keep both sides neutral — anions drift toward the anode, cations toward the cathode. Chemists shorthand the whole cell as anode | anode solution || cathode solution | cathode, where "||" is the salt bridge. Example: Zn | Zn²⁺ || Cu²⁺ | Cu.

Standard cell potential

E°cell = E°cathode − E°anode

Both values are standard *reduction* potentials read straight from the table. Do NOT multiply a potential by the number of electrons or by a balancing coefficient — potential is an intensive property.

Free energy from cell potential

ΔG° = −nFE°

n = moles of electrons transferred, F = 96 485 C·mol⁻¹ (Faraday constant). E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous; E° < 0 ⇒ ΔG° > 0 ⇒ nonspontaneous.

Worked example

Build the Zn/Cu (Daniell) cell from Zn²⁺ + 2e⁻ → Zn (E° = −0.76 V) and Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V). Find E°cell, identify the electrodes, and state ΔG°.

1.The half-reaction with the higher (more positive) reduction potential is reduced — it is the cathode. Cu²⁺/Cu at +0.34 V beats Zn²⁺/Zn at −0.76 V, so copper is the cathode and zinc is the anode (oxidation).
2.Apply E°cell = E°cathode − E°anode using the reduction-potential values as written: E°cell = (+0.34 V) − (−0.76 V).
3.E°cell = +0.34 + 0.76 = +1.10 V. Note we did NOT scale either potential, even though electrons are transferred.
4.Both half-reactions involve 2 electrons, so n = 2 with no scaling needed.
5.ΔG° = −nFE° = −(2)(96 485 C·mol⁻¹)(1.10 V) = −212 267 J ≈ −212 kJ·mol⁻¹.

Answer: E°cell = +1.10 V; anode = Zn, cathode = Cu; ΔG° ≈ −212 kJ·mol⁻¹ (negative, so spontaneous).

Checkpoint

A galvanic cell is built from Ag⁺ + e⁻ → Ag (E° = +0.80 V) and Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V). What is E°cell?

Checkpoint

In the cell notation Fe | Fe²⁺ || Cu²⁺ | Cu, where does oxidation occur and is the cell spontaneous? (Fe²⁺/Fe = −0.44 V, Cu²⁺/Cu = +0.34 V)

Tip

Quick sanity check: in any spontaneous galvanic cell E°cell comes out positive. If you get a negative number, you almost certainly swapped anode and cathode — the electrode with the higher reduction potential is always the cathode.

On the exam

The AP exam frequently tests the "do not multiply potentials" trap. When you scale a half-reaction to balance electrons, you multiply the atoms and the electrons — but never the E° value. Potential is energy per charge, an intensive property that is independent of how much reaction you run.

Answer the 2 checkpoints as you read.