You made it to the last topic section of AP Chemistry. Before jumping in, congratulate yourself for making it this far! Over the AP Chem course, you have learned everything from acids and bases to equilibrium to thermodynamics.
For our final topic of Unit 9 and AP Chemistry, we will discuss nonspontaneous redox reactions and how we can use electrolytic cells to make them happen. This builds on the foundations of how electrolytic cells work, how we can differentiate galvanic (or voltaic) cells from electrolytic cells, and how to use Faraday's Law to make electrolysis calculations.
Review of Electrolytic Cells 🔋
As we learned in the introduction of galvanic and electrolytic cells, an electrolytic cell is a cell in which power from an external source (usually a battery) is used to spur nonspontaneous redox reactions by creating electromotive force. Electrolytic cells require energy because they are thermodynamically unfavorable, and therefore nonspontaneous. This means that a chemical species in an electrolytic reaction that would normally be oxidized is instead reduced and vice versa.
Take a look at an example of an electrolytic cell with Zinc and Copper:
In this image, we see that electrons are being pumped out of Cu to form Cu2+ and transferred to Zn2+ to form Zinc metal. The arrows show the direction of the electron flow, meaning the redox reaction occurring is: Cu + Zn2+ → Cu2+ + Zn.
By using a table of standard reduction potentials, we can calculate the cell potential, Ecell:
First, we write out our half-reactions:
Cu → Cu2+ + 2e- (E = -0.34 V)
Zn2+ + 2e- → Zn (E = -0.76 V)
Adding our E values, we find that Ecell = -1.10 V. Therefore, because of the negative voltage, we must introduce external power to cause the redox reaction to run.
Note on the battery in the cell diagram it must be a battery with a voltage of over 1.10 V. This is because the voltage must be enough to overcome the non-spontaneity of the redox reaction we want to occur (and also the spontaneousness of the reverse reaction which has an E value of +1.10 V). When we connect this battery, the voltage pulls electrons in the direction of the nonspontaneous reaction. Therefore, the mass of the copper electrode will decrease, and the mass of the zinc electrode will increase.
Comparing Electrolytic and Galvanic Cells
There are some important differences between a galvanic cell and an electrolytic cell that are necessary to know and recognize for the AP exam. Examine the following galvanic cell and electrolytic cell side by side:
On the left, the galvanic cell has an anode where oxidation occurs and a cathode where reduction occurs. The reaction occurring is with cadmium and copper. Electrons travel through a wire from the cadmium anode to the copper cathode, causing the cadmium electrode to shrink and the copper electrode to grow. As the electrons travel, a voltmeter measures the electromotive force in volts that the reaction releases. Finally, a salt bridge (with a neutral salt) connects the two half-cells to maintain neutrality.
We can take a closer look at a similar example of a galvanic cell with a zinc and copper reaction below:
For this cell, the copper electrode will grow, and the zinc electrode will shrink, which is typical of an anode and cathode in galvanic cells.
On the right, we have an electrolytic cell. The first notable difference is the direction that the electrons travel. In the galvanic cell, electrons traveled from the cadmium to the copper electrode (and eventually to the Cu2+ to create Cu metal). The electrolytic cell has the opposite occurring. Copper metal oxidizes into Cu2+, and Cd2+ reduces into Cd metal, the inverse reaction of the one in the galvanic cell.
Instead of a voltmeter to measure voltage, a power supply (similar to a battery) is used to produce an electromotive force of at least 0.74 V. Because the reaction occurring in the electrolytic cell is nonspontaneous, it requires an external source of energy of this magnitude to run. The reaction also includes a salt bridge because it is two separate half-cells. In this example, the copper electrode will shrink, and the cadmium electrode will grow.
A major similarity between the two types of cells is that always, always, ALWAYS, oxidation occurs at the anode, and reduction occurs at the cathode. No matter what type the cell is (galvanic or electrolytic), this is the standard method used to label electrodes.
