Move the sliders to see how much mass different currents and times deposit
Controls
Choose an electrolyte
Readings
- Charge passed, Q = It
- 1,200.0C
- Moles of electrons, Q/F
- 0.0124mol
- Cathode product
- Na
- Mass deposited at cathode, m = MIt/(nF)
- 0.286g
- Anode product
- Cl₂
- Mass produced/dissolved at anode
- 0.442g
- Mole ratio of products (cathode : anode)
- 2 : 1
- Right now
- The chloride ion is itself oxidised directly, forming chlorine gas
How to use this simulation
- Start with the defaults: molten NaCl, 2 A, 10 minutes. Press play: a droplet of sodium metal builds up at the cathode on the left, and chlorine gas bubbles rise from the anode on the right.
- Watch the small orange dots on the wires: those are electrons, flowing from the battery’s negative terminal to the cathode, and from the anode to the battery’s positive terminal.
- Switch "Choose an electrolyte" to CuSO₄, then set "Anode material" to copper: now the anode itself slowly erodes, because the copper is dissolving into the solution instead of releasing a gas.
- Still on CuSO₄, switch to an inert electrode: the anode stops eroding, and oxygen bubbles rise from it instead.
- Choose acidified water and change the current: the "mole ratio" reading for H₂ and O₂ always stays 2 : 1, whatever current you pick.
From gold-plated jewellery to refined copper wire
A thin layer of gold on a cheap ring, a shiny chrome bumper that resists rust, a car battery being recharged overnight — all of these run on the same trick: electrolysis. Push a current through the right setup and you can pull a metal ion out of solution and deposit it exactly where you want it, precisely what you’ll watch copper do onto an electrode in this page’s simulation.
On an industrial scale the same idea gets much bigger: extracting aluminium from bauxite, splitting brine into chlorine, hydrogen and sodium hydroxide (the chlor-alkali industry), and refining impure copper into the pure metal used in electrical wiring. The underlying rule behind all of this is simple enough that you can test it yourself in the simulation above.
The question is why passing a current breaks a compound apart at all, which ion goes to which electrode, and how you can predict in advance exactly how much metal or gas will form. This page and the simulation together will let you work that answer out for yourself.
Starting from zero: electrolysis and the electrolyte
An electrolyte is a substance that, when molten or dissolved, conducts electricity and is chemically broken down by it. Solid NaCl does not conduct electricity — its ions are locked into a crystal lattice — but melt it, or dissolve it in water, and the ions become free to move, which is exactly what lets it act as an electrolyte.
Dip two conducting rods or plates into an electrolyte and connect them to a battery, and you have built an electrolytic cell. Whichever electrode is wired to the battery’s negative terminal is called the cathode; the one wired to the positive terminal is the anode. The battery pushes electrons into the cathode, which is why the cathode is the negative electrode in electrolysis — worth remembering, because it is the opposite way round in an ordinary battery.
Positive ions (cations) in the solution are drawn towards the negative cathode, where they pick up electrons and are reduced. Negative ions (anions) are drawn towards the positive anode, where they give up electrons and are oxidised. A simple way to remember it: reduction happens at the cathode, oxidation at the anode — often abbreviated "red cat, an ox".
Electrolysis and an ordinary battery (a galvanic cell) are opposites of each other. In a battery, a spontaneous chemical reaction produces electricity on its own. In electrolysis it runs the other way: the reaction is not spontaneous, so an external current has to force it to happen. That is why electrolysis consumes electrical energy, while a battery supplies it.
