Electrolysis and Faraday's Laws

In short

Pass an electric current through a solution that contains mobile ions, and the ions go somewhere. Positive ions (cations) drift to the negative electrode (cathode); negative ions (anions) drift to the positive electrode (anode). At each electrode, the ion gains or loses enough electrons to become a neutral atom, and plates out — or a molecule of the solvent is broken apart. This process is electrolysis.

Faraday's first law of electrolysis. The mass of substance deposited (or liberated) at an electrode is proportional to the total charge passed:

\boxed{\;m \;=\; Z\,Q \;=\; Z\,I\,t\;}

where Z is the electrochemical equivalent of the substance in kg/C.

Faraday's second law. For the same charge Q passed through different electrolytes, the masses deposited are proportional to each substance's equivalent weight E = M/z (molar mass divided by the ion's valency):

\boxed{\;\frac{m_{1}}{E_{1}} \;=\; \frac{m_{2}}{E_{2}} \;=\; \frac{Q}{F}\;}.

The Faraday constant F = N_{A}\,e \approx 96{,}485\ \text{C/mol} is the charge carried by one mole of electrons. Combine the two laws:

\boxed{\;m \;=\; \frac{M}{zF}\,I\,t\;}

with M the molar mass (kg/mol), z the valency of the ion, and F = 96{,}485 C/mol.

Applications: electroplating gold on Tanishq jewellery, chrome on Royal Enfield silencers, copper refining at Hindustan Copper's plants, chlor-alkali electrolysis to make NaOH and Cl₂, and — on an industrial scale — the Hall-Héroult process that NALCO uses at Angul to make aluminium from bauxite.

Walk into a jeweller's workshop in Zaveri Bazaar in Mumbai. Between the hammer-and-tongs station and the polishing wheels, there is a small glass tank full of a pale-yellow cyanide solution. A silver ring, pre-cleaned and wired up, is dangling from the negative terminal of a 2 V power supply; a lump of pure gold hangs from the positive terminal. The ring disappears into the solution; you leave it for half an hour; you pull it out, rinse it, and it now looks gold. A thin layer of gold — perhaps two microns thick — has plated uniformly onto every surface of the silver. The gold was not melted. It was not painted. It was transported atom by atom from the anode to the cathode, one gold ion at a time, through the solution, driven by the electric current.

That is electrolysis. In this chapter you will learn the one equation that decides how much gold plates in half an hour, how much hydrogen a bank of water-electrolysers makes per minute, how much aluminium NALCO's Angul smelter pours out per day, and why the cell voltages involved — 2 V for gold, 4 V for aluminium — are set not by the current you want, but by the chemistry of the reaction.

The law that runs all of this — mass deposited is proportional to charge passed — was discovered by Michael Faraday in 1832. He did not know about atoms. He did not know about electrons. He did not even have a theory of ions. And yet, just by weighing the masses he got from a known number of coulombs, he pinned down the single most important constant in electrochemistry, to within 0.1% of the value we use today. Faraday's law of electrolysis is one of the most elegant experimental discoveries in physics, and working through it teaches you how to think about conservation of charge applied to matter.

What happens inside the cell

Pour a teaspoon of copper sulphate (CuSO₄) into a beaker of water. The solid dissolves; in solution, each formula unit splits into a positive copper ion and a negative sulphate ion:

\text{CuSO}_{4} \longrightarrow \text{Cu}^{2+} + \text{SO}_{4}^{2-}.

Why: copper sulphate is an ionic compound. Water molecules, which are highly polar, surround each ion and stabilise it — the "solvation shell". The ions then diffuse freely through the solution.

Now dip two copper electrodes into the beaker, connect them to a battery, and switch on. In the solution:

The \text{Cu}^{2+} ions are attracted to the cathode (the negative electrode). When a \text{Cu}^{2+} ion reaches the cathode, it picks up two electrons from the metal surface and becomes a neutral copper atom, which sticks to the electrode: \text{Cu}^{2+} + 2e^{-} \to \text{Cu}. The cathode gets heavier.
At the anode (the positive electrode), the copper of the electrode itself dissolves: \text{Cu} \to \text{Cu}^{2+} + 2e^{-}. The anode gets lighter; the \text{Cu}^{2+} ions it released drift through the solution toward the cathode, replacing the ones that plated out.

