Electrolysis — Faraday's laws (m = ZIt), industrial electrolysis (NaCl, Al): Visual walkthrough

We will use, in order: what charge is, what a coulomb counts, what Faraday's constant does, how ions "spend" electrons, and finally how grams appear. Nothing before it is drawn.

Step 1 — What is charge, and what does an ammeter actually count?

WHAT. Electric current is not a mysterious fluid. It is a counting rate: how many units of electric charge slide past a point each second. The unit of charge is the coulomb (symbol $C$ ). One coulomb is just a fixed number of electrons' worth of charge — a bag of a certain size.

WHY start here. Every later step is going to trade one currency for another (seconds → coulombs → electrons → atoms → grams). We must first pin down the first currency so no symbol sneaks in undefined.

PICTURE. Imagine a wire as a pipe. Every second, a fixed number of little charge-balls flow past the dashed gate. The ammeter reads how fast they pass.

Figure — Electrolysis — Faraday's laws (m = ZIt), industrial electrolysis (NaCl, Al)

So $I \cdot t$ = (balls per second) $\times$ (seconds) = total balls of charge. That is the whole meaning of Step 1.

Step 2 — From coulombs to a count of electrons

WHAT. A coulomb is a chunk of charge, but reactions happen one electron at a time. So we need to convert "coulombs" into "how many moles of electrons". The bridge is Faraday's constant, $F \approx 96, 500 C/mol$ : the charge carried by one mole of electrons.

WHY this number and not another. We need a conversion factor that turns coulombs into moles of electrons — the same way "12 g/mol" turns grams of carbon into moles. $F$ is exactly that factor, but for electron charge. Nothing else in chemistry connects electricity to counting; that is why $F$ must appear.

PICTURE. A big pile of coulombs on the left; we scoop it into standard "mole-of-electrons" buckets, each bucket holding $96, 500 C$ .

Step 3 — How an ion "spends" electrons to become an atom

WHAT. At the cathode, a metal ion $M^{n +}$ grabs electrons and becomes a neutral atom. The little superscript number $n$ tells you how many electrons that ion is short by, so that is exactly how many it must catch.

$M^{n +} + n e^{-} \to M$

WHY this matters. Not every atom costs the same number of electrons. Silver ( $Ag^{+}$ ) costs $1$ ; copper ( $Cu^{2 +}$ ) costs $2$ ; aluminium ( $Al^{3 +}$ ) costs $3$ . The same pile of electrons therefore builds fewer atoms of a high- $n$ metal. This single fact is the heart of Faraday's Second law.

PICTURE. Three checkout lanes. One electron buys one $Ag$ ; two electrons buy one $Cu$ ; three buy one $Al$ .

Step 4 — From a count of atoms to a mass on the scale

WHAT. You cannot weigh "moles" on a balance; you weigh grams. The molar mass $M$ (grams per mole) turns our atom-count into a real number of grams.

WHY. This is the final currency change: counting-units → laboratory-units. Molar mass is the only bridge that does it, so it must enter here.

PICTURE. A pan balance: the pile of counted atoms on the left, its weight in grams read off on the right, with $M$ as the multiplier printed on the beam.

Step 5 — Collapsing everything into one constant $Z$

WHAT. Look at $m = \frac{M}{n F} I t$ . For a given substance, $M$ , $n$ and $F$ never change during the run — only $I$ and $t$ do. So we bundle the three fixed things into one number, the electrochemical equivalent $Z$ .

WHY. It reveals the true shape of the law: mass is simply proportional to $I t$ . Everything chemical about the substance hides inside a single constant. This is Faraday's First law made bare.

PICTURE. The three fixed pieces $M$ , $n$ , $F$ fuse into one labelled brick $Z$ , leaving the clean line $m = Z I t$ .

Step 6 — The edge cases you must never trip over

WHAT & WHY. A formula you trust only in the "nice" case is a trap. Check the corners:

PICTURE. Three tiny scenarios: no current, doubled charge, and a high- $n$ metal.

The one-picture summary

Here is the whole chain of currency-swaps on a single track — seconds to coulombs to electrons to atoms to grams — with the constants that ride each arrow.

Recall Feynman retelling — say it back in plain words

An ammeter just counts charge going by per second, so multiplying current by time gives the total charge in coulombs. Charge is spent one electron at a time, and one bag of $96, 500$ coulombs is exactly one mole of electrons — that is Faraday's constant. Each metal ion is short by $n$ electrons, so it takes $n$ electrons to build one atom; divide the electrons by $n$ to get moles of metal. Multiply by molar mass to turn that count into grams. Bundle the three unchanging pieces — molar mass, $n$ , and $F$ — into one number $Z$ , and the whole thing collapses to $m = Z I t$ : mass is simply proportional to how much electricity you pushed through. No charge, no metal. Double the charge, double the metal. And because different metals cost different numbers of electrons per atom, the same charge builds different masses — which is the second law hiding inside the first.

Recall Quick self-test

Why is $t$ in seconds and not minutes? ::: Because $I$ (amperes) means coulombs per second; $Q = I t$ only gives coulombs if $t$ is in seconds. What does dividing $Q$ by $F$ give you? ::: The number of moles of electrons that flowed. Why do we divide by $n$ ? ::: Because each atom of $M^{n +}$ needs $n$ electrons, so $n$ electrons build only one atom. For the same charge, which deposits more mass, $Ag^{+}$ or $Al^{3 +}$ ? ::: Silver — its chemical equivalent $M / n = 108$ dwarfs aluminium's $9$ .

See also: Redox Reactions and Balancing · Galvanic Cells and Standard Electrode Potentials · Industrial Chemistry