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

Definition Two vocabulary words you must own before Level 1

Electrochemical equivalent $Z$ . The compact form of Faraday's law is written $m = Z \cdot I \cdot t$ . The single letter $Z$ just bundles together the three constants that belong to one specific substance: $Z = \frac{M}{n \cdot F} (grams deposited per coulomb of charge)$ So $m = Z I t$ and $m = \frac{M I t}{n F}$ are the same equation — $Z$ is a nickname for $\frac{M}{n F}$ . Whenever you see $Z$ on this page, mentally expand it to $\frac{M}{n F}$ .

Chemical equivalent (equivalent weight) $= \frac{M}{n}$ . This is the molar mass divided by the number of electrons per formula unit — the mass of substance that reacts with one mole of electrons. Notice $Z = \frac{chemical equivalent}{F}$ : the two ideas are the same ratio $M / n$ , one measured per coulomb, the other per mole of electrons. We use "chemical equivalent" whenever we compare two substances under the same charge (Faraday's second law).

Intuition Sign convention — which way do the electrons go?

Physics defines conventional current $I$ as positive in the direction positive charge would flow — which is opposite to the actual electron motion (electrons are negative). In an electrolysis cell we always quote $I$ as a positive number; it is just the size of the charge flow per second, so $Q = I t$ is always positive and $m$ comes out positive. Look at the figure below and fix these two anchors in your mind:

The cathode is where reduction happens — electrons arrive here from the power supply and are handed to cations ( $M^{n +} + n e^{-} \to M$ ).
The anode is where oxidation happens — electrons are pulled away here, so anions give up electrons ( $2 Cl^{-} \to Cl_{2} + 2 e^{-}$ ).

Mnemonic: "Red Cat, An Ox" — Reduction at the Cathode, Anode is where Oxidation lives. In electrolysis the cathode is the negative terminal (electrons pushed in) and the anode is the positive terminal.

Before we start, three reveal-cards to warm up the vocabulary. Cover the right side:

Charge from current and time

Q = I \cdot t

(amps × seconds = coulombs)

Moles of electrons from charge

\frac{Q}{F} = \frac{I t}{96500}

Electrons per atom for

M^{n +}

exactly

n

— read it off the ion's charge

Electrochemical equivalent

Z

Z = M / (n F)

, grams per coulomb

Chemical equivalent (equivalent weight)

M / n

, grams reacting with one mole of electrons

The whole conversion chain that every problem below walks is drawn here — glance at it whenever you get lost:

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

Level 1 — Recognition

These test whether you can name the pieces and plug into $m = Z I t$ (equivalently $m = \frac{M I t}{n F}$ , since $Z = \frac{M}{n F}$ as defined above) with no twists.

Problem 1.1

A current of $3 A$ flows for $2$ minutes. How much total charge (in coulombs) passes?

Recall Solution

WHAT we need: charge $Q$ . WHY this formula: current is "charge per second," so charge is just current multiplied by how many seconds it flowed. First fix the units — time must be in seconds, not minutes: $t = 2 \times 60 = 120 s$ Then: $Q = I \cdot t = 3 \times 120 = 360 C$

Problem 1.2

In the reaction $Ag^{+} + e^{-} \to Ag$ , how many electrons ( $n$ ) are needed to deposit one silver atom? And for $Al^{3 +} + 3 e^{-} \to Al$ ?

Recall Solution

WHAT $n$ means: it is the number of electrons written in the half-reaction to neutralise one ion, which is the same as the ion's charge number.

For $Ag^{+}$ : the charge is $+ 1$ , so $n = 1$ .
For $Al^{3 +}$ : the charge is $+ 3$ , so $n = 3$ .

You never have to memorise $n$ — just read the superscript on the ion.

Level 2 — Application

Now the full chain: current + time → charge → moles of electrons → moles of substance → mass.

