# Galvanic (voltaic) and electrolytic cells

> AP Chemistry · CED Unit 9: Applications of Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u9-galvanic-and-electrolytic-cells/

This subtopic covers core properties of galvanic and electrolytic electrochemical cells, including electrode identification, standard cell potential calculation, spontaneity relationships, and quantitative electrolysis with Faraday's law, frequently tested on both MCQ and FRQ.

**Prerequisites:** Balancing oxidation-reduction half-reactions; Gibbs free energy and spontaneity rules; Stoichiometric mole calculations

## Learning objectives

- Identify anode and cathode in both galvanic and electrolytic cells
- Calculate standard cell potential from tabulated reduction potentials
- Relate standard cell potential to ΔG° and reaction spontaneity
- Compare key properties of galvanic and electrolytic cells
- Solve quantitative electrolysis problems using Faraday's law

## Core Concepts and Cell Notation

Both galvanic (voltaic) and electrolytic cells are electrochemical cells that separate oxidation and reduction half-reactions to force electron flow through an external wire, rather than direct electron transfer between reactants.

**Galvanic (Voltaic) Cell** — Electrochemical cell that uses a spontaneous redox reaction to convert stored chemical energy into usable electrical energy

*Example:* Disposable batteries, lead-acid car batteries

**Electrolytic Cell** — Electrochemical cell that requires an external source of electrical energy to drive a non-spontaneous redox reaction

*Example:* Electroplating, metal refining, chlorine gas production

Standard notation for all electrochemical cells follows a universal convention: anode (oxidation half-reaction) is placed on the left, cathode (reduction half-reaction) on the right. A single vertical line | denotes a phase boundary, and a double vertical line || denotes a salt bridge or porous disk that separates half-cells and maintains charge neutrality.

> **Memory Hook**
>
> A O C R: **A**node = **O**xidation, **C**athode = **R**eduction. This rule never changes, for any cell type.

**Worked example:** Identify the anode, cathode, direction of electron flow, and state whether the cell below is galvanic or electrolytic if $E^\circ_{\text{cell}} = +1.10\ \text{V}$: $\text{Cu}(s)\ |\ \text{Cu}^{2+}(aq, 1.0\ \text{M})\ ||\ \text{Zn}^{2+}(aq, 1.0\ \text{M})\ |\ \text{Zn}(s)$

1. By standard notation convention, the left half-cell is the anode, so $\text{Cu}(s)$ is the anode, oxidized to $\text{Cu}^{2+}$. The right half-cell is the cathode, so $\text{Zn}^{2+}$ is reduced to $\text{Zn}(s)$, the cathode electrode.
2. Electron flow always goes from anode to cathode through the external wire, so electrons flow from the Cu anode to the Zn cathode.
3. A positive $E^\circ_{\text{cell}}$ means the reaction is spontaneous. Only galvanic cells have spontaneous net reactions, so this is a galvanic cell.
4. Confirm charge: for a galvanic cell, anode is negative and cathode is positive, which matches electron flow from negative to positive.

> **Exam tip:** If asked for the direction of anion flow in the salt bridge, remember anions always go to the anode, regardless of cell type — this is a common low-stakes MCQ trap.

## Electrode Charge and Identification

Only the charge of electrodes reverses between galvanic and electrolytic cells — the reaction at each electrode never changes: oxidation is always at the anode, reduction always at the cathode.

For galvanic cells: the anode produces electrons via oxidation, so it carries a negative charge. Electrons flow from the negative anode through the external wire to the positive cathode, where reduction consumes electrons. The salt bridge maintains charge neutrality: anions flow to the anode to balance positive charge buildup from new cations, and cations flow to the cathode to balance negative charge buildup from consumed cations.

For electrolytic cells: an external battery drives the non-spontaneous reaction, pulling electrons away from the anode and pushing electrons onto the cathode. This gives the anode a positive charge and the cathode a negative charge, but oxidation still occurs at the anode and reduction still at the cathode. Ion flow follows the same rule as galvanic cells: anions flow to the anode, cations flow to the cathode, regardless of cell type.

**Check your understanding**

Which of the following correctly describes an electrolytic cell used to purify copper, where impure copper is the anode and pure copper is deposited on the cathode?

1. Select the correct statement

   - Oxidation occurs at the pure copper cathode, which has a negative charge
   - Oxidation occurs at the impure copper anode, which has a positive charge
   - Reduction occurs at the pure copper cathode, which has a positive charge
   - Reduction occurs at the impure copper anode, which has a negative charge

   *Answer:* Oxidation occurs at the impure copper anode, which has a positive charge

   *Why:* Oxidation is always at the anode, and for electrolytic cells, the anode has a positive charge. Only option B matches both rules.