The net effect: copper is continuously transferred from the anode to the cathode, one atom at a time, with the solution merely serving as a highway for ions. This is the electroplating principle, and the only thing that decides the thickness of the coating is the total number of electrons the external circuit supplies — i.e., the charge passed.

Copper sulphate solution between two copper electrodes. Positive $\text{Cu}^{2+}$ ions drift right toward the cathode (where they deposit as metallic copper); negative $\text{SO}_{4}^{2-}$ ions drift left toward the anode. The current in the external wire is carried by electrons; inside the solution, by both kinds of ions.

Faraday's first law — mass proportional to charge

Faraday discovered, by direct measurement, that the mass of material deposited at an electrode is proportional to the total charge passed:

m \;\propto\; Q \;=\; It \qquad\Longrightarrow\qquad m \;=\; Z\,I\,t

where Z is a proportionality constant called the electrochemical equivalent of the substance, with SI units of kg/C. The first law is really just a statement of conservation of charge combined with the fact that each ion carries a fixed charge: if every ion carries charge ze (where z is its valency and e the elementary charge), then the number of ions that reach the cathode in time t must equal the number of ze chunks of charge delivered, which is

N_{\text{ions}} \;=\; \frac{It}{ze}.

Why: you cannot deposit half an ion. Each ion that arrives takes away z electrons from the cathode surface to become a neutral atom. If the external circuit delivered N_{e} electrons in time t, then exactly N_{e}/z ions could be neutralised.

Each ion, when it becomes a neutral atom on the electrode, contributes a mass equal to the atomic mass M/N_{A} (molar mass divided by Avogadro's number):

m \;=\; N_{\text{ions}}\cdot\frac{M}{N_{A}} \;=\; \frac{It}{ze}\cdot\frac{M}{N_{A}} \;=\; \frac{M}{z\,(N_{A}e)}\,I\,t.

The quantity N_{A}e is the charge carried by one mole of electrons. Give it a name:

\boxed{\;F \;=\; N_{A}\,e \;=\; (6.022\times 10^{23}\ \text{mol}^{-1})(1.602\times10^{-19}\ \text{C}) \;\approx\; 96{,}485\ \text{C/mol}.\;}

This is the Faraday constant. It is the charge in one mole of electrons — the single most important number in electrochemistry.

Putting it together:

\boxed{\;m \;=\; \frac{M}{zF}\,I\,t.\;}

Compared with the empirical first law m = ZIt, we can now identify the electrochemical equivalent:

Z \;=\; \frac{M}{zF}.

Why: the ratio M/(zF) is kg per coulomb (molar mass in kg/mol divided by the charge per mole of electrons times the valency). It is a pure number fixed by the substance, as the first law required.

For copper (M = 0.0635 kg/mol, z = 2 for \text{Cu}^{2+}):

Z_{\text{Cu}} \;=\; \frac{0.0635}{2 \times 96{,}485} \;=\; 3.29\times10^{-7}\ \text{kg/C} \;=\; 0.329\ \text{mg/C}.

So one coulomb of charge deposits about a third of a milligram of copper. One ampere for one hour (3600 C) deposits about 1.18 g. These numbers are exact — Faraday's law is one of the most precise statements in physics, and modern weighing experiments confirm it to better than 1 part in 10⁶.

Faraday's second law — the equivalent-weight rule

Faraday also observed that when the same current is passed in series through several different electrolytic cells (so each cell gets exactly the same charge), the masses deposited at the cathodes are in the ratio of their equivalent weights.

The equivalent weight of a substance in an electrolytic reaction is

E \;=\; \frac{M}{z}

— the molar mass divided by the valency of the ion. It is the mass of substance carried by one mole of electrons.

