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

Before we start, we anchor every symbol used in this bank — nothing below appears until it lives here.

Definition Every symbol in one place (with units)

== $m$ == — mass of substance produced at an electrode, in grams (g).
== $Q$ — total charge== that flowed, in coulombs (C). It is the total count of electrons that passed, and $Q = I \cdot t$ .
== $I$ == — current, in amperes (A). One ampere is one coulomb per second (C/s).
== $t$ == — time the current flowed, in seconds (s).
== $F$ — Faraday's constant==, $F = 96, 500$ C/mol. This is the charge of one mole of electrons — the exchange rate between coulombs and moles of electrons.
== $n$ == — electrons needed per ion (per product unit), read straight off the half-reaction (e.g. $Cu^{2 +} + 2 e^{-} \to Cu$ gives $n = 2$ ). Dimensionless.
== $M$ — molar mass== of the product, in grams per mole (g/mol) — the mass of one mole of its atoms/molecules.
== $Z$ — electrochemical equivalent==, the mass produced per coulomb of charge, in grams per coulomb (g/C), equivalently g/(A·s). It is defined by $Z = \frac{M}{n F}$ .

Now the two forms of Faraday's first law both make sense: the compact one, $m = Z \cdot I \cdot t$ , and the expanded master formula, $m = \frac{M \cdot I \cdot t}{n \cdot F}$ — they are the same statement, since $Z = M / (n F)$ .

Recall The master formula read as a conveyor belt

$m = \frac{M \cdot I \cdot t}{n \cdot F}$ is a four-stage unit conversion. The figure below traces one number through all four stages.

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

Stage 1 — $Q = I \cdot t$ : current × time gives coulombs.
Stage 2 — $\div F$ : coulombs → moles of electrons.
Stage 3 — $\div n$ : moles of electrons → moles of atoms ( $n$ electrons per atom).
Stage 4 — $\times M$ : moles of atoms → grams.

The half-reactions this bank keeps returning to are drawn once here so you can see where the electrons go:

True or false — justify

Faraday's first law

m = Z I t

works only for metals deposited on the cathode.

False. It works for any electrode product — gases like

Cl_{2}

H_{2}

O_{2}

included — as long as you use that product's own

n

and

M

; "deposited" just means "produced at an electrode".

Doubling the current always doubles the mass produced in the same time.

True (for the same substance).

m = Z I t

is linear in

I

, so at fixed

t

and fixed

Z

, twice the current means twice the charge and twice the mass.

Doubling the concentration of

CuSO_{4}

doubles the copper deposited by a fixed charge.

False. Mass depends only on charge passed (

m = Z I t

), not on concentration. Concentration affects how fast you can push current before the cell struggles, not the mass per coulomb.

The same charge passed through molten

NaCl

and through

AlCl_{3}

deposits equal masses of metal.

False. Faraday's second law says masses go as chemical equivalents

M / n

: sodium is

23/1 = 23

, aluminium is

27/3 = 9

, so the same charge deposits far less mass of Al per equivalent than of Na.

If you pass 1 mole of electrons, you always get 1 mole of metal atoms.

False. You get

1/ n

moles. For

Al^{3 +}

(

n = 3

) one mole of electrons yields only

1/3

mole of Al, because each atom needs three electrons.

Electrolysis produces electrical energy from a chemical reaction.

False — that is a galvanic cell (see Galvanic Cells and Standard Electrode Potentials). Electrolysis consumes electrical energy to drive a non-spontaneous reaction; it is the reverse.

In the electrolysis of brine, sodium metal is deposited at the cathode.

False. Water is reduced instead (

2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}

) because reducing

Na^{+}

needs a far larger voltage; you get

H_{2}

gas and

OH^{-}

, not sodium.

The electrochemical equivalent

Z

is the same number for every substance.

False.

Z = M / (n F)

depends on the substance's molar mass

M

and electron count

n

, so silver, copper and aluminium each have a different

Z

Anode is always positive and cathode always negative.

This holds in electrolysis (the external supply forces it), but in a galvanic cell the polarities flip. The reliable rule is the definition itself: oxidation always happens at the anode, reduction always at the cathode, whatever the sign.

Spot the error

"6000 C of charge = 6000 moles of electrons."

Wrong — coulombs are not moles. Divide by

F

6000/96, 500 = 0.062

mol of electrons. Forgetting to divide by

F

is the single most common numeric slip.

"For

Cu^{2 +} + 2 e^{-} \to Cu

, moles of Cu = moles of electrons."

Wrong — you must divide by

n = 2

. Two electrons build one Cu atom, so moles of Cu

= \frac{1}{2} \times

moles of electrons.

"In

m = Z I t

, I plugged in

t = 40

for 40 minutes."

