Surface treatments — anodising, plasma spraying, vapour deposition: Visual walkthrough

This page derives the boxed result from the parent note slowly and visually. If a symbol shows up, we have already drawn it.

Step 1 — What is "charge", and why do we count electrons?

WHAT. We define charge as the running total of electric "stuff" that has flowed.

WHY. In anodising, every bit of oxide is built by electrons leaving aluminium atoms. So if we can count electrons, we can count oxide. Charge is just electrons-in-bulk: count the charge, count the electrons.

PICTURE. Look at the figure: current is a steady stream of charge (blue dots) crossing a line. The area of the blue rectangle — height $I$ , width $t$ — is the total charge $Q$ . Widen the rectangle (more time) or raise it (more current) and you enclose more charge.

Figure — Surface treatments — anodising, plasma spraying, vapour deposition

Step 2 — From electrons to moles of oxide: the half-reaction

WHAT. This chemical equation is a counting recipe. It says: to build one formula-unit of $A l_{2} O_{3}$ , exactly 6 electrons must be stripped away.

WHY this number 6? Each aluminium atom goes from charge $0$ to $+ 3$ , losing 3 electrons. There are two Al atoms per $A l_{2} O_{3}$ , so $2 \times 3 = 6$ electrons per oxide unit. This "6" is not magic — it is why a 6 sits in the denominator of the final formula. Change the metal and you change this number.

PICTURE. The figure shows a little "electron budget": 6 electrons flow out on the left, and in exchange one oxide block is stacked on the right. It is a fixed exchange rate — 6 electrons buy 1 oxide.

Step 3 — Faraday's constant: converting coulombs into moles of electrons

WHAT. We have $Q$ in coulombs (Step 1). We want moles of electrons. Divide by $F$ : $n_{e^{-}} = \frac{Q}{F} [\frac{C}{C/mol} = mol]$

WHY this tool and not a plain number? A coulomb is a huge pile of individual electrons ( $\sim 6.24 \times 1 0^{18}$ of them). Counting one at a time is hopeless. $F$ lets us jump straight to moles of electrons — the unit chemistry actually uses to weigh things. This is why we invoke Faraday's Laws of Electrolysis here.

PICTURE. The figure is a two-column converter: pour coulombs in the top, divide by $F$ , moles of electrons come out the bottom. A worked tick shows $96485 C \to 1 mol e^{-}$ .

Step 4 — Chaining the two conversions: moles of oxide

WHAT. We stack two conversions: coulombs $\to$ moles of electrons (Step 3), then moles of electrons $\to$ moles of oxide (Step 2 said "divide by 6").

WHY the 6 lands in the denominator. Each oxide unit costs 6 electrons, so the number of oxide units is the electron count divided by 6. More electrons-per-product means fewer products per coulomb — hence 6 sits below the line.

But not every electron builds oxide. Some current is "wasted" — for example evolving oxygen gas or side reactions. We call the useful fraction the ==current efficiency $η$ == (a number between 0 and 1). Multiply by it: $n_{oxide} = \frac{Q}{6 F} η$

$η = 1$ → every electron builds oxide (ideal).
$η = 0.6$ → only $60%$ does; the rest is lost.

PICTURE. A pipe of electrons splits: a fraction $η$ goes into the "oxide-building" branch, the rest $(1 - η)$ leaks off as gas. Only the $η$ branch counts.

Step 5 — From moles to mass, then to volume

WHAT. Turn moles of oxide into a physical chunk of oxide — first its mass, then how much space it fills. $m_{oxide} = n_{oxide} M = \frac{M Q η}{6 F} V_{oxide} = \frac{m _{oxide}}{ρ} = \frac{M Q η}{6 F ρ}$

WHY these two tools. Anodising doesn't hand us moles — it hands us a layer you can measure with a micrometer. Thickness is a length, and length comes from volume ÷ area. To get volume we must first get mass ( $\times M$ ), then convert mass to volume ( $\div ρ$ ).

PICTURE. A ladder of three rungs — moles → mass (×M) → volume (÷ρ) — each rung labelled with the operation that climbs it.

Step 6 — Spreading the volume over the area to get thickness

WHAT. The oxide sits as a thin flat film on the part's surface. Volume = area × thickness, so thickness = volume ÷ area. We used $Q = I t$ from Step 1 to write charge as current×time.