The second law is then:

\frac{m_{1}}{m_{2}} \;=\; \frac{E_{1}}{E_{2}} \qquad\text{(same }Q\text{ through each cell)}.

Combining with the first law:

m \;=\; \frac{E}{F}\,Q \;=\; \frac{M}{zF}\,It.

The two "laws" collapse into one. The first law captures the proportionality to charge; the second law captures the proportionality to equivalent weight. Together, they give the full formula.

Example: four cells in series

Pass the same 1 ampere for 30 minutes through four cells in series. The deposited masses are:

Ion	M (g/mol)	z	E = M/z (g)	mass deposited in 1800 C
Ag⁺ (silver)	108.0	1	108.0	(108/96485)\cdot1800 = 2.014 g
Cu²⁺ (copper)	63.5	2	31.75	(31.75/96485)\cdot1800 = 0.592 g
Au³⁺ (gold)	197.0	3	65.67	(65.67/96485)\cdot1800 = 1.225 g
Al³⁺ (aluminium)	27.0	3	9.00	(9.00/96485)\cdot1800 = 0.168 g

The masses are in the ratio of the equivalent weights: Ag : Cu : Au : Al = 108 : 31.75 : 65.67 : 9.00. That is the second law in action — verify by reading down the last column.

Deriving the formula from first principles — the ion-by-ion picture

Here is the derivation in one clean pass, because it is worth having the chain of reasoning under one roof.

Step 1. The current I flowing through the cell is the rate at which charge crosses any cross-section of the cell. In time t, the total charge is Q = It.

Step 2. The external circuit supplies electrons. The number of electrons is

N_{e} \;=\; \frac{Q}{e} \;=\; \frac{It}{e}.

Why: the elementary charge is the charge of one electron. Dividing the total charge by e gives the number of electrons — but be careful, this is only the count if we track only one direction of flow, which is exactly what the cathode does.

Step 3. Each ion arriving at the cathode uses z electrons to become a neutral atom. So the number of atoms deposited is

N_{\text{atoms}} \;=\; \frac{N_{e}}{z} \;=\; \frac{It}{ze}.

Why: a z-valent ion carries net charge +ze, which is neutralised by z electrons. Divide the electrons by z to get atoms.

Step 4. Each atom has mass M/N_{A}, so the total mass deposited is

m \;=\; N_{\text{atoms}}\cdot\frac{M}{N_{A}} \;=\; \frac{It}{ze}\cdot\frac{M}{N_{A}} \;=\; \frac{M\,I\,t}{z\,(N_{A}e)} \;=\; \frac{M\,I\,t}{z\,F}.

Step 5. Collect this into either the "first law" form or the explicit form:

\boxed{\;m \;=\; Z\,I\,t \;=\; \frac{M}{zF}\,I\,t.\;}

The only constants are M (a chemistry table), z (the ion's valency), and F (a universal constant). Once you know the ion and its charge, you know how much mass deposits per coulomb.

Watching deposition happen

The cumulative mass deposited grows linearly with time (at constant current):

m(t) \;=\; Z\,I\,t.

For a gold-plating bath at 0.5 A, the electrochemical equivalent of gold is Z_{\text{Au}} = M_{\text{Au}}/(zF) = 0.197/(3\cdot 96485) = 6.81\times 10^{-7} kg/C, so the mass grows at

\frac{dm}{dt} \;=\; ZI \;=\; (6.81\times10^{-7})(0.5) \;=\; 3.4\times10^{-7}\ \text{kg/s} \;=\; 0.34\ \text{mg/s}.

Cumulative mass of gold deposited on the cathode during a 0.5 A electroplating run. The deposition is strictly linear in time — no saturation, no exponential rise — because every coulomb passed deposits the same fixed amount of gold. After 60 minutes (one hour), about 20 mg of gold has plated out.

Explore the deposition-mass calculator

Drag the current slider to see how the cumulative mass deposited after a 1-hour run changes.

Drag the red dot to change the plating current between 20 mA and 80 mA. The curve shows the cumulative mass of gold versus time in hours; the readout gives the total plated after 1 hour. Notice that the mass is *perfectly* linear in $I$ — doubling the current exactly doubles the gold, as Faraday's first law requires.