The following comparison emphasizes additional differences between the cell types:
Faraday’s Law and Electrolysis Problems ⚡
An essential type of problem regarding electrolytic cells is calculating the mass of a metal that accumulates at an electrode. To do this, we want to complete a chain of conversions using the definition of an amp (1A = 1C/s) and Faraday’s constant (1 mol e- = 96485 coulombs), represented as I = q / t, where I is the current in amps, q is charge in coulombs, and t is time in seconds. Ultimately, using Faraday's law, we can calculate the amount of charge flow based on changes in the amounts of reactants and products in an electrochemical cell.
Take a look at this example problem:
Determine the mass of chromium that can be produced when a solution of Cr(NO3)2 is electrolyzed for 60 minutes with a current of 15 amps.
To begin with, we need to convert from amps to coulombs:
60 minutes * 60s/min * 15 C/s = 54000 coulombs
Next, we can use this value of charge to convert to moles of electrons:
54000 coulombs * 1 mol e-/96485C = 0.559 mol e-
Then, we can use stoichiometry to convert from mol e- to moles of chromium. After that, we use the molar mass of chromium (52.0 g/mol) to convert to grams:
0.559 mol e- * (1 mol Cr/2 mol e-) * (52g/mol Cr) = 14.55g Cr produced.
This sort of dimensional analysis can also be done in one long chain as follows:
Prepare to solve for any of these unknowns: the number of electrons transferred, the mass of material deposited on or removed from an electrode, current, time elapsed, or the charge of ionic species. A problem could ask about finding the time necessary to produce a certain mass or the amperage required for a given time to produce a mass. To complete problems like these, you can simply manipulate the chain we completed previously.
For example, if a problem asked for a time but gave a mass, you could start with the mass, convert it to moles, then convert it to moles of electrons, then convert it to charge, and then to amperage.
We just need to do proper unit cancellation. Look at the units here:
Now you know how to use Faraday's constant to solve electrolysis problems!
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| cell potential | The electrical potential difference between the anode and cathode of an electrochemical cell, which drives the spontaneous redox reaction. |
| concentration cell | An electrochemical cell in which the two half-cells contain the same chemical species but at different concentrations. |
| equilibrium | The state in which the forward and reverse reaction rates are equal, resulting in constant concentrations or partial pressures of reactants and products. |
| Le Châtelier's principle | A principle stating that when a system at equilibrium is disturbed, the system shifts to counteract the disturbance and re-establish equilibrium. |
| Nernst equation | The equation E = E° − (RT/nF) ln Q that relates cell potential to standard cell potential and the reaction quotient under nonstandard conditions. |
| nonstandard conditions | Electrochemical conditions where concentrations of active species differ from 1 M, pressures differ from 1 atm, or temperature differs from 25°C. |
| reaction quotient | A value calculated using the same expression as the equilibrium constant but using current (non-equilibrium) concentrations or partial pressures. |
| spontaneous electron flow | The natural movement of electrons from the anode to the cathode in an electrochemical cell driven by the cell potential. |
| standard cell potential | The cell potential (E°) measured under standard conditions where all concentrations are 1 M, pressure is 1 atm, and temperature is 25°C. |
Frequently Asked Questions
What is electrolysis and how does it actually work?
Electrolysis is using an external power source to force a non-spontaneous redox reaction. The cathode (negative) is where reduction happens (cations gain e–), the anode (positive) is where oxidation happens (anions or metal lose e–). Current (I) moves electrons through the external circuit while ions carry charge in the electrolyte. Faraday’s law connects what you measure (current, time) to chemical change. Key equations: - I = q / t (current = charge / time) - q (Coulombs) = I × t - moles of electrons transferred = q / F (F = 96,485 C·mol–1) - moles of product formed = q / (n · F), where n = electrons per mole in the balanced half-reaction - mass deposited = (molar mass × q) / (n · F) So to predict electroplating or amount reacted, write the half-reaction, find n, compute q from I and t, then use q/(nF). This is exactly what AP Topic 9.11 tests—practice these stoichiometric steps (see the Topic 9.11 study guide on Fiveable: https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and try problems at https://library.fiveable.me/practice/ap-chemistry.
Why does electrolysis need electricity to make chemical reactions happen?
