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

This is the practice companion to the parent note on Faraday's laws. There we built the master formula. Here we break it against every kind of question an exam can fire at you. Before the examples, a small map so you always know which "case" you are standing in.

Let me define every symbol once, plainly, so nothing appears unearned:

The scenario matrix

Every electrolysis calculation lands in one of these cells. The six examples below hit all of them.

Cell	What is hidden / the twist	Example
A · Find mass	Given $I,t$ → find $m$	Ex 1
B · Find time	Given $m,I$ → find $t$	Ex 2
C · Find current	Given $m,t$ → find $I$	Ex 3
D · Same-charge ratio	Cells in series, compare two metals (2nd law)	Ex 4
E · Gas at STP	Product is a gas → volume, not grams	Ex 5
F · Degenerate / zero & sign of electrode	Zero time, and which electrode (anode vs cathode)	Ex 6
G · Real-world + efficiency twist	Industrial cell, current efficiency < 100%	Ex 7

A word on "sign" for chemistry (there are no negative masses): the analogue of a quadrant here is which electrode and which direction electrons flow. Reduction happens at the cathode (electrons arrive, ions gain them); oxidation at the anode (electrons leave). Getting this wrong flips your $n$ or your product entirely — Ex 6 makes you face it head-on.

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Forecast: 30 minutes is short and Al needs 3 electrons each — guess: a small fraction of a gram, well under 1 g. Hold that thought.

1. Convert time to seconds. $t=30\times 60=1800$ s. Why this step? The ampere is coulombs per second; if $t$ is in minutes, $It$ is nonsense.
2. Read $n$ off the half-reaction. ${\text{Al}}^{3+}+3e^{-}\to \text{Al}$ shows $n=3$. Why this step? Each aluminium atom needs 3 electrons; skip this and the mass is 3× too big.
3. Total charge. $Q=It=3.0\times 1800=5400$ C. Why this step? This is the raw number of "electron-units" we pushed through.
4. Apply the formula. $m=\frac{MIt}{nF}=\frac{27\times 5400}{3\times 96500}=\frac{145800}{289500}=0.5037$ g. Why this step? $\frac{M}{nF}$ is the exchange rate turning coulombs into grams of Al.

Answer: $\approx 0.504$ g.

Verify: Moles of electrons $=\frac{5400}{96500}=0.0560$ mol. Divide by 3 → $0.01865$ mol Al. Times 27 → $0.504$ g. Matches, and it is well under 1 g as forecast. Units: $\frac{(\text{g/mol})(\text{C})}{(\text{C/mol})}=\text{g}$. ✓

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Forecast: Silver only needs 1 electron and current is modest, so this should take a good while — guess "over an hour."

1. Rearrange the master formula for $t$. $t=\frac{mnF}{MI}$. Why this step? Time is the hidden slot; solve for it algebraically before plugging numbers.
2. Identify $n=1$. From ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$. Why this step? Silver is the classic $n=1$ metal — always confirm, never assume.
3. Substitute. $t=\frac{5.00\times 1\times 96500}{108\times 2.0}=\frac{482500}{216}=2233.8$ s. Why this step? Straight arithmetic once the equation is solved for $t$.
4. Convert to minutes. $2233.8/60=37.2$ min. Why this step? Human-readable, and confirms it is under one hour — my forecast was slightly off.

Answer: $\approx 2234$ s $\approx 37.2$ min.

Verify: Run it forward. $m=\frac{108\times 2.0\times 2233.8}{1\times 96500}=\frac{482500}{96500}=5.00$ g. Returns the input. ✓

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Forecast: $12.7$ g is exactly one-fifth of $63.5$, i.e. $0.2$ mol Cu, in one hour. That is a lot of atoms → expect a sizeable current, tens of amps.

1. Solve for $I$. $I=\frac{mnF}{Mt}$. Why this step? Current is now the hidden slot.
2. Get $t$ in seconds. $t=3600$ s. And $n=2$. Why this step? Copper(II) needs two electrons; hour → seconds so amperes come out clean.
3. Substitute. $I=\frac{12.7\times 2\times 96500}{63.5\times 3600}=\frac{2451100}{228600}=10.72$ A. Why this step? Delivers the answer directly.

Answer: $\approx 10.7$ A.

Verify: Moles Cu $=\frac{12.7}{63.5}=0.200$; electrons $=0.400$ mol $=0.400\times 96500=38600$ C; over $3600$ s gives $\frac{38600}{3600}=10.7$ A. ✓

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Forecast: Gold is heavy (197) but needs 3 electrons per atom. Silver is light but needs only 1. The "cost" per unit mass is $M/n$: Ag gives $108/1=108$, Au gives $197/3\approx 66$. So the same electrons make less mass of Au than of Ag. Guess: under $2.16$ g.

1. Use Faraday's second law. In series the charge is identical, so masses go as chemical equivalents: $\frac{m_{\text{Au}}}{m_{\text{Ag}}}=\frac{M_{\text{Au}}/n_{\text{Au}}}{M_{\text{Ag}}/n_{\text{Ag}}}.$ Why this step? Series ⇒ same $Q$; comparing two cells cancels $F$ and $t$ entirely — no need for the current.
2. Compute the equivalents. $\frac{197}{3}=65.67$ for Au; $\frac{108}{1}=108$ for Ag. Why this step? The equivalent $M/n$ is "grams deposited per mole of electrons" — the fair comparison.
3. Solve. $m_{\text{Au}}=2.16\times \frac{65.67}{108}=2.16\times 0.6080=1.313$ g. Why this step? Scales silver's mass by the equivalent ratio.

Answer: $\approx 1.31$ g of gold.