Worked examples

Example 1: Gold-plating a pair of earrings at Tanishq

A pair of silver earrings (total surface area 12 cm²) is gold-plated in a cyanide bath. The current is kept at 0.50 A and the plating continues for 25 minutes. Gold has atomic mass 197 g/mol and the gold ion in this bath is \text{Au}^{3+}. Gold's density is 19.3 g/cm³. Find the total mass of gold deposited and the average thickness of the gold layer.

Gold plating schematic. $\text{Au}^{3+}$ ions migrate from the gold anode, through the bath, onto the silver earring at the cathode.

Step 1. Total charge.

Q \;=\; I\,t \;=\; 0.50 \times (25 \times 60) \;=\; 0.50 \times 1500 \;=\; 750\ \text{C}.

Step 2. Mass of gold deposited.

m \;=\; \frac{M}{zF}\,Q \;=\; \frac{0.197}{3 \times 96{,}485} \times 750 \;=\; \frac{0.197 \times 750}{289{,}455} \;\approx\; 5.1\times10^{-4}\ \text{kg} \;=\; 0.51\ \text{g}.

Why: plug into Faraday's combined law directly. The molar mass of gold is 0.197 kg/mol, the valency of \text{Au}^{3+} is 3, and F = 96{,}485 C/mol.

Step 3. Volume of gold.

V \;=\; \frac{m}{\rho_{\text{Au}}} \;=\; \frac{0.51}{19.3}\ \text{cm}^{3} \;\approx\; 0.0264\ \text{cm}^{3}.

Step 4. Average thickness over 12 cm² of surface.

d \;=\; \frac{V}{A} \;=\; \frac{0.0264\ \text{cm}^{3}}{12\ \text{cm}^{2}} \;=\; 2.2\times10^{-3}\ \text{cm} \;=\; 22\ \mu\text{m}.

Why: the gold forms a roughly uniform layer on all surfaces, so dividing the volume by the total surface area gives the mean thickness. In reality, the thickness is somewhat non-uniform — sharper edges plate thicker — but 22 µm is the area-averaged value.

Result. The earrings gain 510 mg of gold in 25 minutes at 0.5 A, plating to an average thickness of 22 micrometres — a sturdy plating by jewellery standards (jewellers typically aim for 5 to 25 µm depending on the piece). At today's gold price of about ₹6000 per gram, that is ₹3060 worth of gold per pair.

What this shows. A jeweller running a bath at a fixed current and a fixed time gets a reproducible, measured mass of gold on every piece. This predictability — not the chemistry, but the physics of Faraday's law — is what makes electroplating an industrial process rather than an art.

Example 2: Hydrogen from water electrolysis

A water-electrolysis cell, operated at 2.1 V and drawing 15 A for 1 hour, produces hydrogen gas at the cathode by the reaction 2\text{H}^{+} + 2e^{-} \to \text{H}_{2}. Find (a) the mass of hydrogen produced and (b) the volume of hydrogen at STP (0 °C, 1 atm).

Step 1. Total charge.

Q \;=\; 15 \times 3600 \;=\; 54{,}000\ \text{C}.

Step 2. Mass of hydrogen (monoatomic mass 1.008 g/mol, released as H₂ with z = 1 per H⁺).

m \;=\; \frac{M_{H}}{zF}\,Q \;=\; \frac{0.001008}{1 \times 96{,}485} \times 54{,}000 \;=\; 5.64\times10^{-4}\ \text{kg} \;=\; 0.564\ \text{g}.

Why: per H⁺ ion, one electron is needed to make half an H₂ molecule. The molar mass that matters is the atomic mass of H (not H₂), and the valency is 1.

Step 3. Number of moles.

n_{H_{2}} \;=\; \frac{m}{2M_{H}} \;=\; \frac{0.564}{2 \times 1.008} \;\approx\; 0.280\ \text{mol}.

Why: two atoms of hydrogen combine into one molecule of H₂. Divide the moles of H atoms by 2 to get moles of H₂ gas.

