AP Chemistry Galvanic Cells and Cell Potential

Anode, cathode, E°cell, and what the sign means, three full worked examples.

A galvanic cell problem usually asks for the same few things in the same order: which half-reaction is the oxidation and which is the reduction, what the standard cell potential is, and whether the reaction is thermodynamically favored. Here is that sequence, worked on real half-cells.

The Galvanic Cell Rules and the E°cell Formula

In every electrochemical cell, oxidation happens at the anode and reduction happens at the cathode. In a galvanic cell, electrons travel through the external wire from the anode to the cathode, while a salt bridge lets ions move to keep each half-cell electrically balanced. Given standard reduction potentials, the cell potential is E°_cell = E°_cathode − E°_anode, with both values taken exactly as listed for the reduction half-reaction. Identifying which species is oxidized and which is reduced works the same way as in oxidation number problems: the species whose oxidation number increases is oxidized.

Worked Example 1: Finding E°cell for a Zinc-Copper Cell

A cell pairs Zn²⁺/Zn (E° = −0.76 V) with Cu²⁺/Cu (E° = +0.34 V). Identify the anode and cathode, write the overall reaction, and find E°_cell.

The more positive reduction potential belongs to copper, so Cu²⁺ is reduced at the cathode and zinc is oxidized at the anode.
Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

E°_cell = E°_cathode − E°_anode = (+0.34) − (−0.76) = +1.10 V

Electrons flow from the zinc electrode to the copper electrode through the wire, the zinc electrode loses mass as it dissolves, and the copper electrode gains mass as copper plates onto it. The half-reaction balancing here is the same skill covered in the half-reaction method.

Worked Example 2: Choosing the Anode and Cathode for a Silver-Copper Cell

A cell pairs Ag⁺/Ag (E° = +0.80 V) with Cu²⁺/Cu (E° = +0.34 V). Find E°_cell and ΔG°.

Silver has the more positive reduction potential, so silver is reduced (cathode) and copper is oxidized (anode):
Anode: Cu(s) → Cu²⁺(aq) + 2e⁻
Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)
E°_cell = (+0.80) − (+0.34) = +0.46 V

The silver half-reaction was doubled to balance the two electrons from copper, but its E° stays +0.80 V, cell potential is not multiplied by a coefficient. The doubling only matters for n = 2 electrons transferred:
ΔG° = −nFE° = −(2)(96,485 C/mol)(0.46 V) = −88,800 J/mol = −88.8 kJ/mol

Worked Example 3: Recognizing a Reaction That Is Not Thermodynamically Favored

Does Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s) run spontaneously as a galvanic cell?

As written, zinc is reduced (cathode) and copper is oxidized (anode):
E°_cell = E°_cathode − E°_anode = (−0.76) − (+0.34) = −1.10 V

The negative value (exactly the reverse of Example 1) means the reaction is not thermodynamically favored as written. It would only occur if an external power source forced it, which is an electrolytic cell. A spontaneous galvanic cell always has E°_cell > 0.

Common Galvanic Cell Mistakes

Flipping the anode and cathode. Oxidation is always the anode and reduction is always the cathode, even in an electrolytic cell.
Reversing the sign of the anode's potential before subtracting. Use E°_cathode − E°_anode with both values as listed; flipping the sign and also subtracting double-reverses it.
Multiplying E° by the number of electrons or by a balancing coefficient. E° is not scaled; only n changes when converting to ΔG°.
Calling a negative E°_cell a galvanic cell. A galvanic cell needs a positive E°_cell; a negative one describes an unfavored reaction.
Spending time labeling electrodes positive or negative. That labeling is explicitly excluded; knowing which electrode is the anode and which is the cathode is what's asked.

This same favored-or-not reasoning applies across the rest of AP Chemistry too -- for every other free tool and guide on this site, start from the AP Chem Score Calculator.

Frequently Asked Questions

What is a galvanic cell?

A galvanic (voltaic) cell uses a thermodynamically favored redox reaction to generate electrical energy. The oxidation and reduction half-reactions happen in separate half-cells, connected by an external wire for electrons and a salt bridge for ions.

Which electrode is the anode and which is the cathode?

Oxidation always happens at the anode and reduction always happens at the cathode, in every kind of cell. In a galvanic cell, electrons flow through the external wire from the anode to the cathode.

How do you calculate E°cell?

E°cell = E°cathode − E°anode, using the standard reduction potential of each half-reaction exactly as listed. The half-reaction with the more positive reduction potential is the one that proceeds as a reduction (the cathode).

Do you multiply E° by the coefficient when you double a half-reaction?

No. Cell potential is an intensive quantity, like temperature or concentration, so scaling a half-reaction to balance electrons does not change its E° value. The number of electrons transferred (n) only matters when you convert E° to ΔG° with ΔG° = −nFE°.

What does the sign of E°cell tell you?

A positive E°cell means the reaction as written is thermodynamically favored (negative ΔG°) and can run as a galvanic cell. A negative E°cell means it is not favored as written; running it would require an external power source, which is an electrolytic cell.

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