## Standard Cell Potential and Spontaneity

Standard reduction potentials ($E^\circ_{\text{red}}$) are tabulated for all common half-reactions, measured relative to the standard hydrogen electrode (SHE), which is assigned $E^\circ_{\text{red}} = 0\ \text{V}$.

$$E^\circ_{\text{cell}} = E^\circ_{\text{red (cathode)}} - E^\circ_{\text{red (anode)}}$$

An equivalent form is $E^\circ_{\text{cell}} = E^\circ_{\text{red (cathode)}} + E^\circ_{\text{ox (anode)}}$, where $E^\circ_{\text{ox (anode)}} = -E^\circ_{\text{red (anode)}}$, so both formulas give the same result. A key relationship between cell potential and Gibbs free energy is:

$$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$

where $n$ = moles of electrons transferred in the balanced overall reaction, and $F$ = Faraday's constant ($96500\ \text{C/mol }e^-$). From this formula, we get the spontaneity rule: if $E^\circ_{\text{cell}}$ is positive, $\Delta G^\circ$ is negative, and the reaction is spontaneous (galvanic cell). If $E^\circ_{\text{cell}}$ is negative, $\Delta G^\circ$ is positive, and the reaction is non-spontaneous (requires external voltage, so electrolytic cell).

> **note**
>
> $E^\circ$ is an intensive property, meaning it does not depend on the amount of reactant. You never multiply $E^\circ$ values by coefficients when balancing electrons.

**Worked example:** Given $E^\circ_{\text{red (Fe}^{2+}/\text{Fe)}} = -0.44\ \text{V}$ and $E^\circ_{\text{red (Br}_2/\text{Br}^-)} = +1.07\ \text{V}$, write the balanced spontaneous reaction, calculate $E^\circ_{\text{cell}}$, and find $\Delta G^\circ$.

1. For a spontaneous reaction, $E^\circ_{\text{cell}}$ must be positive. The half-reaction with the higher (more positive) $E^\circ_{\text{red}}$ is the cathode (reduction), so $\text{Br}_2$ is reduced, and $\text{Fe}$ is oxidized (anode).
2. Write half-reactions:
3. $$\text{Reduction: } \text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq) \\ \text{Oxidation: } \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-$$
4. Electrons are already balanced, so $n = 2$; we do not need to scale $E^\circ$ values. Calculate $E^\circ_{\text{cell}}$:
5. $$E^\circ_{\text{cell}} = 1.07\ \text{V} - (-0.44\ \text{V}) = 1.51\ \text{V}$$
6. Calculate $\Delta G^\circ$:
7. $$\Delta G^\circ = -nFE^\circ_{\text{cell}} = -(2\ \text{mol }e^-)(96500\ \text{C/mol }e^-)(1.51\ \text{V}) = -291430\ \text{J} = -291\ \text{kJ}$$
8. The negative $\Delta G^\circ$ confirms the reaction is spontaneous, as expected.

> **Exam tip:** If you struggle with sign errors when calculating $E^\circ_{\text{cell}}$, flip the sign of the anode's reduction potential to get oxidation potential, then add to the cathode's reduction potential. This eliminates subtracting negative numbers.

## Faraday's Law of Quantitative Electrolysis

Electrolytic cells are used to drive non-spontaneous redox reactions for industrial applications. We can relate the amount of product produced to the current passed through the cell and time of electrolysis using Faraday's law of electrolysis.

The core relationship is that total charge ($Q$, measured in coulombs, C) equals current ($I$, measured in amperes, A, where $1\ \text{A} = 1\ \text{C/s}$) multiplied by time ($t$, measured in seconds):

$$Q = I \times t$$

Total moles of electrons transferred is then $n_{e^-} = \frac{Q}{F} = \frac{I t}{F}$. We use the stoichiometry of the reduction half-reaction to relate moles of electrons to moles of product, then convert moles to mass or volume as needed. For example, to produce 1 mole of Al from $\text{Al}^{3+}$, you need 3 moles of electrons.

**Worked example:** How many grams of copper metal are plated from a $\text{Cu}^{2+}$ solution when a constant current of 2.10 A is passed through the cell for 1.50 hours? Molar mass of Cu = 63.55 g/mol, $F = 96500\ \text{C/mol }e^-$.