Problem 2.1

A current of $4 A$ passes through molten $AlCl_{3}$ for $30$ minutes. What mass of aluminium is deposited? ( $M_{Al} = 27$ , $n = 3$ )

Recall Solution

Step 1 — time to seconds (WHY: formula needs SI): $t = 30 \times 60 = 1800 s$ Step 2 — plug into Faraday's law (WHY: it converts coulombs straight to grams of this substance): $m = \frac{M \cdot I \cdot t}{n \cdot F} = \frac{27 \times 4 \times 1800}{3 \times 96500}$ Step 3 — arithmetic: $m = \frac{194400}{289500} = 0.671 g$ Sanity check the chain: charge $= 4 \times 1800 = 7200$ C → $7200/96500 = 0.0746$ mol $e^{-}$ → each Al eats 3 electrons → $0.0746/3 = 0.0249$ mol Al → $0.0249 \times 27 = 0.671$ g. ✓

Problem 2.2

How long (in seconds) must a $2 A$ current flow to deposit $5 g$ of copper from $CuSO_{4}$ ? ( $M_{Cu} = 63.5$ , $n = 2$ )

Recall Solution

WHAT we're solving for: time $t$ . So rearrange $m = \frac{M I t}{n F}$ to make $t$ the subject. WHY rearrange rather than guess: every quantity except $t$ is known, so algebra hands us the answer directly. $t = \frac{m \cdot n \cdot F}{M \cdot I} = \frac{5 \times 2 \times 96500}{63.5 \times 2}$ $t = \frac{965000}{127} = 7598 s \approx 2.11 hours$

Level 3 — Analysis

Here you must compare substances or reason about what happens where — Faraday's second law and electrode assignment.

Problem 3.1 (Second Law)

The same charge is passed through cells of $AgNO_{3}$ and $CuSO_{4}$ in series. If $2.16 g$ of silver is deposited, how much copper is deposited? ( $M_{Ag} = 108, n_{Ag} = 1$ ; $M_{Cu} = 63.5, n_{Cu} = 2$ )

Recall Solution

WHY the second law applies: "same charge, same time, cells in series" means the same number of electrons flows through both. The masses then split in the ratio of their chemical equivalents $\frac{M}{n}$ (the equivalent weight defined at the top of the page — grams per mole of electrons). $\frac{m _{Cu}}{m _{Ag}} = \frac{M _{Cu} / n _{Cu}}{M _{Ag} / n _{Ag}} = \frac{63.5/2}{108/1} = \frac{31.75}{108}$ $m_{Cu} = 2.16 \times \frac{31.75}{108} = 0.635 g$ Meaning: silver is "cheap" in electrons (1 per atom) so a given charge deposits a lot of it; copper demands 2 electrons per atom, so the same electrons build fewer, and lighter-per-electron, copper — hence less mass.

Problem 3.2 (Electrode reasoning)

During electrolysis of aqueous $NaCl$ , why does hydrogen gas appear at the cathode instead of sodium metal? Write the cathode half-reaction that actually occurs.

Recall Solution

WHY not sodium: at the cathode (where reduction happens — electrons arrive from the supply), whichever species is easier to reduce wins. Water is far easier to reduce than $Na^{+}$ — depositing sodium would demand a much larger voltage because $Na^{+}$ clings to its ionic form. So water grabs the electrons first. The half-reaction is: $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ The leftover $Na^{+}$ pairs with the new $OH^{-}$ to give $NaOH$ in solution — this is exactly the chlor-alkali logic from the parent note. See the related idea of preferred discharge in Galvanic Cells and Standard Electrode Potentials.

Problem 3.3 (Competing anode reactions)

At the anode of a brine cell, two oxidations compete: chloride $\to$ chlorine, and water $\to$ oxygen. On paper, water is thermodynamically easier to oxidise, yet real cells still pour out $Cl_{2}$ . Explain why, and state what happens if the brine is made very dilute.