Key terms in electrochemistry
Get the vocabulary straight before the formulas; definition questions in exams come straight from this table.
| Term | What it means | Example |
|---|---|---|
| Electrolyte | A substance that conducts electricity, and is broken down by it, when molten or dissolved | Molten NaCl, CuSO₄ solution |
| Electrolysis | Using a current to break an electrolyte down into its elements | NaCl → Na + Cl₂ |
| Cathode | The electrode wired to the battery’s (−) terminal, where reduction happens | Na⁺ + e⁻ → Na |
| Anode | The electrode wired to the battery’s (+) terminal, where oxidation happens | 2Cl⁻ → Cl₂ + 2e⁻ |
| Faraday constant, F | The total charge on one mole of electrons | 96,485 coulombs/mol |
| Charge, Q | Current multiplied by time | Q = It |
| Spectator ion | An ion present in solution that takes no part in the electrode reaction | SO₄²⁻ in CuSO₄ |
Faraday's law: from charge to mass
Michael Faraday observed that the mass deposited or liberated at an electrode is proportional to the total charge passed, and that this mass also depends on how many electrons each ion of that element needs. Those two observations are usually written as two separate laws, but for problem-solving they combine neatly into one equation.
Q = Itcharge = current × time
n(electrons) = Q/Fmoles of electrons passed
m = MIt/(nF)Faraday's law: M is molar mass, n electrons per ion
What the Faraday constant actually is
The Faraday constant is simply the total charge on one mole of electrons. Each electron carries 1.602 × 10⁻¹⁹ coulombs, and a mole contains Avogadro's number of particles (6.022 × 10²³). Multiplying the two gives F = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ ≈ 96,485 coulombs per mole.
Four electrolyte set-ups, side by side
The table below lines up all four states the simulation offers (molten NaCl, CuSO₄ with a copper anode, CuSO₄ with an inert anode, and acidified water). Notice that the chloride ion is the only anion actually oxidised itself; in the other three, it is water that gets oxidised (or the copper electrode itself), while the sulfate ion is left as a spectator.
| Electrolyte | Cathode product | Anode product | Mole ratio (cathode : anode) |
|---|---|---|---|
| Molten NaCl | Na | Cl₂ | 2 : 1 |
| CuSO₄ solution — inert anode | Cu | O₂ | 2 : 1 |
| CuSO₄ solution — copper anode | Cu | Cu (↓) | 1 : 1 |
| Acidified water | H₂ | O₂ | 2 : 1 |
Try it yourself in the simulation
Predict each result before you press play, then check.
- With molten NaCl at 2 A and 10 minutes, read "Mass deposited at cathode" in the readings panel, then check it against the hand calculation in Problem 1 below.
- Switch to CuSO₄ and toggle between a copper and an inert anode: the cathode keeps depositing copper at the same rate either way — only the anode’s behaviour changes.
- Double the current: the mass deposited at the cathode doubles too (see Problem 7), because m is directly proportional to I.
- With acidified water, change the current to anything you like and watch the "mole ratio" reading for H₂ and O₂ — it always stays 2 : 1.
- Push the time slider up and watch the bubbles or the plating keep growing, because the mass is directly proportional to t as well.
Solved problems
Each solution first finds the charge Q = It, then applies Faraday's law m = MIt/(nF).
Problem 1: molten NaCl (the simulation's own defaults)
Current I = 2 A, time t = 10 minutes = 600 s. Charge Q = It = 2 × 600 = 1,200 coulombs. Moles of electrons = Q/F = 1,200/96,485 = 0.01244 mol.
At the cathode, sodium (M = 23, n = 1): m = MIt/(nF) = 23 × 1,200 / (1 × 96,485) = 0.286 g. At the anode, chlorine gas (M = 71, n = 2): m = 71 × 1,200 / (2 × 96,485) = 0.442 g. Mole ratio Na : Cl₂ = 2 : 1.
Problem 2: refining copper (a copper anode in CuSO₄)
Current 2 A, time 30 minutes = 1,800 s. Charge Q = 3,600 C, moles of electrons = 0.03731 mol.
Cu²⁺ + 2e⁻ → Cu (n = 2, M = 63.5): mass deposited at the cathode = 1.185 g. The anode’s copper dissolves at exactly the same rate, 1.185 g, since both electrodes share the same n. Mole ratio 1 : 1 — one ion’s loss is instantly balanced by the other’s gain, which is why the solution’s blue colour barely changes.
