Preparation, properties, uses of dihydrogen: Visual walkthrough

This is the picture-story behind one number on the parent page: when you push a current $I$ through water for a time $t$ , exactly how much dihydrogen ( $H_{2}$ ) bubbles up? We will build the machinery of Electrolysis and Faraday's Laws from nothing — no formula assumed — and re-derive Worked Example 3 (5 A for 2 hours) so you can see every gram appear.

We are looking closely at one method from the parent: Preparation, properties, uses of dihydrogen → Electrolysis of Water.

Step 1 — What is a "current", really? Counting the marching electrons

WHAT. Before any chemistry, we define the two words on our meter: charge and current.

An electron is a tiny particle carrying a fixed lump of negative electric charge. We measure charge in coulombs (symbol $C$ ). One electron carries an unimaginably small charge, so instead of counting electrons one by one, we count coulombs.

Current $I$ is how fast charge flows past a point — coulombs per second. The unit is the ampere (amp, $A$ ): $1 A = 1 C/s$ .

WHY multiply? If water flows out of a tap at $I$ litres every second, after $t$ seconds you have $I \times t$ litres in the bucket. Charge is identical: rate $\times$ time $=$ total. We multiply because both a faster flow and a longer wait independently add more.

PICTURE. The wire is a pipe; each chalk-white dot is a packet of charge drifting past the meter. Count the dots that cross the dashed line in time $t$ — that count is $Q$ .

Figure — Preparation, properties, uses of dihydrogen

Step 2 — Where the electrons go: the cathode reaction

WHAT. We now watch what the arriving electrons do. In water electrolysis the negative electrode is the cathode (where reduction — gain of electrons — happens; see Redox Reactions). Water molecules queue up and collect electrons:

$2 water molecules 2 H_{2} O + 2 electrons in 2 e^{-} ⟶ 1 gas molecule out H_{2} + 2 leftover hydroxides 2 OH^{-}$

$2 H_{2} O$ ::: the raw material queuing at the plate.
$2 e^{-}$ ::: two electrons must be absorbed — this "2" is the hero of the whole page.
$H_{2}$ ::: exactly one molecule of dihydrogen is born.
$2 OH^{-}$ ::: hydroxide ions left behind (they make the solution basic near the cathode).

WHY this reaction and not another? Hydrogen in water is in the $+ 1$ state; to become neutral gas $H_{2}$ each H atom must gain one electron. Two H atoms in one $H_{2}$ molecule $\Rightarrow$ 2 electrons per molecule. This count is fixed by chemistry, not by us.

PICTURE. Two electrons flow into the plate; one $H_{2}$ bubble floats off. The little "×2" tag on the electron arrow is the exchange rate we are about to use.

Step 3 — The exchange rate: how many coulombs build one $H_{2}$ ?

WHAT. We have a crowd of coulombs ( $Q$ ) arriving, and a rule ( $z = 2$ electrons per molecule). We now need a bridge between "coulombs" and "molecules". That bridge is the Faraday constant.

WHY we need it. $Q$ is measured in coulombs, but the electron rule is measured in countable electrons. One electron's charge is $1.602 \times 1 0^{- 19} C$ — hopeless to count directly. So chemists bundle electrons into moles (a mole $= 6.022 \times 1 0^{23}$ objects; see Hydrogen - Introduction and Isotopes for the mole idea in context).

PICTURE. A staircase of three units — coulombs → moles of electrons → moles of $H_{2}$ — with the conversion factor written on each step ( $\div F$ , then $\div z$ ).

Moles of electrons that arrived: $\frac{Q}{F}$ — total charge divided by charge-per-mole.
Moles of $H_{2}$ made: divide again by $z$ , because each molecule swallowed $z$ electrons.

Step 4 — Assembling Faraday's law

WHAT. Chain the two divisions from Step 3 into one formula.

$n_{H_{2}} = \frac{Q}{z F} = \frac{I t}{z F}$

$n_{H_{2}}$ ::: moles of dihydrogen produced — the answer we want.
$Q = I t$ ::: total charge (Step 1) substituted in.
$z$ ::: electrons per molecule $= 2$ (Step 2).
$F$ ::: coulombs per mole of electrons $= 96485$ (Step 3).

WHY it looks like this. Numerator = how much charge you sent. Denominator = how much charge one molecule costs. Total supply divided by cost-per-item = number of items. That is literally "money ÷ price = quantity."