Electrolysis needs electricity because you’re forcing a non-spontaneous redox reaction to occur by pushing electrons where they wouldn’t go on their own. In a galvanic cell spontaneous electron flow makes ΔG negative; in electrolysis the cell’s Ecell is negative, so ΔG would be positive. Supplying an external voltage (work) reverses that and drives oxidation at the anode and reduction at the cathode. Faraday’s laws connect the electrical side to chemical change: charge Q (coulombs) = I·t, moles of e– = Q/F (Faraday constant), and stoichiometry tells you how many moles of product form per n electrons transferred. So current and time directly control how much material is deposited or released (electroplating/electrodeposition)—exactly what Topic 9.11 focuses on (I = q/t and Faraday constant). For a refresher see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and more practice problems at (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between electrolysis and a regular battery?
A regular battery (galvanic cell) produces electrical energy from a spontaneous redox reaction: chemical energy → electrical work. In a battery the anode is oxidized, the cathode reduced, E°cell is positive, and electrons flow through the external circuit from anode → cathode. Electrolysis is the opposite: you force a non-spontaneous redox reaction by applying an external voltage. Electrons are driven into the cathode (reduction) and pulled from the anode (oxidation); E°cell is effectively negative for the spontaneous direction so you must supply energy. For AP Chem (Topic 9.11) the key difference is also how you account for charge and stoichiometry: in electrolysis you use Faraday’s laws (I = q/t, q = n·F) to calculate moles of electrons transferred and mass deposited/removed at an electrode (electrodeposition/electroplating). On the exam you might be asked to compute charge (C), current (A), time (s), n (electrons), or mass using the Faraday constant. Review the Topic 9.11 study guide here (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain Faraday's law in simple terms because I'm totally lost?
Faraday’s law in plain terms: it links electricity (charge) to how much chemical change happens in electrolysis. Every electron that flows is part of a redox stoichiometry—so if a half-reaction needs n electrons per mole of product, the total moles of product = moles of electrons transferred ÷ n. Key steps/equations: - Charge (q, in coulombs) = current (I, A) × time (t, s). (I = q/t) - Moles of electrons = q / F, where F (Faraday constant) ≈ 96,485 C·mol⁻¹. - Moles of product = (moles e⁻) / n (n = electrons per mole from the balanced half-reaction). Quick example: 193,000 C passed through Ag+ → Ag (n = 1). Moles e⁻ = 193,000 / 96,485 ≈ 2.00 mol → 2.00 mol Ag deposited. Use this on the AP to connect number of electrons, mass deposited (electroplating), current, time, and ionic charge. For a focused study guide, see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I calculate how much metal gets deposited during electroplating?
Use Faraday’s law: mass deposited m = (I · t · M) / (n · F). Steps: 1. Write the cathode half-reaction to find n (electrons per metal atom). 2. Convert current and time to total charge Q = I·t (amps·seconds = coulombs). 3. Moles of electrons = Q / F (F = 96,485 C·mol⁻¹). 4. Moles of metal deposited = (moles e⁻) / n. 5. Mass = moles metal × molar mass M. Example: plating Cu²⁺ to Cu (n = 2) with I = 2.00 A for 30.0 min (1800 s): Q = 3600 C; moles e⁻ = 3600/96485 = 0.0373 mol; moles Cu = 0.0373/2 = 0.01865 mol; mass = 0.01865×63.55 = 1.19 g. On the AP, you’ll be asked to relate electrons, charge, current and time (I = q/t) and show stoichiometry (n) from the half-reaction. Also note real cells may have <100% Faradaic efficiency (less actual mass). For extra practice, check the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x), the full Unit 9 page (https://library.fiveable.me/ap-chemistry/unit-9), and practice problems (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about the relationship between current, time, and charge - what's the formula?
The relationship is I = q / t, so rearranged q = I × t. - I is current in amperes (A), q is charge in coulombs (C), t is time in seconds (s). 1 A = 1 C/s. - For electrolysis use Faraday’s law: moles of electrons = q / F, where F ≈ 96,485 C·mol⁻¹ (Faraday constant). Then moles of product = (moles e⁻) / n, where n is electrons transferred per mole in the half-reaction. Quick example: a 2.0 A current for 3.0 min (180 s) gives q = 2.0×180 = 360 C. Moles e⁻ = 360 / 96485 ≈ 3.73×10⁻³ mol e⁻. If the half-reaction uses 2 e⁻ per product (n=2), product formed = 1.87×10⁻³ mol. This I = q/t formula is explicitly in the AP CED for Topic 9.11 (electrolysis/Faraday’s law). For more practice and the topic study guide, see Fiveable’s electrolysis study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and the AP Chem practice sets (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when they say "number of electrons transferred" in electrolysis?
