Best examples of electrolytic cell examples and calculations for students

Real-world examples of electrolytic cell examples and calculations

Let’s start where most students actually learn best: concrete, real-world setups. Every example of an electrolytic cell has three moving parts:

• An external power source (battery or DC supply)
• Two electrodes (anode = oxidation, cathode = reduction)
• An electrolyte containing ions that can carry charge

In all of these examples of electrolytic cell examples and calculations, the pattern is the same:

Current → Charge → Moles of electrons → Moles of substance → Mass or volume

That’s Faraday’s laws in action.

Example of copper refining: purifying impure copper anodes

Electrolytic refining of copper is one of the best examples of industrial electrolytic cells.

Setup

• Anode: impure copper slab (Cu with Ag, Au, Zn, etc.)
• Cathode: thin sheet of pure copper
• Electrolyte: acidic CuSO₄ solution
• Power supply: DC source forcing electrons into the cathode

Key half-reactions
At the anode (oxidation):

Cu(s) → Cu²⁺(aq) + 2 e⁻

At the cathode (reduction):

Cu²⁺(aq) + 2 e⁻ → Cu(s)

So the net result is: impure copper dissolves at the anode, pure copper plates out at the cathode.

Sample calculation: mass of copper refined

Suppose a refinery runs a cell at 5000 A for 24 hours. How much copper is deposited on the cathode?

1. Convert time to seconds
   24 h × 3600 s/h = 86,400 s

2. Calculate total charge (Q) using Q = I·t
   Q = 5000 A × 86,400 s = 4.32 × 10⁸ C

3. Convert charge to moles of electrons using Faraday’s constant (F ≈ 96,485 C/mol e⁻)
   nₑ = Q / F = (4.32 × 10⁸ C) / (96,485 C/mol) ≈ 4,477 mol e⁻

4. Relate moles of electrons to moles of Cu
   Each Cu²⁺ needs 2 e⁻ to become Cu(s):

n(Cu) = nₑ / 2 ≈ 4,477 / 2 ≈ 2,238.5 mol Cu

5. Convert moles to mass (M(Cu) ≈ 63.55 g/mol)
   m(Cu) ≈ 2,238.5 mol × 63.55 g/mol ≈ 1.42 × 10⁵ g ≈ 142 kg

So this electrolytic cell produces about 142 kg of refined copper in one day. This is one of the best examples of electrolytic cell examples and calculations because it matches real industrial numbers.

For more background on copper refining and electrochemistry in materials, you can explore resources from engineering programs such as MIT OpenCourseWare.

Electroplating as everyday examples of electrolytic cell examples and calculations

If you’ve ever seen “gold-plated” jewelry or chrome-plated car parts, you’ve seen the result of an electrolytic cell. Electroplating is one of the most common examples of electrolytic cell examples and calculations that show up on exams.

Typical gold-plating setup

• Cathode: metal object to be plated (e.g., a ring)
• Anode: gold metal or inert electrode
• Electrolyte: solution containing Au³⁺ ions
• Power source: low-voltage DC

Cathode half-reaction:

Au³⁺(aq) + 3 e⁻ → Au(s)

Sample calculation: thickness of plating

A jeweler plates a ring with gold using a current of 0.50 A for 30.0 minutes. Assume 100% efficiency and Au³⁺ ions. How much gold mass is deposited?

1. Time in seconds: 30.0 min × 60 s/min = 1800 s

2. Charge: Q = I·t = 0.50 A × 1800 s = 900 C

3. Moles of electrons: nₑ = Q / F = 900 / 96,485 ≈ 9.33 × 10⁻³ mol e⁻

4. Moles of Au: each Au³⁺ needs 3 e⁻
   n(Au) = nₑ / 3 ≈ (9.33 × 10⁻³) / 3 ≈ 3.11 × 10⁻³ mol

5. Mass of Au (M ≈ 197.0 g/mol):
   m(Au) ≈ 3.11 × 10⁻³ mol × 197.0 g/mol ≈ 0.612 g

With density and surface area, you could continue this example of an electrolytic cell calculation to find coating thickness. That’s often the next step in engineering design.

Water electrolysis: hydrogen and oxygen production

Splitting water into hydrogen and oxygen is one of the classic examples of electrolytic cell examples and calculations, and it’s highly relevant to current energy research.

