Galvanic (voltaic) and electrolytic cells — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Cell notation conventions, anode/cathode identification, standard cell potential calculation, ΔG and E°cell relationships, spontaneity criteria, comparisons of galvanic and electrolytic cells, and Faraday’s law of electrolysis for quantitative calculations.

You should already know: Oxidation-reduction half-reaction balancing, Gibbs free energy and spontaneity criteria, stoichiometric mole calculations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Galvanic (voltaic) and electrolytic cells?

This topic is part of Unit 9: Applications of Thermodynamics, which accounts for 14-18% of the total AP Chemistry exam score; this subtopic makes up roughly one-third of Unit 9, or ~5-6% of your total exam score. It appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with thermodynamics and stoichiometry questions.

Both galvanic (voltaic) and electrolytic cells are electrochemical cells that separate oxidation and reduction half-reactions to force electron transfer through an external wire, rather than direct electron transfer between reactants. The core difference is spontaneity: galvanic (voltaic) cells rely on spontaneous redox reactions to convert stored chemical energy into usable electrical energy—this is how all disposable and rechargeable batteries work. Electrolytic cells use an external source of electrical energy to drive non-spontaneous redox reactions, and are used for industrial processes like electroplating, metal refining, and production of pure gases like chlorine.

Standard notation for all electrochemical cells follows the convention: anode on the left, cathode on the right, single vertical line | for phase boundaries, and double vertical line || for a salt bridge or porous disk that separates the two half-cells.

2. Core Cell Structure and Electrode Identification

All electrochemical cells follow one universal rule that never changes: oxidation (loss of electrons) always occurs at the anode, and reduction (gain of electrons) always occurs at the cathode. This is true for both galvanic and electrolytic cells—confusion about charge on electrodes comes from the fact that charge reverses between the two cell types, even though the anode/cathode reaction identity stays the same.

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

For electrolytic cells: an external battery drives the reaction, pulling electrons away from the anode and pushing them onto the cathode. This gives the anode a positive charge and the cathode a negative charge, but oxidation still happens 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.

Standard cell notation always puts the anode half-reaction on the left and cathode on the right, so you can identify electrodes directly from the notation.

Worked Example

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

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 $\text{Zn}(s)$, which is 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 $E_{\text{cell}}^{\circ }$ 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 whether the cell is galvanic or electrolytic. This is one of the most common low-stakes MCQ questions you will see.

3. Standard Cell Potential and Spontaneity

Standard reduction potentials ($E_{\text{red}}^{\circ }$) are tabulated for all common half-reactions, measured relative to the standard hydrogen electrode (SHE), which is assigned a value of $E_{\text{red}}^{\circ }=0\text{V}$. To calculate the standard cell potential for any electrochemical cell, we use the formula: $$E_{\text{cell}}^{\circ }=E_{\text{red (cathode)}}^{\circ }-E_{\text{red (anode)}}^{\circ }$$ An equivalent form is $E_{\text{cell}}^{\circ }=E_{\text{red (cathode)}}^{\circ }+E_{\text{ox (anode)}}^{\circ }$, where $E_{\text{ox (anode)}}^{\circ }=-E_{\text{red (anode)}}^{\circ }$, so both formulas give the same result.

A key relationship between cell potential and Gibbs free energy (the spontaneity criterion) is: $$\Delta G^{\circ }=-nF{E}_{\text{cell}}^{\circ }$$ 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 universal spontaneity rule: if $E_{\text{cell}}^{\circ }$ is positive, $\Delta G^{\circ }$ is negative, and the reaction is spontaneous (galvanic cell). If $E_{\text{cell}}^{\circ }$ is negative, $\Delta G^{\circ }$ is positive, and the reaction is non-spontaneous (requires an external voltage, so it is an electrolytic cell).

Importantly, $E^{\circ }$ is an intensive property: it does not depend on the amount of reactant, so you never multiply $E^{\circ }$ values by coefficients when balancing electrons.

Worked Example

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

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

Exam tip: If you struggle with sign errors when calculating $E_{\text{cell}}^{\circ }$, use the second formula: flip the sign of the anode's reduction potential to get the oxidation potential, then add it to the cathode's reduction potential. This eliminates the need to subtract a negative number, which is where most sign errors happen.

4. Quantitative Electrolysis and Faraday's Law

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

The core relationship is that total charge ($Q$, measured in coulombs, C) passed through the cell 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 $n_{e^{-}}=\frac{Q}{F}=\frac{It}{F}$. 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 Al³⁺, you need 3 moles of electrons, since ${\text{Al}}^{3+}+3e^{-}\to \text{Al}(s)$.

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): $1.50\text{h}\times 3600\text{s/h}=5400\text{s}$.
2. Calculate total charge: $Q=It=(2.10\text{A})(5400\text{s})=11340\text{C}$.
3. Calculate moles of electrons: $n_{e^{-}}=\frac{11340\text{C}}{96500\text{C/mol }{e}^{-}}=0.1175\text{mol }{e}^{-}$.
4. The reduction half-reaction is ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}(s)$, so moles of Cu = $0.1175\text{mol }{e}^{-}\times \frac{1\text{mol Cu}}{2\text{mol }{e}^{-}}=0.05875\text{mol Cu}$.
5. Convert to mass: $0.05875\text{mol}\times 63.55\text{g/mol}=3.73\text{g Cu}$.