Recall Solution

The two candidate anode half-reactions: $2 Cl^{-} \to Cl_{2} + 2 e^{-} vs. 2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ WHY chlorine wins in concentrated brine — two effects stacked:

Overpotential. The oxygen reaction is kinetically sluggish on the electrode: it needs a large extra voltage (an overpotential) beyond its ideal value before it runs at a useful rate. Chlorine has a much smaller overpotential, so at practical currents chloride discharges first even though oxygen looks better "on paper."
Concentration. Concentrated brine floods the anode with $Cl^{-}$ ions; the sheer supply of chloride pushes its discharge (a mass-action, concentration effect). What dilution does: in very dilute NaCl, chloride is scarce, its concentration advantage vanishes, and the anode switches over to producing oxygen instead: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ So the product at an electrode is decided by three things together — thermodynamic ease (electrode potential), overpotential (kinetics), and concentration. The figure contrasts the two regimes.

The quantitative version of "how concentration shifts electrode potential" is the Nernst Equation; the driving-force bookkeeping behind "thermodynamically easier" is in Thermodynamics of Electrochemical Cells.

Common mistake L3 trap: flipping the equivalent ratio upside-down (and trusting thermodynamics alone)

In the second-law ratio it is horribly tempting to write $\frac{m _{Cu}}{m _{Ag}} = \frac{n _{Cu}}{n _{Ag}} \times \frac{M _{Ag}}{M _{Cu}}$ or to put $n$ on top ("more electrons → more mass, right?"). It feels right because "copper needs more electrons." Fix: the chemical equivalent is $\frac{M}{n}$ — molar mass over valency. A higher $n$ shrinks the equivalent, so it goes in the denominator. Always write $equivalent = \frac{M}{n}$ first, then take the plain ratio of equivalents.

A second, deeper trap: predicting electrode products from electrode potentials alone. Students confidently say "water oxidises before chloride, so brine gives oxygen" — which contradicts every real chlor-alkali plant. Fix: always ask the three questions together — potential (thermodynamics), overpotential (kinetics), and concentration. Chlorine wins on the last two despite losing the first.

Level 4 — Synthesis

Multi-step, industrial-scale, unit-juggling problems that combine everything.

Problem 4.1 (Chlor-alkali, gas volume)

A chlor-alkali cell runs at $100 A$ for $5$ hours. What mass of $Cl_{2}$ is produced, and what volume does it occupy at STP ( $22.4 L/mol$ )? Anode: $2 Cl^{-} \to Cl_{2} + 2 e^{-}$ , so $n = 2$ , $M_{Cl_{2}} = 71$ .

Recall Solution

Step 1 — time to seconds: $t = 5 \times 3600 = 18000 s$ Step 2 — mass of chlorine (WHY $n = 2$ : making one $Cl_{2}$ molecule frees 2 electrons): $m = \frac{M I t}{n F} = \frac{71 \times 100 \times 18000}{2 \times 96500} = \frac{127800000}{193000} = 662.2 g$ Step 3 — convert mass to moles, then to volume (WHY: STP volume is per mole, not per gram): $moles = \frac{662.2}{71} = 9.327 mol$ $V = 9.327 \times 22.4 = 208.9 L$

Problem 4.2 (Hall–Héroult, energy check)

A Hall–Héroult cell delivers $200, 000 A$ for $12$ hours. (a) How much aluminium is produced? (b) If the cell operates at $4.5 V$ , how many kWh of electrical energy did it use, and what is the energy per kg of Al? ( $M_{Al} = 27, n = 3$ )

Recall Solution

(a) Mass of aluminium. $t = 12 \times 3600 = 43200 s$ $m = \frac{27 \times 200000 \times 43200}{3 \times 96500} = \frac{233280000000}{289500} = 805803 g \approx 805.8 kg$ (b) Energy used. WHY this formula: electrical energy = charge × voltage $= (I t) \times V$ in joules; divide by $3.6 \times 1 0^{6}$ to get kWh. $Q = I t = 200000 \times 43200 = 8.64 \times 1 0^{9} C$ $E = Q \cdot V = 8.64 \times 1 0^{9} \times 4.5 = 3.888 \times 1 0^{10} J$ $E_{kWh} = \frac{3.888 \times 1 0 ^{10}}{3.6 \times 1 0 ^{6}} = 10800 kWh$ Energy per kg: $\frac{10800 kWh}{805.8 kg} = 13.4 kWh/kg$ This lands close to the real-world $\sim 15 kWh/kg$ from the parent note — the small gap is real cells' inefficiency (heat losses, side reactions). See Industrial Chemistry and Thermodynamics of Electrochemical Cells for where that lost energy goes.