Overall reaction (in acidic solution):

2 H₂O(l) → 2 H₂(g) + O₂(g)

Half-reactions (one common representation):

Cathode (reduction):

2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻(aq)

Anode (oxidation):

4 OH⁻(aq) → O₂(g) + 2 H₂O(l) + 4 e⁻

Sample calculation: volume of hydrogen gas

An electrolyzer for a small lab runs at 2.0 A for 1.0 hour. Approximate the volume of H₂ gas produced at 25 °F (about 298 K) and 1 atm, assuming ideal behavior and 100% efficiency.

1. Time: 1.0 h = 3600 s

2. Charge: Q = 2.0 A × 3600 s = 7200 C

3. Moles of electrons: nₑ = 7200 / 96,485 ≈ 7.46 × 10⁻² mol e⁻

4. Relate electrons to H₂: from the cathode half-reaction, 2 e⁻ → 1 mol H₂
   n(H₂) = nₑ / 2 ≈ 3.73 × 10⁻² mol

5. Volume using ideal gas law (at ~298 K, 1 atm, use ~24.5 L/mol as a reasonable classroom value):
   V(H₂) ≈ 3.73 × 10⁻² mol × 24.5 L/mol ≈ 0.91 L

So this electrolytic cell produces just under 1 liter of hydrogen gas in an hour at 2.0 A.

Research into high-efficiency water electrolysis for hydrogen fuel is actively discussed in materials science and energy courses; U.S. Department of Energy materials at energy.gov are a good entry point for updated context.

Molten salt electrolysis: sodium and aluminum as classic examples

Some of the best examples of electrolytic cell examples and calculations come from molten salts, where water isn’t present to interfere.

Example: Molten NaCl

Overall reaction:

2 NaCl(l) → 2 Na(l) + Cl₂(g)

Cathode: Na⁺(l) + e⁻ → Na(l)
Anode: 2 Cl⁻(l) → Cl₂(g) + 2 e⁻

Imagine passing 10,000 C through molten NaCl. How many grams of sodium form?

1. Moles of electrons: nₑ = 10,000 / 96,485 ≈ 0.104 mol e⁻

2. From the cathode half-reaction, 1 e⁻ → 1 Na⁺ reduced
   n(Na) = 0.104 mol

3. Mass of Na (M ≈ 22.99 g/mol):
   m(Na) ≈ 0.104 × 22.99 ≈ 2.39 g

This is a simple, clean example of an electrolytic cell calculation with a 1:1 electron-to-metal ratio.

Example: Aluminum production (Hall–Héroult process)

Modern aluminum production uses molten Al₂O₃ dissolved in cryolite. The simplified overall reaction is:

2 Al₂O₃(l) → 4 Al(l) + 3 O₂(g)

That’s 12 electrons per 2 Al₂O₃, or 3 electrons per Al atom.

Say a cell runs at 150,000 A (this is realistic for industry) for 1.0 hour. Estimate the mass of aluminum produced.

1. Time: 1.0 h = 3600 s

2. Charge: Q = 150,000 A × 3600 s = 5.40 × 10⁸ C

3. Moles of electrons: nₑ = 5.40 × 10⁸ / 96,485 ≈ 5,597 mol e⁻

4. Moles of Al: 3 e⁻ per Al
   n(Al) = 5,597 / 3 ≈ 1,866 mol

5. Mass of Al (M ≈ 26.98 g/mol):
   m(Al) ≈ 1,866 × 26.98 ≈ 50,300 g ≈ 50.3 kg

Industrial cells run continuously, so daily production is far larger, but this gives a clean example of an electrolytic cell calculation you can adapt in exams.

Using cell potentials in examples of electrolytic cell examples and calculations

So far, we’ve focused on mass and moles. But another layer in advanced examples of electrolytic cell examples and calculations is cell potential.

For a spontaneous galvanic cell, the cell potential E° is positive. For an electrolytic cell, you’re forcing a non-spontaneous reaction, so you must apply a voltage at least equal to the negative of that E° (plus extra for overpotential and resistance).

Example: Electrolysis of aqueous NaCl vs water

In aqueous NaCl, several reactions are possible. Standard reduction potentials (at 25 °C, 1 M) include:

• 2 H₂O + 2 e⁻ → H₂ + 2 OH⁻ E° = −0.83 V (as written for reduction of water to H₂)
• 2 Cl⁻ → Cl₂ + 2 e⁻ E°(reduction of Cl₂) = +1.36 V, so oxidation of Cl⁻ is −1.36 V

In practice, because of overpotentials, water reduction to H₂ often wins at the cathode, while Cl⁻ oxidation to Cl₂ happens at the anode. This gives you a realistic example of an electrolytic cell where thermodynamics and kinetics both matter.