1. Convert time to seconds (required for current units):
2. $$1.50\ \text{h} \times 3600\ \text{s/h} = 5400\ \text{s}$$
3. Calculate total charge:
4. $$Q = I t = (2.10\ \text{A})(5400\ \text{s}) = 11340\ \text{C}$$
5. Calculate moles of electrons:
6. $$n_{e^-} = \frac{11340\ \text{C}}{96500\ \text{C/mol }e^-} = 0.1175\ \text{mol }e^-$$
7. The reduction half-reaction is $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$, so moles of Cu are:
8. $$0.1175\ \text{mol }e^- \times \frac{1\ \text{mol Cu}}{2\ \text{mol }e^-} = 0.05875\ \text{mol Cu}$$
9. Convert to mass:
10. $$0.05875\ \text{mol} \times 63.55\ \text{g/mol} = 3.73\ \text{g Cu}$$

> **Exam tip:** Always convert time to seconds before calculating charge. AP exam questions frequently give time in minutes or hours to test unit conversion recall.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Claiming the anode is always negatively charged, regardless of cell type
  - Why it fails: Students memorize charge only for galvanic cells and forget it reverses for electrolytic cells, where the external battery pulls electrons from the anode to make it positive
  - Correct: Remember: oxidation at the anode always, charge depends on cell type: galvanic = anode negative, electrolytic = anode positive
- **Wrong:** Multiplying standard reduction potentials by reaction coefficients when balancing electron transfer
  - Why it fails: Students confuse intensive properties ($E^\circ$) with extensive properties ($\Delta G$, enthalpy) that do scale with reaction size
  - Correct: Only adjust $\Delta G$ when scaling half-reactions; leave $E^\circ$ values unchanged regardless of coefficients
- **Wrong:** Stating that electrons flow through the salt bridge to complete the circuit
  - Why it fails: Students mix up ion flow and electron flow when recalling how the circuit is completed
  - Correct: Electrons flow only through the external wire; anions and cations flow through the salt bridge to maintain charge neutrality
- **Wrong:** Forgetting to account for ion charge when calculating product mass in electrolysis, leading to a wrong mole ratio
  - Why it fails: Students rush from moles of electrons straight to mass without writing the half-reaction
  - Correct: Always write the reduction half-reaction for your product first to get the correct mole ratio of electrons to product
- **Wrong:** Claiming a negative $E^\circ_{\text{cell}}$ is impossible for any electrochemical cell
  - Why it fails: Students associate all electrochemical cells with spontaneous galvanic cells that produce voltage
  - Correct: Negative $E^\circ_{\text{cell}}$ is expected for electrolytic cells; you just need an external voltage larger than $|E^\circ_{\text{cell}}|$ to drive the non-spontaneous reaction
- **Wrong:** Swapping anode and cathode in cell notation, putting the cathode on the left
  - Why it fails: Students confuse the order of notation with charge sign
  - Correct: Always follow the standard rule: anode (oxidation) left, cathode (reduction) right in cell notation

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Universal Anode/Cathode | Oxidation = Anode, Reduction = Cathode | True for both cell types |
| Galvanic Cell Charge | Anode = (-), Cathode = (+) | Spontaneous, produces electrical energy |
| Electrolytic Cell Charge | Anode = (+), Cathode = (-) | Non-spontaneous, requires external energy |
| Cell Notation | Anode (left) $\mid\mid$ Cathode (right) | Single \| = phase boundary, double \|\| = salt bridge |
| Standard Cell Potential | $E^\circ_{\text{cell}} = E^\circ_{\text{red (cathode)}} - E^\circ_{\text{red (anode)}}$ | Intensive: do not multiply by coefficients |
| ΔG° / E° Relationship | $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ | $F = 96500\ \text{C/mol }e^-$, $n$ = moles e⁻ |
| Spontaneity Criterion | $E^\circ_{\text{cell}} > 0$ = spontaneous | Positive = galvanic, negative = electrolytic |
| Faraday's Law Charge | $Q = I \times t$ | $I$ in A, $t$ in s, $Q$ in C |

## What's next

This subtopic is a core foundation for all electrochemistry content in AP Chemistry Unit 9, and connects closely to more advanced topics like non-standard cell potential and the Nernst equation. Electrochemistry questions on the AP exam regularly combine the rules and calculations you learned here with concepts from thermodynamics and stoichiometry, especially in multi-part free response questions. Mastery of this content will make more advanced electrochemistry topics much easier to understand.

- [Unit 9: Applications of Thermodynamics Overview](https://www.owlsprep.com/study/ap-chemistry-u9-overview/)
- [Cell Potential and Free Energy](https://www.owlsprep.com/study/ap-chemistry-u9-cell-potential-and-free-energy/)
- [Cell potential under nonstandard conditions](https://www.owlsprep.com/study/ap-chemistry-u9-cell-potential-under-nonstandard-conditions/)

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