Exam tip: Always convert time to seconds before calculating charge. AP exam questions frequently give time in minutes or hours to test if you remember the unit conversion.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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? A) Oxidation occurs at the pure copper cathode, which has a negative charge B) Oxidation occurs at the impure copper anode, which has a positive charge C) Reduction occurs at the pure copper cathode, which has a positive charge D) Reduction occurs at the impure copper anode, which has a negative charge

Worked Solution: First, recall the universal rule: oxidation always at anode, reduction always at cathode. This eliminates options A and D, which swap oxidation/reduction for anode/cathode. Next, for electrolytic cells, anode is positive and cathode is negative. Option C claims the cathode is positive, which is incorrect. Only option B matches all rules: oxidation at the impure anode, which has a positive charge for an electrolytic cell. The correct answer is B.

Question 2 (Free Response)

A student constructs an electrochemical cell with a cadmium half-cell (Cd electrode in 1.0 M Cd(NO₃)₂) and a lead half-cell (Pb electrode in 1.0 M Pb(NO₃)₂). Given $E_{{\text{red (Cd}}^{2+}/\text{Cd)}}^{\circ }=-0.40\text{V}$, $E_{{\text{red (Pb}}^{2+}/\text{Pb)}}^{\circ }=-0.13\text{V}$: (a) Write the balanced net ionic equation for the spontaneous reaction and calculate $E_{\text{cell}}^{\circ }$. (b) Calculate $\Delta G^{\circ }$ for the reaction in kJ, and identify if the cell is galvanic or electrolytic. (c) The student wants to reverse this reaction using an external power source. What is the minimum voltage the power source must supply to drive the reverse reaction? Justify your answer.

Worked Solution: (a) For spontaneous reaction, higher $E_{\text{red}}^{\circ }$ is reduction (cathode). Pb²⁺ is reduced, Cd is oxidized. Half-reactions: Reduction: ${\text{Pb}}^{2+}(aq)+2e^{-}\to \text{Pb}(s)$ Oxidation: $\text{Cd}(s)\to {\text{Cd}}^{2+}(aq)+2e^{-}$ Balanced net reaction: ${\text{Pb}}^{2+}(aq)+\text{Cd}(s)\to \text{Pb}(s)+{\text{Cd}}^{2+}(aq)$ $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=-0.13\text{V}-(-0.40\text{V})=0.27\text{V}$

(b) n = 2 mol e⁻: $\Delta G^{\circ }=-nF{E}_{\text{cell}}^{\circ }=-(2)(96500)(0.27)=-52110\text{J}=-52\text{kJ}$ Positive $E_{\text{cell}}^{\circ }$ and negative $\Delta G^{\circ }$ means the reaction is spontaneous, so this is a galvanic cell.

(c) The reverse reaction has $E_{\text{cell}}^{\circ }=-0.27\text{V}$. The external power source must supply a minimum of 0.27 V to overcome the non-spontaneous cell potential and drive the reverse reaction.

Question 3 (Application / Real-World Style)

A small electroplating business wants to plate 5.00 g of silver onto a piece of jewelry from an Ag⁺ solution. The plating machine operates at a constant current of 0.750 A. How many minutes will it take to plate the required mass of silver? Molar mass Ag = 107.87 g/mol, F = 96500 C/mol e⁻.

Worked Solution:

1. Calculate moles of Ag: $n(\text{Ag})=\frac{5.00\text{g}}{107.87\text{g/mol}}=0.04635\text{mol Ag}$.
2. Reduction half-reaction: ${\text{Ag}}^{+}+e^{-}\to \text{Ag}(s)$, so $n_{e^{-}}=0.04635\text{mol }{e}^{-}$.
3. Total charge Q = $n_{e^{-}}F=(0.04635\text{mol})(96500\text{C/mol})=4472.78\text{C}$.
4. Time t = $\frac{Q}{I}=\frac{4472.78\text{C}}{0.750\text{A}}=5964\text{s}$. Convert to minutes: $\frac{5964\text{s}}{60\text{s/min}}=99.4\text{minutes}$. In context: Plating 5 grams of silver at 0.75 A takes just under 100 minutes, which is consistent with small-scale commercial electroplating operation.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the core foundation for all electrochemistry content in AP Chemistry. Next, you will apply the relationships between $E_{\text{cell}}^{\circ }$, $\Delta G$, and concentration to derive and use the Nernst equation, which allows calculation of cell potentials under non-standard conditions, including for concentration cells that generate voltage from concentration differences alone. Without mastering anode/cathode identification, $E_{\text{cell}}^{\circ }$ calculation, and the $\Delta G$-$E^{\circ }$ relationship from this chapter, working with non-standard cell potentials will be extremely difficult, as all Nernst equation problems build directly on these fundamentals. This topic also connects back to core thermodynamics of spontaneous processes, and feeds into real-world AP exam questions on batteries, electrolytic refining, and corrosion.

• The Nernst equation and non-standard cell potentials
• Gibbs free energy and spontaneity
• Applications of electrochemistry