Error — the formula needs SI units, so time must be in seconds:

t = 40 \times 60 = 2400

s. Amperes are coulombs per second, so minutes give an answer 60× too small.

"Overall chlor-alkali reaction:

2 NaCl \to 2 Na + Cl_{2}

Wrong for aqueous brine. Because water reduces preferentially, the real products are

Cl_{2}

H_{2}

and

NaOH

2 NaCl + 2 H_{2} O \to Cl_{2} + H_{2} + 2 NaOH

. Only molten NaCl gives sodium metal.

"In the Hall–Héroult cell, oxygen just bubbles off the anode as

O_{2}

Incomplete — the

O_{2}

immediately attacks the hot carbon anode:

C + O_{2} \to CO_{2}

. That is why carbon anodes burn away and must be replaced.

"We use cryolite because it reduces the aluminium."

Wrong — cryolite does no reducing. It is a molten solvent that dissolves

Al_{2} O_{3}

and drops the working temperature from ~2072 °C to ~1000 °C while conducting current. The electrons do the reducing.

"For the half reaction

2 Cl^{-} \to Cl_{2} + 2 e^{-}

, use

n = 1

because chlorine is

- 1

Wrong —

n

is electrons per product unit

Cl_{2}

, and two electrons appear, so

n = 2

. Read

n

off the balanced half-reaction, not off the ion's charge alone.

Why questions

Why does

m

depend on

M / n

and not just on

M

Because charge counts electrons, and an ion needing more electrons (

n

large) uses up more charge per atom. Dividing

M

n

converts "grams per mole of atoms" into "grams per mole of electrons", which is what a coulomb actually buys.

Why is aluminium extracted by electrolysis rather than by heating its ore with carbon?

Aluminium bonds oxygen so strongly that carbon cannot pull it off at practical temperatures. Electrolysis supplies electrons directly at high enough "electrical pressure" (voltage) to force

Al^{3 +} + 3 e^{-} \to Al

(linked to Thermodynamics of Electrochemical Cells).

Why does molten aluminium sink to the bottom of the cell even though metals feel "heavier"?

It is denser than the surrounding molten cryolite, so it settles below the electrolyte where it can be tapped off — density relative to the bath is what matters, not everyday intuition.

Why are diaphragm or membrane barriers needed in the chlor-alkali cell?

To keep

Cl_{2}

(anode) and

NaOH

(cathode region) apart; if they mix they react to form bleach (

NaOCl

), ruining both products (see Industrial Chemistry).

Why is aluminium smelting sited next to cheap hydroelectric power?

It draws ~15 kWh per kg of Al — huge because you must supply 3 moles of electrons per mole of metal continuously at high current. Electricity is the dominant cost, so cheap power decides where plants are built.

Why does the same charge deposit more mass of silver than of copper?

Silver needs only 1 electron per atom while copper needs 2, so a fixed pile of electrons builds twice as many Ag atoms as Cu atoms — and its equivalent mass

M / n = 108/1

far exceeds copper's

63.5/2

Edge cases

If the current

I

is zero, what mass is produced?

Zero. With

I = 0

Q = I t = 0

, so no electrons flow and

m = Z I t = 0

— no charge, no chemistry, no matter how long you wait.

What happens to the mass produced if the circuit is broken for half the run time?

Only the time while current actually flowed counts. Effective

t

is the "current-on" time, so a break halves the charge and roughly halves the mass; wall-clock time alone is misleading.

Two cells (CuSO₄ then AgNO₃) are wired in series. Do they receive the same charge?

Yes — in series the same current flows through both for the same time, so identical charge passes. Their deposited masses then differ purely by chemical equivalent (Faraday's second law), which is exactly how series cells are used to compare

M / n

At 100 % current efficiency vs real cells: does

m = Z I t

over- or under-predict the real mass?

It over-predicts, because real cells lose some charge to side reactions (e.g.

O_{2}

H_{2}

evolution, heating). Actual mass = predicted mass × current efficiency (< 1), so

m = Z I t

is an ideal upper bound.

For a gas product like

Cl_{2}

, does "mass deposited" even make sense in

m = Z I t

Yes —

m

is the mass of

Cl_{2}

produced, whether it stays put or bubbles away. Use

M = 71

and

n = 2

; the gas escaping does not change how much was made per coulomb.

Molten NaCl vs aqueous NaCl at the cathode — same product?

No. Molten NaCl has no water, so

Na^{+}

is reduced to sodium metal; aqueous NaCl has water, which is reduced first to

H_{2}

. The presence of water is the deciding factor.

Recall One-line self-test

What does the factor $\frac{1}{n F}$ physically convert? ::: Total charge in coulombs into moles of atoms — $\div F$ gives moles of electrons, $\div n$ gives moles of atoms.