This deep dive rebuilds the parent topic's worked examples as a visual derivation. Everything rests on ideas from 2.7.01-Oxidation-numbers-and-redox-definition (what oxidation is) and 2.7.04-Galvanic-cells-and-cell-notation (why anode and cathode are separated).
WHAT. A wire carrying current is a pipe with tiny charged marbles — electrons — flowing through it. Current, written I, is simply how much charge flows past one spot each second.
WHY this idea first. Before we can connect a battery's chemistry to a number of amps, we must know what an amp is made of. An amp is not magic — it is a counting rate of electrons.
PICTURE. Look at the cross-section of the wire below. In one second, a whole crowd of electrons streams past the dashed line.
Q=amps=C/sI×secondst
Q on the left is the total charge that passed.
I tells you the rate (coulombs each second).
t is how long you let it flow. Multiply rate × time → total. Exactly like "60 km/h for 2 h = 120 km."
WHAT. We divide the total charge Q by the charge of one electron e. The answer is a plain integer: how many electrons walked through.
WHY divide? Because Q is measured in coulombs, but each electron carries only e coulombs. Asking "how many electrons?" is the same as asking "how many 1.6×10−19 C packets fit inside Q?" — and "how many of X fit inside Y" is always a division.
PICTURE. Imagine pouring the big bucket of charge Q into tiny electron-sized cups. The number of cups you fill is your electron count.
Ne−=eQ=eIt
Ne− = number of electrons (just a count, no units).
WHAT. Each metal atom releases a fixed number of electrons when it oxidizes. For zinc that number is 2, because zinc becomes Zn2+:
Zn(s)→Zn2+(aq)+two electrons per atom2e−
WHY we can't skip this. Electrons are the currency, but atoms are the goods. One zinc atom "pays" with 2 electrons. So the number of atoms is not the number of electrons — it is the electron count divided by the electrons-per-atom. Miss this and every mass answer is off by a factor of 2 (or 3, for aluminium, etc.).
PICTURE. Below, each zinc atom stands at a turnstile handing over exactly 2 electrons before it can leave the metal.
Natoms=zNe−,z=electrons released per atom
Natoms = number of metal atoms consumed.
z = the charge number — read it straight off the half-reaction (z=2 for Zn, z=2 for Pb and PbO₂, z=1 for Li).
WHAT. Atoms are too many to count one by one, so chemists bundle them into a mole — a fixed pack of NA=6.022×1023 particles. Dividing our atom count by NA gives moles.
WHY a mole? Because the periodic table gives us mass per mole, not mass per atom. To use those handy molar masses, we must first speak in moles. The mole is the translator between "number of particles" and "grams on a scale."
PICTURE. A giant crate labelled "1 mole = 6.022×1023 atoms." We scoop our atoms into crates; the number of crates is our moles.
n=NANatoms
n = moles of substance.
NA = Avogadro's number=6.022×1023 — the pack size.
WHAT. Nothing new — we feed the lead-acid cathode reaction into the very same chain. The half-reaction
PbO2+4H++SO42−+2e−→PbSO4+2H2O
tells us z=2 electrons per PbO2, and MPbO2=239.2g/mol.
WHY show a second case. To prove the recipe is universal: pick the half-reaction, read off z, plug into m=zFItM. Only the numbers change.
PICTURE. The starter motor gulps 200A for 3s — a huge current for a tiny time — funnelled through the identical pipeline.
\;\Rightarrow\;
m = \frac{600}{2\times 96485}\times 239.2 \approx 0.744\ \text{g of PbO}_2.$$
Out of ~1 kg of PbO₂ per cell, cranking the engine burns **under 0.1\%** — which is exactly why your car starts again tomorrow.
> [!intuition] Why track only PbO₂, when *both* electrodes change mass?