Verify: Charge from silver: $\frac{2.16}{108}=0.0200$ mol $e^{-}$ → $1930$ C. For gold, moles Au $=\frac{0.0200}{3}=0.006\overline{6}$; times 197 $=1.313$ g. Independent route agrees, and it is under $2.16$ g as forecast. ✓

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Forecast: $9650$ C is exactly one-tenth of $F$, so $0.1$ mol of electrons. Two electrons per ${\text{Cl}}_2$ → $0.05$ mol gas → about $1.1$ L.

1. Moles of electrons. $n_{e^{-}}=\frac{Q}{F}=\frac{9650}{96500}=0.100$ mol. Why this step? Charge always converts to electrons first — the universal middle step.
2. Moles of gas via $n=2$. Each ${\text{Cl}}_2$ costs 2 electrons: $n_{{\text{Cl}}_2}=\frac{0.100}{2}=0.0500$ mol. Why this step? Here $n$ is "electrons per molecule of product," not per atom — read the half-equation.
3. Convert moles of gas to volume. $V=0.0500\times 22.4=1.12$ L. Why this step? For gases we finish in litres using the molar volume, not grams — that is the whole twist of this cell.

Answer: $\approx 1.12$ L of ${\text{Cl}}_2$ at STP.

Verify: $0.05$ mol $\times 22.4$ L/mol $=1.12$ L, and $0.05$ mol $\times 71$ g/mol $=3.55$ g if you wanted mass instead. Consistent. ✓

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Forecast: (a) Zero seconds means zero charge means literally nothing — the formula must return $0$. (b) 2 electrons per ${\text{H}}_2$, so $0.2$ mol ${\text{H}}_2\approx 0.4$ g.

1. Part (a): plug $t=0$. $m=\frac{M}{nF}I\cdot 0=0$ g. Why this step? It confirms the formula behaves sanely at the boundary — no charge, no chemistry. Every physical formula should survive its zero case.
2. Part (b): confirm the electrode. "Cathode" = reduction = electrons gained. The given half-equation consumes electrons, so it is correctly a cathode reaction with $n=2$ per ${\text{H}}_2$. Why this step? If you mistake this for the anode you would grab ${\text{Cl}}_2$ ($M=71$) instead of ${\text{H}}_2$ ($M=2$) — a wildly wrong answer. The electrode label is the sign information in electrochemistry.
3. Moles of gas. $n_{{\text{H}}_2}=\frac{0.400}{2}=0.200$ mol. Why this step? Two electrons build one ${\text{H}}_2$ molecule.
4. Mass. $m=0.200\times 2.0=0.400$ g. Why this step? Finish in grams as asked.

Answer: (a) $0$ g. (b) $0.400$ g of ${\text{H}}_2$.

Verify: (a) trivially $0$. (b) $0.400$ mol $e^{-}$ → half as many ${\text{H}}_2$ = $0.200$ mol → $0.4$ g; the same $0.400$ mol $e^{-}$ at the anode would give $0.200$ mol ${\text{Cl}}_2=14.2$ g — proof that the electrode choice dominates the answer. ✓

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Forecast: Without losses this monster current for a day should give around a tonne of aluminium; the $92\%$ shaves a bit off. Guess: roughly $700$–$750$ kg.

1. Total charge. $t=24\times 3600=86400$ s; $Q_{\text{total}}=100000\times 86400=8.64\times 10^9$ C. Why this step? Always start from raw charge.
2. Apply efficiency. Only $92\%$ does useful work: $Q_{\text{useful}}=0.92\times 8.64\times 10^9=7.9488\times 10^9$ C. Why this step? The efficiency is the real-world twist — the formula assumes every electron counts, so we correct the charge before using it.
3. Master formula with $Q_{\text{useful}}$. $m=\frac{M{Q}_{\text{useful}}}{nF}=\frac{27\times 7.9488\times 10^9}{3\times 96500}$. Why this step? Same formula, honest charge.
4. Compute. $m=\frac{2.1462\times 10^{11}}{2.895\times 10^5}=7.413\times 10^5$ g $=741.3$ kg. Why this step? Convert grams to kilograms for a sensible industrial figure.

Answer: $\approx 741$ kg of aluminium.

Verify: Useful electrons $=\frac{7.9488\times 10^9}{96500}=8.237\times 10^4$ mol; divide by 3 → $2.746\times 10^4$ mol Al; times 27 → $7.41\times 10^5$ g $=741$ kg. Sits inside the forecast band, and equals $0.92\times$ the ideal $806$ kg. ✓

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Recall Quick self-test (reveal after guessing)

Which cell of the matrix does a "compare masses in two series cells" problem belong to? ::: Cell D — Faraday's second law, charge is shared, current and time cancel. Why must you convert time to seconds first? ::: Because $1$ A $=1$ C/s; charge $Q=It$ is only in coulombs when $t$ is in seconds. A product is a gas — what final unit do you use and what constant? ::: Litres at STP, using $22.4$ L/mol. What does $n$ mean in $m=\frac{M}{nF}It$? ::: Electrons needed per formula unit, read straight off the balanced half-reaction. At the cathode, are electrons gained or lost? ::: Gained — cathode = reduction.

Where to go next

• Force powers the reverse of a battery — see Galvanic Cells and Standard Electrode Potentials for spontaneous cells.
• How much voltage you need (and why ${\text{H}}_2\text{O}$ beats ${\text{Na}}^{+}$) is a potentials question: Nernst Equation and Thermodynamics of Electrochemical Cells.
• Getting $n$ from a half-equation relies on Redox Reactions and Balancing.
• Industrial scale-up: Industrial Chemistry; the flip side (metals wasting away) is Corrosion and Electrochemical Protection.