Problem 3: CuSO₄ with an inert anode
Current 1.5 A, time 20 minutes = 1,200 s. Charge Q = 1,800 C.
The cathode still deposits 0.592 g of copper. But with no copper at the anode, water is oxidised instead, releasing oxygen gas (O₂, n = 4, M = 32): 0.149 g. Mole ratio Cu : O₂ = 2 : 1.
Problem 4: acidified water — the H₂ : O₂ ratio
Current 3 A, time 500 s. Charge Q = 1,500 C.
At the cathode, H₂ (n = 2, M = 2): 0.016 g. At the anode, O₂ (n = 4, M = 32): 0.124 g. Mole ratio H₂ : O₂ = 2 : 1 — the same charge always makes twice as many moles of H₂, because each H₂ molecule needs only 2 electrons while O₂ needs 4.
Problem 5 (beyond the three presets): silver electroplating
Plating silver onto jewellery uses Ag⁺ + e⁻ → Ag (n = 1, M = 108). Running 0.5 A for 30 minutes passes a charge of Q = 900 C.
Mass of silver deposited: m = 108 × 900 / (1 × 96,485) = 1.007 g. The same formula works for any metal beyond the three presets — only M and n change.
Problem 6: what the Faraday constant means, in seconds
Question: at 5 A, how long does it take to pass exactly 1 mole of electrons (one Faraday of charge)?
t = F/I = 96,485 / 5 = 19,297 s ≈ 5.36 hours. That is the real meaning of the Faraday constant — the exact amount of charge it takes to shift one mole of electrons.
Problem 7: double the current, double the mass
Compare how much copper deposits at the cathode of a CuSO₄ cell with an inert anode, running 1 A versus 4 A, both for the same 10 minutes (600 s).
At 1 A, m = 0.197 g. At 4 A, m = 0.790 g. 0.790 ÷ 0.197 ≈ 4, exactly the ratio of the currents — because Faraday's law makes m directly proportional to I.
Common mistakes
Avoiding these keeps both definition questions and calculation questions from losing marks.
- Assuming the cathode is positive in electrolysis. It is negative; the anode is positive. It is the opposite way round in an ordinary battery, which is exactly why the two get mixed up.
- Swapping reduction and oxidation. Reduction always happens at the cathode, oxidation always at the anode, in every electrochemical cell.
- Assuming the sulfate ion reacts at the anode. In CuSO₄ or acidified water, sulfate is only a spectator; water is oxidised instead (or the copper electrode itself, if one is used).
- Plugging the wrong number in for n in Faraday's law. n is the number of electrons per ion or molecule of product (2 for Cu²⁺, 1 for Na⁺, 4 for O₂) — not the total charge or the current.
- Leaving time in minutes when using Faraday's law. Always convert to seconds before substituting into m = MIt/(nF).
- Assuming doubling the current also requires doubling the time to double the mass. In fact I and t are each independently proportional to mass (m ∝ It), so doubling either one alone already doubles the mass.
Electrolysis in everyday life
Electrolysis is not only an exam topic — it sits behind plenty of things, from heavy industry to a piece of jewellery.
- Electroplating: a thin layer of gold or silver on cheaper jewellery, or chromium and nickel plating on car parts to resist rust.
- Copper refining: electrolysing impure copper as the anode deposits pure copper at the cathode — exactly the purity needed for electrical wiring.
- Aluminium extraction: molten alumina (from bauxite) is electrolysed to extract aluminium metal (the Hall–Héroult process), because aluminium is too reactive to be extracted by ordinary chemical reduction.
- The chlor-alkali industry: electrolysing brine (salty water) produces chlorine gas, hydrogen gas and sodium hydroxide together, the starting materials behind soap and bleach manufacturing.
- Charging a car battery: while charging, the battery behaves exactly like an electrolytic cell, taking in outside electricity to reverse its own internal chemical reaction.
Exam tips
Electrolysis and Faraday’s laws appear in nearly every general-chemistry and electrochemistry unit. Definition questions want the exact roles of cathode and anode; calculation questions want the charge worked out first, then the formula applied.