PICTURE. The formula drawn as a supply-over-cost fraction: the flood of coulombs on top, the price-tag $z F$ underneath.

Step 5 — Watch the numbers appear (Worked Example 3)

WHAT. Push $I = 5 A$ for $t = 2 hours$ . Find the mass and STP volume of $H_{2}$ .

Step 5a — charge. Convert hours to seconds ( $2 \times 3600 = 7200 s$ ), then multiply: $Q = I t = 5 \times 7200 = 36000 C .$ Why seconds? Because $1 A = 1 C/s$ — the amp already assumes seconds.

Step 5b — moles. Divide by $z F$ : $n_{H_{2}} = \frac{36000}{2 \times 96485} = 0.1866 mol .$

Step 5c — mass. $m = 0.1866 \times 2 = 0.373 g .$

Step 5d — volume at STP. One mole of any gas fills $22.4 L$ at STP: $V = 0.1866 \times 22.4 = 4.18 L .$

PICTURE. A number-line flow: $36000 C \to 0.1866 mol \to 0.373 g \to 4.18 L$ , each arrow labelled with the operation that produced it.

Step 6 — The edge cases you must never trip over

WHAT. Every input can go to a strange value. Walk each one.

$t = 0$ or $I = 0$ (no current, or no wait). Then $Q = 0 \Rightarrow n = 0$ . Nothing happens — zero electrons, zero gas. The formula correctly gives $0$ .
Wrong $z$ . If you accidentally use $z = 1$ you get double the true hydrogen. The $z = 2$ is not decoration; it comes from the two H atoms in $H_{2}$ . Forgetting it is the single most common exam slip.
Voltage too low. Faraday's law tells you how much gas per coulomb, but says nothing about whether current flows. Below the minimum $E_{cell} = 1.23 V$ (parent note) essentially no current passes, so $I \approx 0$ and again $n \approx 0$ . In practice you need $1.8$ – $2.0 V$ to beat overpotential and resistance.
Oxygen at the anode. The same charge $Q$ also makes $O_{2}$ , but there $z = 4$ (four electrons per $O_{2}$ ). Same $Q$ , different $z \Rightarrow$ half as many moles of $O_{2}$ as $H_{2}$ — the classic $2 : 1$ volume ratio.

WHY show this. The law is a single fraction; misreading any symbol silently corrupts the answer. Knowing where it breaks is knowing you have really understood it.

PICTURE. Two side-by-side test tubes over the electrodes: the $H_{2}$ tube fills twice as fast as the $O_{2}$ tube, because $z_{H_{2}} = 2$ but $z_{O_{2}} = 4$ for the same coulombs.

The one-picture summary

Every arrow of this page, chained end to end: current and time make charge; charge divided by the electron price $z F$ makes moles; moles times molar mass make grams; moles times $22.4$ make litres.

Recall Feynman retelling — say it like a story

Imagine electricity as a delivery van of electrons. The van's speed is the current $I$ ; how long it drives is $t$ ; so the total cargo it dumps at the negative plate is $Q = I t$ coulombs. Now, at that plate there's a factory rule: it takes exactly two electrons to weld together one hydrogen molecule. But we can't count single electrons — they're absurdly tiny — so we bundle them into moles, and one mole of electrons is a fixed pile worth $96485$ coulombs, called $F$ . To find how many molecule-batches we can afford, take the cargo $Q$ , divide by $F$ to get moles of electrons, then divide by the " $2$ " rule to get moles of $H_{2}$ : that's $n = \frac{I t}{z F}$ . Multiply by $2 g/mol$ and you have grams; multiply by $22.4 L/mol$ and you have litres. Send $5$ amps for two hours ( $36000$ coulombs) and out comes about a third of a gram of hydrogen, filling roughly $4.2$ litres. Turn the current off and the whole story freezes: no cargo, no gas. That's the entire law — supply of charge, divided by the price of a molecule.

Quick self-check

Same current, half the time

half the charge, so half the hydrogen —

n

scales linearly with

t

Why is

z = 2

for hydrogen?

One

H_{2}

has two H atoms, each needing one electron to go from

+ 1

0

What does

F

convert?

Coulombs of charge into moles of electrons (

96485 C = 1

mol e

^{-}

Related machinery: Redox Reactions · Fuel Cells and Hydrogen Economy (runs this same reaction in reverse) · Electrolysis and Faraday's Laws.