“Number of electrons transferred” (n) just means how many electrons move in the redox change for every mole of reactant or product in the half-reaction. In electrolysis you write the half-reaction (e.g., Cu2+ + 2 e– → Cu) and read off n = 2—two electrons are consumed to deposit one Cu atom (per formula unit, or per mole when scaled). Faraday’s law then links that n to charge and mass: total charge q = (moles of product) × n × F (F ≈ 96485 C·mol–1). Combined with I = q/t, you can convert current and time into moles of electrons and therefore moles (or mass) of material plated or removed. On the AP exam you’ll be asked to use half-reactions, stoichiometry of e–, and Faraday’s constant to calculate charge, mass, current, or time (Topic 9.11). For practice and step-by-step examples, check the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and lots of practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you figure out how long to run electrolysis to get a specific amount of product?
You use Faraday’s law and I = q/t. Steps: 1) Write the half-reaction to find n (electrons transferred per mole of product). 2) Convert the desired mass (or volume/particles) of product to moles. 3) Moles of electrons = moles product × n. 4) Total charge q = (moles of electrons) × F (Faraday constant = 96,485 C·mol⁻¹). 5) Solve t = q / I (use given current I). Example: to deposit 0.100 mol Cu (Cu2+ + 2e– → Cu), n = 2 so moles e– = 0.200 mol; q = 0.200 × 96485 = 19297 C; at 2.00 A, t = 19297 / 2.00 = 9648.5 s (~2.68 h). On the AP exam you’ll be expected to connect stoichiometry, moles of electrons, charge, current and time (CED 9.11.A). For a quick review see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and practice more problems at (https://library.fiveable.me/practice/ap-chemistry).
Why do we need to know the charge of ions for Faraday's law calculations?
You need the ion’s charge because Faraday’s law links moles of electrons (which carry charge) to moles of substance. The charge (oxidation state) tells you n, the number of electrons exchanged per ion in the half-reaction. Faraday’s constant F = 96,485 C·mol⁻¹ e⁻ converts moles of electrons to coulombs, so q = n · F · (moles of ions reacted). For example, reducing 1.00 mol Cu²⁺ to Cu(s) requires 2 mol e⁻ → q = 2 · F ≈ 193,000 C. If you used the wrong ionic charge you’d get the wrong n and so the wrong charge, current or mass deposited. This is exactly what the CED expects you to do for 9.11.A (number of electrons transferred, moles of electrons, mass deposited, current/time, charge of ionic species). For review, see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
What happens at each electrode during electrolysis and how do I remember which is which?
At each electrode in electrolysis: reduction occurs at the cathode (gain of e−) and oxidation occurs at the anode (loss of e−). In electrolysis an external power source forces a non-spontaneous redox: the cathode is negatively charged (it supplies electrons to cations, e.g., M^n+ + n e− → M(s)) and the anode is positively charged (it pulls electrons from anions or metal atoms, e.g., 2 Cl− → Cl2(g) + 2 e− or M → M^n+ + n e−). Memory tricks: “Red Cat, An Ox” (reduction at the cathode; anode is oxidation) or pair with “cathode attracts cations.” For calculations use Faraday’s law: Q = I·t, moles e− = Q / F (F ≈ 96,485 C·mol−1), then stoichiometry to find mass deposited/removed. This is exactly what the CED asks you to apply (I = q/t and Faraday’s constant). For review and worked examples see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and more practice (https://library.fiveable.me/practice/ap-chemistry).
How does electroplating work and what's the connection to Faraday's law?