Level 5 — Mastery

Reverse problems, efficiency corrections, and multi-cell reasoning — the exam's hardest tier.

Problem 5.1 (Current efficiency)

A copper-refining cell is supposed to deposit copper by $Cu^{2 +} + 2 e^{-} \to Cu$ . A current of $10 A$ runs for $1$ hour, but only $10.5 g$ of copper is actually deposited. What is the current efficiency (actual mass ÷ theoretical mass)? ( $M_{Cu} = 63.5, n = 2$ )

Recall Solution

Step 1 — theoretical mass (what Faraday says should deposit if every electron did copper work): $t = 3600 s, m_{theory} = \frac{63.5 \times 10 \times 3600}{2 \times 96500} = \frac{2286000}{193000} = 11.845 g$ Step 2 — efficiency (WHY < 100%: some current is stolen by side reactions like $H_{2}$ evolution — exactly the competing-reaction idea from Problem 3.3): $η = \frac{m _{actual}}{m _{theory}} = \frac{10.5}{11.845} = 0.8864 = 88.6 %$

Problem 5.2 (Two cells, unknown metal)

The same charge is passed through a silver cell and a cell containing an unknown metal $X^{2 +}$ . The silver cell deposits $1.08 g$ of Ag ( $M_{Ag} = 108, n = 1$ ); the X cell deposits $0.2965 g$ of X, and X uses $n = 2$ . Find the molar mass of X and identify it.

Recall Solution

Step 1 — how many moles of electrons flowed (from the silver, our "coulometer"): Silver uses 1 electron per atom, so moles of $e^{-}$ = moles of Ag: $mol e^{-} = \frac{1.08}{108} = 0.01 mol$ Step 2 — moles of X (WHY divide by 2: each X atom needs 2 electrons): $mol X = \frac{0.01}{2} = 0.005 mol$ Step 3 — molar mass of X: $M_{X} = \frac{0.2965}{0.005} = 59.3 g/mol$ That is essentially the atomic mass of nickel ( $58.7$ ). See Redox Reactions and Balancing for how the $n = 2$ half-reaction $Ni^{2 +} + 2 e^{-} \to Ni$ is built.

Problem 5.3 (Reverse-engineer the charge)

An electroplating job deposited $0.4407 g$ of chromium from $Cr^{3 +}$ ( $M_{Cr} = 52, n = 3$ ). What total charge (in coulombs) was required?

Recall Solution

WHY start from mass: we run the chain backwards — mass → moles of Cr → moles of electrons → charge. $mol Cr = \frac{0.4407}{52} = 0.008475 mol$ Each Cr needs 3 electrons: $mol e^{-} = 3 \times 0.008475 = 0.025425 mol$ $Q = mol e^{-} \times F = 0.025425 \times 96500 = 2453.5 C$

Recall Quick self-test recap (fold before exam)

One coulometer metal that makes mol-electrons = mol-atoms ::: silver, because $n = 1$ Electrochemical equivalent $Z$ in symbols ::: $Z = M / (n F)$ , grams per coulomb Chemical equivalent (equivalent weight) in symbols ::: $M / n$ , grams per mole of electrons Formula for current efficiency ::: $η = m_{actual} / m_{theory}$ , always $\leq 1$ Electrical energy from charge and voltage ::: $E = Q V = (I t) V$ joules Why $H_{2}$ not $Na$ at the cathode in aqueous NaCl ::: water is easier to reduce than $Na^{+}$ Three things that decide the electrode product ::: potential, overpotential, concentration Which electrode is reduction, which is oxidation ::: Red Cat, An Ox — reduction at cathode, oxidation at anode