If a textbook gives you E° values and asks for the minimum applied voltage, you:

• Write the half-reactions in the direction they occur in the cell
• Add their potentials (remember, oxidation half-reactions use the negative of their tabulated reduction potential)
• Take the magnitude of the negative E° as the minimum required external voltage, then note that real cells need a bit more due to overpotential and resistance.

For detailed tables and data, standard reduction potentials are widely available in general chemistry texts and online references. Many university chemistry departments, such as those at Harvard University, publish useful data tables for students.

Rechargeable batteries as modern examples of electrolytic cell behavior

Rechargeable batteries give you a nice hybrid: during discharge they act as galvanic cells; during charging they behave as electrolytic cells.

Example: Lithium-ion battery charging

During charging, an external power source forces Li⁺ ions to move back into the graphite anode. While the detailed half-reactions depend on the specific chemistry, conceptually you can still use the same style of examples of electrolytic cell examples and calculations.

Say you have a 3.0 Ah lithium-ion cell. How many moles of electrons move during a full charge?

1. Ampere-hour to coulombs: 3.0 Ah × 3600 C/(A·h) = 10,800 C

2. Moles of electrons: nₑ = 10,800 / 96,485 ≈ 0.112 mol e⁻

If you know the stoichiometry of Li⁺ insertion (e.g., roughly 1 e⁻ per Li⁺), you can estimate how many moles of Li⁺ move and relate that to capacity fade and cycle life.

Battery safety and health impacts of materials (like cobalt) are discussed in public health and environmental chemistry; the U.S. Environmental Protection Agency at epa.gov often provides accessible overviews.

Putting it together: strategy for electrolytic cell calculations

Students often memorize formulas but still get lost in longer questions. The best way to approach any of these examples of electrolytic cell examples and calculations is to use a consistent strategy:

• Step 1 – Identify the species changing: Which ions are being reduced at the cathode? Which are being oxidized at the anode?
• Step 2 – Write half-reactions with electrons shown explicitly.
• Step 3 – Connect current and time to charge using Q = I·t.
• Step 4 – Convert charge to moles of electrons using Faraday’s constant.
• Step 5 – Use half-reaction coefficients to link moles of electrons to moles of substance.
• Step 6 – Convert moles to mass, volume, or concentration as the problem demands.
• Step 7 – If cell potentials are involved, use E° values to check spontaneity and estimate minimum applied voltage.

Once you’ve worked through several real examples of electrolytic cell examples and calculations like copper refining, electroplating, water splitting, and molten salt electrolysis, exam problems start to feel like variations on a theme instead of brand-new puzzles.

FAQ: common questions about electrolytic cell examples and calculations

What are the most common examples of electrolytic cell processes in industry?

Common industrial examples include copper refining, aluminum production (Hall–Héroult process), chlor-alkali production (making Cl₂, H₂, and NaOH from brine), and electroplating for corrosion protection and aesthetics. These are not just textbook cases; they’re major parts of the global metals and chemicals economy.

Can you give a simple example of calculating mass in an electrolytic cell?

Yes. Suppose a current of 2.0 A passes through a CuSO₄ solution for 40 minutes, and copper plates out at the cathode. Convert 40 minutes to seconds, find Q = I·t, convert Q to moles of electrons, divide by 2 (because Cu²⁺ needs 2 e⁻), then multiply by copper’s molar mass. That’s exactly the style of example of an electrolytic cell calculation you’ll see on most general chemistry exams.

How are examples of electrolytic cell examples and calculations used in environmental or health contexts?

Electrochemical methods are used for wastewater treatment, heavy metal removal, and even some medical device manufacturing (like surface finishing of implants). While the detailed chemistry may differ, the same calculation framework applies: current, time, charge, and moles. For health-related materials and exposure issues, agencies like the National Institutes of Health and CDC provide accessible summaries.

Why do some examples of electrolytic cells focus on overpotential and efficiency?

Real electrolytic cells rarely operate at the neat thermodynamic voltage predicted by E°. You need extra voltage to overcome activation barriers, ohmic resistance, and mass transport limits. That extra voltage is called overpotential, and it cuts into energy efficiency. In modern energy systems, like water electrolysis for hydrogen, those efficiency losses matter a lot, which is why newer research and 2024–2025 trends focus heavily on better catalysts and cell designs.

Are there good online resources with more worked examples of electrolytic cell examples and calculations?

Yes. Many university general chemistry courses post problem sets and solutions online. Look for .edu resources from U.S. or U.K. universities. Combined with data tables and explanatory notes from sites like Harvard’s chemistry department and federal agencies, you can build your own library of practice problems based on the examples described here.