Electroplating is electrodeposition: you put the part you want plated at the cathode so metal cations in solution (like Ag+) are reduced and stick to it, while the anode oxidizes (often dissolving metal to replace ions). The key link to Faraday’s law is stoichiometry between charge and moles of electrons. Total charge q = I·t (current × time). Moles of electrons = q / F (Faraday constant ≈ 96,485 C·mol⁻¹). Use the half-reaction to find n (electrons per metal ion). Mass deposited m = (M × q)/(n × F) or m = (M × I × t)/(n × F), where M = molar mass of metal. Example: plating 1.0 mol Ag+ (n = 1) requires 96,485 C. Don’t forget real cells can be less than 100% Faradaic (coulombic) efficiency if side reactions occur. Practice these calculations for the AP task in 9.11.A. For a clear review and extra problems, see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and more practice at (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how to use stoichiometry with electrolysis problems - can someone help?
Think of electrolysis problems as regular stoichiometry but with electrons as a reagent. Use these steps every time: 1. Write the half-reaction for the species being deposited/removed and identify n (electrons transferred per mole). 2. Convert the given mass or moles of product/reactant to moles. 3. Use stoichiometry to get moles of electrons: moles product × (n mol e− / 1 mol product). 4. Convert moles e− to charge: q = (moles e−) × F, where F ≈ 96,485 C·mol−1. 5. Relate charge, current, and time with I = q/t (so t = q/I if time is asked). Example: depositing 1.00 mol Ag (Ag+ + e− → Ag) requires 1.00 mol e− → q = 1.00×96,485 = 96,485 C. Also note Faradaic (coulombic) efficiency—real cells may deposit less than theoretical. This is exactly what AP Topic 9.11 asks you to calculate (9.11.A). For extra practice and worked examples see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and lots of practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between the amount of charge and the amount of current in these problems?
Charge (q) is how much electric stuff actually moved—measured in coulombs. Current (I) is the rate that charge flows—coulombs per second, aka amperes. They’re related by I = q/t. In electrolysis/Faraday problems you usually: - Use stoichiometry to find moles of electrons transferred (n from half-reaction). - Convert moles e− to charge: q = n · F (Faraday constant ≈ 96485 C·mol−1). - If you know time, find current: I = q / t. Or if you know current and want mass deposited, find q = I·t then moles e− = q/F. Example: 0.010 mol e− → q = 0.010·96485 ≈ 965 C. If that passed in 100 s, I = 965/100 = 9.65 A. On the AP exam you’ll be asked to tie moles of electrons, charge (C), current (A), and time (s) to electrode mass or reaction stoichiometry (see Topic 9.11 study guide for worked examples) (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x). For more practice, try the AP problem set (https://library.fiveable.me/practice/ap-chemistry).
How do I know how many electrons are involved in an electrolysis reaction?
You find the number of electrons by writing and balancing the half-reaction(s) for the species being oxidized or reduced. Balance atoms first, then add electrons to balance charge. The number of electrons added is n (moles e– per mole reaction). Example: Cu2+ + 2 e– → Cu(s) so n = 2; Ag+ + e– → Ag(s) so n = 1. In electrolysis/Faraday problems use n to connect moles of product to moles of electrons: moles e– = n × moles of product. Then convert to charge q = (moles e–) × F (Faraday’s constant ≈ 96485 C·mol–1) and to current/time with I = q/t. On the AP exam you’ll be asked to calculate number of electrons transferred and charge or mass deposited—so always show the balanced half-reaction and the stoichiometric link to moles of product (CED 9.11.A). For a quick review, see the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why does the mass of material deposited depend on the current and time?
Because current is a flow of charge, the total charge passed equals I·t (q = I t). Faraday’s law links that charge to how many moles of electrons moved: moles e− = q / F (F = 96,485 C·mol−1). The chemical stoichiometry tells you how many electrons (n) reduce one mole of the metal ion, so moles of metal deposited = (q / F) / n. Converting to mass gives: m = (M · I · t) / (n · F), where M is the metal’s molar mass. So higher current or longer time = more charge q, more electrons transferred, and therefore more mass plated. The number of electrons n (from the half-reaction) and the Faraday constant set the proportionality. This is exactly what the AP CED expects you to use when calculating deposited mass or charge (Topic 9.11). For a short walkthrough and practice problems, check the Topic 9.11 study guide (https://library.fiveable.me/ap-chemistry/unit-9/electrolysis-faradays-law/study-guide/v39gKqGDszHzsS5opd6x) and more practice at (https://library.fiveable.me/practice/ap-chemistry).