Step 4. Volume at STP (1 mol of ideal gas at STP = 22.4 L).

V \;=\; n\,V_{m} \;=\; 0.280 \times 22.4 \;\approx\; 6.27\ \text{L}.

Result. In 1 hour at 15 A, the cell makes 0.56 g of hydrogen — about 6.3 L at STP. The energy input was VIt = 2.1 \times 15 \times 3600 = 1.13\times10^{5} J, or 1.13\times10^{5}/(3.6\times10^{6}) = 0.031 kWh of electricity. At ₹5 per kWh, the electricity cost is ₹0.16 for 6.3 L of H₂ — the basic arithmetic of green hydrogen, before you add overpotential losses (real industrial cells run at 2.0 V, not the thermodynamic 1.23 V, because of kinetic barriers). An Indian green-hydrogen plant's economics is entirely dominated by whether the electricity is cheap.

What this shows. Faraday's law tells you exactly how much hydrogen per coulomb, and therefore exactly how much per rupee of electricity. It is the first-pass model for sizing any electrolyser plant.

Example 3: NALCO's aluminium smelter — a single pot in Angul

NALCO's Hall-Héroult aluminium smelter in Angul, Odisha runs each electrolytic pot at about 350 kA and 4.3 V. The ion being reduced is \text{Al}^{3+} (molar mass 27 g/mol). Assuming 90% current efficiency (10% of the current is wasted in side reactions), how many kilograms of aluminium does one pot produce per day?

Simplified Hall-Héroult aluminium-reduction pot. The carbon-lined steel pot acts as the cathode; carbon blocks dip in from above as the anodes; molten cryolite (Na₃AlF₆) dissolves alumina (Al₂O₃); aluminium ions are reduced at the cathode and collect as a liquid pool at the bottom, tapped out once per day.

Step 1. Total charge per day (86 400 s).

Q \;=\; I\,t \;=\; 350{,}000 \times 86{,}400 \;=\; 3.024 \times 10^{10}\ \text{C}.

Step 2. Apply the 90% efficiency: the charge that actually reduces Al³⁺ is

Q_{\text{eff}} \;=\; 0.9 \times 3.024 \times 10^{10} \;=\; 2.722 \times 10^{10}\ \text{C}.

Why: in a real Hall-Héroult cell, about 10% of the current is consumed by side reactions — chiefly the re-oxidation of molten aluminium by CO₂ gas generated at the anode. Only 90% of the coulombs pay for product.

Step 3. Mass of aluminium.

m \;=\; \frac{M}{zF}\,Q_{\text{eff}} \;=\; \frac{0.027}{3 \times 96{,}485} \times 2.722 \times 10^{10} \;=\; 2540\ \text{kg}.

Step 4. Electrical energy used.

E \;=\; V\,I\,t \;=\; 4.3 \times 350{,}000 \times 86{,}400 \;=\; 1.30 \times 10^{11}\ \text{J} \;=\; 36{,}130\ \text{kWh}.

Per kg of aluminium produced:

\frac{E}{m} \;=\; \frac{36{,}130}{2540} \;\approx\; 14.2\ \text{kWh/kg}.

Result. One NALCO pot produces 2.54 tonnes of aluminium per day and consumes 14.2 kWh per kilogram of output. NALCO operates about 960 pots at Angul, giving a daily output of roughly 2440 tonnes — or 890 000 tonnes per year.

What this shows. Primary aluminium is concentrated electricity in metallic form. The reason Indian smelters cluster near captive coal-fired power plants (NALCO Angul, Vedanta Jharsuguda, Hindalco Hirakud) is not transportation of bauxite but transportation of electricity — the economics are entirely set by power price, and Faraday's law is the conversion factor between coulombs and kilograms of metal.

