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

WHY it matters: This is how we extract aluminum from ore (Al₂O₃), produce chlorine and NaOH from brine, electroplate metals, and refine copper. Without electrolysis, modern metallurgy and chemical industry wouldn't exist.

Faraday's First Law — The Quantitative Link

$m \propto Q$

Since $Q = I \cdot t$ (current × time), we get:

$m = Z \cdot I \cdot t$

where:

• $m$ = mass deposited/liberated (grams)
• $Z$ electrochemical equivalent (g/C) — mass deposited per coulomb
• $I$ = current (amperes)
• $t$ = time (seconds)

Step 1: What does current measure? Current $I$ is charge flow per second: $I = \frac{Q}{t}$, so $Q = I \cdot t$ coulombs.

Step 2: How does charge relate to moles of electrons? One mole of electrons Faraday's constant $F = 96,500$ C/mol (approximately). If charge $Q$ passes, moles of electrons = $\frac{Q}{F} = \frac{I \cdot t}{F}$

Step 3: How do electrons relate to mass? Consider a metal ion $M^{n+} + ne^- \rightarrow M$

• $n$ electrons deposit1 atom of $M$
• 1 mole of $M$ requires $n$ moles of electrons
• Moles of $M$ deposited = $\frac{\text{moles of } e^-}{n} = \frac{I \cdot t}{n \cdot F}$

Step 4: Convert moles to mass $m = \text{moles} \times M_{\text{molar}} = \frac{I \cdot t}{n \cdot F} \times M$

Step 5: Define electrochemical equivalent $Z = \frac{M}{n \cdot F}$

This is the mass deposited per coulomb. Therefore:

$\boxed{m = Z \cdot I \cdot t = \frac{M}{n \cdot F} \cdot I \cdot t}$

WHY this formula works: Each electron that flows reduces/oxidizes one unit of charge on an ion. More current → more electrons → more atoms deposited. The factor $\frac{M}{nF}$ converts "coulombs" into "grams of this specific substance."

Faraday's Second Law — Comparing Different Substances

$\frac{m_1}{m_2} = \frac{M_1/n_1}{M_2/n_2} = \frac{Z_1}{Z_2}$

WHY: The same number of electrons must produce proportional amounts of different substances, scaled by how many electrons each needs and its atomic mass.

Solution:

• $I = 2.5$ A
• $t = 40 \times 60 = 2400$ s
• $F = 96,500$ C/mol
• $n = 2$ (since $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$)

$m = \frac{M \cdot I \cdot t}{n \cdot F} = \frac{63.5 \times 2.5 \times 2400}{2 \times 96500}$

$m = \frac{381,000}{193,000} = 1.97 \text{ g}$

Why this step? We're counting: (2.5 C/s)(2400 s) = 6000 C total. That's $\frac{6000}{96500} = 0.0622$ mol $e^-$. Since 2 electrons make 1 Cu atom, we get $\frac{0.0622}{2} = 0.0311$ mol Cu = $0.0311 \times 63.5 = 1.97$ g.

Solution: $\frac{m_{\text{Cu}}}{m_{\text{Ag}}} = \frac{M_{\text{Cu}}/n_{\text{Cu}}}{M_{\text{Ag}}/n_{\text{Ag}}} = \frac{63.5/2}{108/1} = \frac{31.75}{108}$

$m_{\text{Cu}} = 1.08 \times \frac{31.75}{108} = 0.317 \text{ g}$

Why? Same electrons passed. Ag needs 1 electron per atom (lighter per electron), Cu needs 2 electrons per atom (heavier per electron but needs more electrons), so less mass of Cu is deposited.

Industrial Electrolysis Applications

1. Chlor-Alkali Process (Electrolysis of Brine)

Setup: Brine flows through an electrolytic cell with inert electrodes (often Ti/graphite).

At the Anode (oxidation): $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ Chloride ions lose electrons → chlorine gas bubles up.

At the Cathode (reduction): $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ Water is reduced (not Na⁺, because water is easier to reduce) → hydrogen gas + hydroxide ions.

Overall Reaction: $2\text{NaCl} + 2\text{H}_2\text{O} \xrightarrow{\text{electrolysis}} \text{Cl}_2 + \text{H}_2 + 2\text{NaOH}$

WHY these products?

• Cl⁻ is easier to oxidize than water at the anode (lower oxidation potential)
• Water is easier to reduce than Na⁺ at the cathode (Na⁺ would require much higher voltage)
• The Na⁺ and OH⁻ left behind form NaOH solution

Industrial Importance:

• Chlorine: PVC plastics, disinfectants, solvents
• Sodium hydroxide: Paper, soap, chemical manufacturing
• Hydrogen: Ammonia synthesis, fuel

Types of Cells:

1. Diaphragm cell: Porous barrier prevents Cl₂ and NaOH from mixing (would form NaOCl)
2. Membrane cell: Ion-exchange membrane (modern, more efficient)
3. Mercury cell: Uses Hg cathode (being phased out due to toxicity)

Solution: Reaction: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ For each Cl₂ molecule, we need 2 electrons, so $n = 2$.

$t = \frac{m \cdot n \cdot F}{M \cdot I} = \frac{10,000 \times 2 \times 96,500}{71 \times 50}$

$t = \frac{1,930,000,000}{3,550} = 543,661 \text{ s} = 151 \text{ hours}$

Why? 10 kg = 10,000/71 = 140.85 mol Cl₂. Each mol needs 2 mol electrons = 281.7 mol $e^-$. That's $281.7 \times 96,500 = 27,184,050$ C. At 50 A, time = 27,184,050/50 = 543,681 s ≈ 151 hours.