A worked exam-style question
Question: a cell electrolyses CuSO₄ solution using inert platinum electrodes, with a current of 1.5 A for 20 minutes.
(a) Define an electrolyte. (b) Write the half-equations at the cathode and anode. (c) Calculate the mass of copper deposited at the cathode. (d) "Using a copper anode instead of an inert one would have produced no gas at the anode" — justify this statement.
Answer to (c): m = MIt/(nF) = 63.5 × 1,800 / (2 × 96,485) = 0.592 g. Answer to (d): with a copper anode, the copper itself is oxidised instead of sulfate or water (Cu → Cu²⁺ + 2e⁻), dissolving into the solution rather than releasing any gas.
Revision: the one-screen summary
The night before an exam, this list plus the three lines of Faraday’s law above is all you need to revisit.
- Electrolysis: an external current forces a non-spontaneous reaction; the cathode is negative (reduction), the anode is positive (oxidation).
- Faraday's law: Q = It, moles of electrons = Q/F, m = MIt/(nF); F = 96,485 coulombs/mol.
- In molten NaCl the chloride ion is itself oxidised; in aqueous CuSO₄ and acidified water, water is usually oxidised instead, with sulfate as a spectator.
- Electrolysing CuSO₄ with a copper anode dissolves the anode instead of releasing gas — the principle behind copper refining.
- In acidified water, H₂ and O₂ always form in a 2 : 1 mole ratio, because H₂ needs n = 2 and O₂ needs n = 4.
- Mass is directly proportional to both current and time (m ∝ It); doubling either one alone doubles the deposited mass.
Frequently asked questions
What is electrolysis?
Electrolysis is the process of passing an external electric current through an electrolyte — a molten or dissolved ionic compound — to break it down chemically into its elements. It is a non-spontaneous process, so it only runs while electrical energy is supplied.
Which electrode is the cathode and which is the anode?
The cathode is wired to the battery’s negative terminal, and it is where cations pick up electrons and are reduced. The anode is wired to the positive terminal, and it is where anions give up electrons and are oxidised.
What is Faraday's law of electrolysis?
m = MIt/(nF). Here m is the mass deposited, M is molar mass, I is the current in amperes, t is the time in seconds, n is the number of electrons per ion, and F = 96,485 coulombs per mole is the Faraday constant.
What is the Faraday constant, and what is its value?
The Faraday constant is the total electric charge carried by one mole of electrons, roughly 96,485 coulombs per mole. It comes from multiplying Avogadro’s number by the charge on a single electron.
What forms when molten NaCl is electrolysed?
Sodium metal forms at the cathode (Na⁺ + e⁻ → Na) and chlorine gas forms at the anode (2Cl⁻ → Cl₂ + 2e⁻). This is the industrial method used to produce pure sodium metal.
Why do copper and inert electrodes give different results in CuSO₄ solution?
Both deposit copper at the cathode. But a copper anode simply dissolves into the solution (releasing no gas), while an inert anode oxidises water instead, releasing oxygen gas, because an inert electrode does not react itself.
Why do H₂ and O₂ always form in a 2 : 1 ratio when acidified water is electrolysed?
Forming H₂ needs only 2 electrons per molecule (2H⁺ + 2e⁻ → H₂), while forming O₂ needs 4 (2H₂O → O₂ + 4H⁺ + 4e⁻). For the same charge passed, that always makes twice as many moles of H₂ as O₂.
What's the difference between electrolysis and an ordinary battery (a galvanic cell)?
A galvanic cell uses a spontaneous chemical reaction to generate electricity on its own. An electrolytic cell runs the opposite way: the reaction is not spontaneous, so an external current is needed to force it — which is also why the cathode and anode carry opposite charges in the two kinds of cell.
If you double the current, does the deposited mass double too?
Yes. Faraday's law makes m directly proportional to I for a fixed time, so doubling the current exactly doubles the deposited mass. The same is true for doubling the time instead.
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