Common confusions

"The ions carry the current in the external circuit." No — in the external metallic wire, the current is carried by free electrons. Inside the electrolyte, the current is carried by the motion of positive ions in one direction and negative ions in the other. The two currents are connected at the electrode surfaces by the chemistry: electrons in the wire become ions in the solution (or vice versa) via the electrode reaction.
"Faraday's first law says m \propto I." Not quite — it says m \propto It, i.e., mass is proportional to charge. If you double the current but halve the time, the mass is unchanged. What matters is coulombs, not amperes.
"Faraday's constant is the charge of one ion." No. F is the charge of one mole of electrons. The charge of one ion with valency z is ze (where e \approx 1.6\times 10^{-19} C). The Faraday is e scaled up by Avogadro's number.
"The deposition rate is the same at every point on the cathode." Only on average. In reality, sharp edges and points have higher local current density and plate thicker; deep recesses plate thinner. This is why well-designed plating baths use baffles, auxiliary anodes, and agitation to even out the current distribution — but Faraday's law always gives the average.
"Water electrolysis works at 1.23 V." Thermodynamically yes, but practically no. Real cells need 1.8-2.1 V to overcome kinetic barriers at the electrodes (the "overpotential"). The extra voltage beyond 1.23 V appears as Joule heat in the cell and is lost — which is why efficient electrolysers are still an active engineering problem in 2026.
"The ammeter reading times the time gives the charge through the cathode." Yes, if current efficiency is 100%. In practice, side reactions (corrosion of the anode, oxidation of the solvent, gas bubbles that short-circuit part of the cell) mean that only a fraction of the current actually deposits product. Commercial gold plating is usually > 95% efficient; aluminium smelters run at about 90%; some complex anodisation processes are only 60%.

If you came here for exam-level electrolysis, you have what you need. What follows covers (a) the thermodynamic minimum voltage of a cell and where kinetic overpotentials come from, (b) the Hall-Héroult process in more detail, (c) Faraday's law in biology (nerve impulses), and (d) the link to Avogadro's number.

Minimum voltage of an electrolysis cell

The minimum voltage required to drive an electrolysis reaction is set by thermodynamics. For any cell reaction,

\Delta G \;=\; -n\,F\,E_{\text{cell}}

where \Delta G is the Gibbs free energy change of the reaction, n is the number of electrons transferred per formula unit, and E_{\text{cell}} is the standard cell potential. If \Delta G > 0 (the reaction is non-spontaneous), then E_{\text{cell}} < 0, and to drive the reaction you must apply a voltage of at least -E_{\text{cell}}.

For water electrolysis, \Delta G = +237.1 kJ/mol (at 25 °C, 1 atm), and n = 2 per water molecule split. So

E_{\min} \;=\; \frac{\Delta G}{nF} \;=\; \frac{237{,}100}{2 \times 96{,}485} \;=\; 1.229\ \text{V}.

This is the minimum voltage below which water will not electrolyse at all. In practice, you need more than this because of overpotentials at the electrodes — kinetic barriers to the electron-transfer steps. For oxygen evolution at a platinum anode, the overpotential is about 0.3 V; for hydrogen evolution at a platinum cathode, about 0.05 V. Total practical voltage: about 1.6 V, and with real cathode and anode materials (and with a finite cell current), the number climbs to 1.8-2.1 V. The difference between 1.23 V (thermodynamic) and 2.0 V (actual) is dissipated as heat, making the electrolyser hot.

The Hall-Héroult process

Aluminium's high chemical stability makes it impossible to reduce economically from an aqueous solution — the water would preferentially electrolyse first, giving hydrogen, not aluminium. The Hall-Héroult process (1886, independently discovered by Charles Martin Hall in the US and Paul Héroult in France) uses instead a molten salt electrolyte: sodium hexafluoroaluminate (cryolite, Na₃AlF₆) with alumina (Al₂O₃) dissolved in it at 950 °C. There is no water. At the cathode,

\text{Al}^{3+} + 3e^{-} \longrightarrow \text{Al (liquid)}.

At the carbon anode,

2\text{O}^{2-} + \text{C (solid)} \longrightarrow \text{CO}_{2} + 4e^{-}.