> During discharge the **anode** also reacts ($\text{Pb} + \text{SO}_4^{2-} \to \text{PbSO}_4 + 2e^-$), so *both* plates gain a white $\text{PbSO}_4$ coat and the acid is used up too. So why did we compute only the PbO₂? Because **the same electrons flow through every part of the series circuit** — the count $n_{e^-}=Q/F$ is *identical* at both electrodes. Once you know it for one half-reaction, the others follow by their own $z$ and $M$: the anode consumes $0.00311\,\text{mol}$ of Pb ($M=207.2$, so $\approx 0.644\,\text{g}$), and $2\times0.00311\,\text{mol}$ of $\text{H}_2\text{SO}_4$ vanishes. We track **one reactant** simply because the question asked for one — the belt handles any of them from the same charge $Q$.
---
## Step 7 — Edge and degenerate cases (never leave a gap)
**WHAT & WHY.** A recipe is only trustworthy if it survives the weird inputs. Here is every corner:
![[deepdives/dd-chemistry-2.7.08-d2-s07.png]]
- **$I = 0$ (switch off).** Then $Q = 0 \Rightarrow m = 0$. No current, no chemistry consumed. The battery just *sits* — this is why an unused cell still slowly dies (side reactions), but the *useful* mass loss is zero.
- **$z = 1$ (lithium-ion).** $\text{Li} \to \text{Li}^+ + 1e^-$. Because we divide by a *smaller* $z$, the **same charge moves more atoms** — one of the reasons Li stores so much per gram. See [[2.7.05-Standard-electrode-potentials]] for why Li's potential is so favourable too.
- **Recharging (secondary cell).** During charge the electrons are *forced backwards*, so $m$ is mass **regenerated**, not consumed. The arithmetic is identical; only the sign of "gain vs loss" flips. Primary cells (dry cell) can't do this — the products won't reverse.
- **Voltage vs mass are separate questions.** This whole page computes **how much stuff** (mass) reacts. It says *nothing* about the **voltage** — that comes from electrode potentials, $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$. A cell can be near-empty (voltage sagging) yet still push current. Mass = *how much*; voltage = *how hard*.
- **"Theoretical vs real."** Our $m$ is the *ideal* amount. Real cells suffer overpotential, incomplete reaction, and side products, so a real flashlight runs ~8 h not the ideal ~11 h. The formula sets the *ceiling*.
---
## The one-picture summary
![[deepdives/dd-chemistry-2.7.08-d2-s08.png]]
The entire derivation is one conveyor belt:
$$\boxed{\,I\,t \;\xrightarrow[\ \div F\ ]{}\; n_{e^-} \;\xrightarrow[\ \div z\ ]{}\; n_{\text{substance}} \;\xrightarrow[\ \times M\ ]{}\; m\,}$$
```mermaid
flowchart LR
A["Current I times time t"] -->|"Q equals I t"| B["Charge Q in coulombs"]
B -->|"divide by F"| C["moles of electrons"]
C -->|"divide by z"| D["moles of substance"]
D -->|"times M"| E["mass in grams"]
```
> [!recall]- Feynman retelling — say it back in plain words
> A wire is a pipe of electrons, and **current** just counts how many pass per second (by convention the current arrow points the *opposite* way to the electrons, but the count is the same). Multiply that rate by the time and you know the **total charge** that flowed. Every mole of electrons is exactly **96,485 coulombs** — that's the Faraday constant — so dividing charge by $F$ tells you **how many moles of electrons** moved. But each atom hands over a fixed number of electrons ($z$: two for zinc and lead, one for lithium), so I divide by $z$ to get **moles of the actual chemical**. Finally I multiply by the **molar mass** off the periodic table to land on **grams**. Amps in, grams out — and it's the *same* four-step belt whether it's a flashlight eating zinc, a car burning lead dioxide, or a phone shuttling lithium. The same electrons pass through *every* electrode, so once I know the charge I can find the mass change at *any* plate by using its own $z$ and $M$. Switch it off and $I=0$ so nothing is consumed; run it backwards and you *regenerate* the reactants instead. That, and nothing more, is battery arithmetic.
> [!mnemonic] Remember the belt
> **"Q ÷ F ÷ z × M"** — *"Quit Fussing, Zebras Munch"* — Charge, Faraday, charge-number, Molar mass, in that order.
Related: [[3.1.12-Gibs-free-energy-and-spontaneity]] connects the *voltage* side to energy, and [[2.7.09-Corosion-and-prevention]] is the same electron-counting applied to unwanted oxidation.