2. Extraction of Aluminum (Hall-Héroult Process)

WHY not just heat Al₂O₃? Aluminum is too reactive—it holds onto oxygen too tightly. Chemical reduction would require an even more reactive metal. Electrolysis is the only practical method.

Setup:

• Electrolyte: Al₂O₃ dissolved in molten cryolite (~1000°C)
  • Pure Al₂O₃ melts at 2072°C (too high, too expensive)
  • Cryolite lowers melting point to ~1000°C and conducts electricity
• Cathode: Carbon-lined steel cell (the lining acts as cathode)
• Anode: Multiple carbon (graphite) rods suspended into the bath

At the Cathode (reduction): $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ Molten aluminum collects at the bottom (denser than cryolite, despite being a metal).

At the Anode (oxidation): $2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-$

The oxygen immediately reacts with the carbon anode: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$

WHY carbon anodes? They're cheap, conduct well, and the CO₂ reaction "consumes" the anode slowly, so anodes must be replaced periodically—but this prevents O₂ buildup.

Overall Reaction: $2\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2$

Energy Cost: Aluminum production is extremely energy-intensive—about 15 kWh per kg Al. This is why Al smelters are built near cheap hydroelectric power.

Solution: $m = \frac{M \cdot I \cdot t}{n \cdot F} = \frac{27 \times 150,000 \times (24 \times 3600)}{3 \times 96,500}$

$m = \frac{27 \times 150,000 \times 86,400}{289,500} = \frac{349,920,000,000}{289,500} = 1,208,839 \text{ g} \approx 1209 \text{ kg}$

Why? Charge = $150,000 \times 86,400 = 1.296 \times 10^{10}$ C. That's $\frac{1.296 \times 10^{10}}{96,500} = 134,300$ mol $e^-$. Since 3 electrons make 1 Al atom, we get $\frac{134,300}{3} = 44,767$ mol Al = $44,767 \times 27 = 1,209$ kg.

Common Mistakes & How to Fix Them

Why it feels right: The2+ charge and the 2 electrons are the same number.

What's actually happening: $n$ is the number of electrons transferred in the half-reaction, not just the charge. For $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, we do need 2 electrons, so $n = 2$. But for $\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$, $n = 3$. Always write the half-reaction to count electrons.

Fix: Write the reduction/oxidation half-equation first, count the electrons explicitly.

Why it feels right: We often measure time in minutes or hours in practice.

What's actually happening: The formula requires $I$ in amperes (C/s) and $t$ in seconds. If you use minutes, your answer will be off by a factor of 60.

Fix: Always convert time to seconds before calculation. $t_{\text{seconds}} = t_{\text{minutes}} \times 60 = t_{\text{hours}} \times 3600$.

Why it feels right: Same charge seems like it should do the same work.

What's actually happening: Faraday's Second Law says the ratio depends on $M/n$. Light metals with high charge (like Al, $M/n = 27/3 = 9$) deposit less mass than heavy metals with low charge (like Ag, $M/n = 108/1 = 108$) for the same charge.

Fix: Always account for both molar mass and valency when comparing different substances.

Why it feels right: We're electrolyzing a sodium salt, sodium should come out.

What's actually happening: Electrode potential determines what actually gets reduced. Water has a less negative reduction potential than Na⁺, meaning it's easier to reduce water than Na⁺. The cell takes the easier path unless forced with very high voltage.

$\text{Na}^+ + e^- \rightarrow \text{Na} \quad E^\circ = -2.71 \text{ V (very negative—hard)}$ $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \quad E^\circ = -0.83 \text{ V (easier)}$

Fix: Remember that the species with the less negative (more positive) reduction potential gets reduced first at the cathode in aqueous solution.

Connections & Extensions

• Galvanic Cells and Standard Electrode Potentials — Electrolysis is the reverse of galvanic cells; we apply voltage to reverse the spontaneous reaction
• Nernst Equation — Electrode potentials determine which species actually reacts at each electrode
• Thermodynamics of Electrochemical Cells — Minimum voltage needed for electrolysis = $-E_{\text{cell}}^\circ$ of the reverse galvanic reaction
• Redox Reactions and Balancing — Every electrolysis is a redox reaction forced by external voltage
• Industrial Chemistry — Chlor-alkali and aluminum extraction are multi-billion dollar industries
• Corrosion and Electrochemical Protection — Sacrificial anodes use electrolysis principles in reverse to protect structures

Now, different metals need different numbers of workers per atom. Aluminum is stuborn—it needs 3 workers per atom. Silver is easy—just 1 worker per atom. So the same amount of work makes more silver than aluminum.

In factories, we use this trick to make aluminum (for soda cans and airplanes) and chlorine (for cleaning pools). We dissolve rocks with aluminum in them, heat it up really hot, and zap it with HUGE amounts of electricity. The aluminum melts and sinks to the bottom where we collect it. It takes so much electricity that they build these factories next to big dams!

For $Z = M/(nF)$: "My Naughty Friend" (Molar mass, valency, Faraday constant)

For products of chlor-alkali: "Children Hate Naughty behavior" → Cl₂, H₂, NaOH

For aluminum extraction: "Carbon Anodes Consume Oxygen, Creating CO₂" → At the anode, C + O₂ → CO₂

#flashcards/chemistry

What is electrolysis? :: Using electrical energy to force a non-spontaneous redox reaction to occur.

What is the Hall-Héroult process? :: The industrial electrolytic process for extracting aluminum from alumina (Al₂O₃) dissolved in molten cryolite.

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)