WHY divide by $A$ . The same volume of oxide spread over a bigger face makes a thinner film — like the same blob of butter over a bigger slice of toast. So area lives in the denominator.

Term-by-term, right where each appears:

$M$ — heavier oxide per mole → thicker film → top.
$I t$ — more charge → more electrons → more oxide → top.
$η$ — more of that charge is useful → top.
$6$ — 6 electrons per oxide unit; costly product → bottom.
$F$ — big charge-per-mole → many coulombs per mole → bottom.
$ρ$ — denser oxide packs into less volume → thinner → bottom.
$A$ — larger surface dilutes the film → bottom.

PICTURE. A slab of known volume being flattened across a rectangular plate; a caliper reads off the height $d$ . Doubling $A$ visibly halves the caliper reading.

Step 7 — Plug in the worked numbers

WHAT. We invert the boxed formula for the one unknown we can dial (time) and evaluate.

WHY invert? The plant operator sets $I$ , knows $A$ , and wants a target $d$ . The only free variable is the clock. So we solve for $t$ .

PICTURE. A "recipe card": knobs we set (blue) feed into the formula, the timer (orange) is the output. A bar shows $\approx 378 s$ .

Step 8 — Edge and degenerate cases (never let the reader fall off a cliff)

WHAT. We check the formula at its extremes so it never surprises you.

Case	What the formula does	Does it make physical sense?
$η \to 0$	$d \to 0$	Yes — no useful current, no oxide, ever.
$t = 0$	$d = 0$	Yes — nothing has flowed yet.
$A \to$ large	$d \to$ small	Yes — same oxide spread thinner.
$I$ doubled	$d$ doubled (same $t$ )	Yes — twice the charge, twice the oxide.
$η > 1$	impossible	Efficiency cannot exceed 1; a flag that inputs are wrong.

WHY include this. A formula you can't stress-test is a formula you don't understand. Each limit above matches intuition — that is our confidence check.

PICTURE. Thickness-vs-time: the ideal straight line (constant $η$ ) versus the real curve that bends over as $η$ decays.

The one-picture summary

Everything above collapses into a single conversion chain: current × time → charge → electrons → oxide moles → mass → volume → thickness, with $η$ , $6$ , $F$ , $M$ , $ρ$ , $A$ each entering at exactly one link.

Recall Feynman: the whole walkthrough in plain words

We wanted the thickness of the crust we grow on aluminium. Here's the story with no symbols: electricity is a stream of tiny charged beads; run it for a while and you've delivered a known pile of beads (that's charge = current × time). Chemistry has a fixed price list: it takes exactly six beads to build one brick of oxide, and Faraday's number tells us how many beads are in a "mole-sized bag." So: divide the pile by the bag size to get bags of beads, divide by six to get bricks of oxide, but throw away the fraction of beads that leaked off as gas (that's efficiency $η$ ). Now weigh those bricks (multiply by molar mass), see how much space they take (divide by density), and finally smear that lump of oxide flat over the part's face — spread over a bigger face, it's thinner (divide by area). What pops out is $d = M I t η / (6 F ρ A)$ . Every letter is just one of those honest steps.

Connections

Faraday's Laws of Electrolysis — supplies $F$ and the charge-to-moles conversion (Step 3)
Corrosion and Passivation — the oxide we grow is engineered passivation
Aluminium Alloys in Aerospace — the substrate whose surface we are thickening
Adhesion and Surface Roughness — porous anodic oxide keys mechanically to later coats
Parent topic

Charge from steady current

Q = I \times t

(coulombs)

Electrons per oxide unit and why it is in the denominator

6, because each oxide unit costs 6 electrons so oxide count = electrons ÷ 6

Role of Faraday's constant in the derivation

converts coulombs to moles of electrons (

n_{e} = Q / F

)

What

η

corrects for

fraction of charge that actually builds oxide (rest lost as gas)

Why area

A

sits in the denominator

same oxide volume over larger area gives a thinner film

Time to grow 20 µm (worked case)

\approx 378

\approx 6.3

min

Formula for anodised thickness

d = M I t η / (6 F ρ A)