The carbon anodes are consumed — every tonne of aluminium produced chews up about 400 kg of carbon anode — and must be replaced every few weeks. This is a feature, not a bug: the carbon contributes part of the reduction energy, lowering the required voltage from 2.23 V (if only electricity were used) to about 1.8 V.

The thermodynamic minimum for Al production is 1.8 V, and the actual 4.3 V used in industrial pots is the overpotential plus the IR drop across the molten-salt resistance. The difference between 1.8 V and 4.3 V — about 2.5 V — is Joule-dissipated in the pot, which is good news, because that heat is precisely what keeps the cryolite molten. A Hall-Héroult pot is self-heating, and the 350 kA current must be on continuously — let a pot cool and the cryolite solidifies, and the only way to restart is to break up the solid mess.

Faraday's law in nerve conduction

Your nervous system sends signals by temporarily opening ion channels in the membranes of nerve cells. When a sodium channel opens, Na⁺ ions flow into the cell, carrying charge. The same Faraday-law arithmetic applies: the charge that flows in is Q = (n_{\text{ions}})\cdot e, and the number of ions is set by the number of channels open and the duration. A typical action potential involves about 10^{6} Na⁺ ions flowing into each square micrometre of membrane over about 1 ms — a current density of about 10 mA/cm².

The reverse process (pumping Na⁺ back out against the electrochemical gradient) uses the sodium-potassium ATPase pump. It runs at 100 cycles per second per pump, each cycle moving three Na⁺ out and two K⁺ in. Faraday's law tells you the electrical current this pump generates; it does not by itself tell you the rate of pumping (that is biochemistry, set by ATP concentration and temperature). But once you know the rate, Faraday gives you the ion flux.

Avogadro's number from electrolysis

One beautiful experiment — still sometimes done in undergraduate labs — determines Avogadro's number from Faraday's law. Here is how:

Electrolyse a solution of known composition (say, copper sulphate) at a measured current for a measured time.
Weigh the mass of copper deposited.
From Faraday's law, m = M\,I\,t/(zF), so F = M\,I\,t/(zm). You now know F to about 0.1%.
Measure the elementary charge e by a Millikan-style experiment (oil drop with a known field balancing gravity).
Avogadro's number is N_{A} = F/e.

This is the historical route by which Faraday's constant and Avogadro's number were nailed to 3-4 significant figures by the end of the 19th century, long before atoms could be "seen" directly. The method is still accurate to better than 1 part in 10⁵ and competes with X-ray silicon-lattice-density measurements as one of the most precise ways of measuring N_{A}.

The link to redox reactions

Every electrolysis is a redox reaction: reduction at the cathode (gaining electrons) and oxidation at the anode (losing electrons). The valency z in Faraday's law is exactly the number of electrons transferred per ion in the electrode reaction. For Cu²⁺ → Cu, z=2. For Au³⁺ → Au, z=3. For 2H⁺ → H₂, z=1 per H⁺ (but 2 per H₂ molecule). Writing out the half-reaction forces you to get z right, which is the most common source of factor-of-two errors in exam problems.

Spontaneous redox reactions (a zinc-copper voltaic cell, say) run in the opposite direction: they generate an EMF, not consume it. Electrolysis is redox driven uphill by an external source. Faraday's law is the same in both directions — it just tells you how many atoms cross per coulomb, regardless of whether the coulombs are supplied by a battery or supplied to one.

Where this leads next

Heating Effect of Current — Joule's Law — the I^{2}R Joule heat in an electrolyte is the reason overpotentials matter, and why electrolysers run hot.
Thermoelectricity — the Seebeck, Peltier, and Thomson effects at metallic junctions, which are the thermal siblings of electrolytic charge transfer.
EMF and Internal Resistance — the reverse of electrolysis: spontaneous redox reactions in a voltaic cell, and the V = \varepsilon - Ir that governs them.
Electric Current, Drift Velocity and Mobility — the microscopic picture of ionic conduction in solutions, the direct analogue of electronic conduction in metals.
Electrical Energy and Power — the energy bookkeeping that decides the rupee cost of a kilogram of aluminium or a kilogram of gold plated.