Faraday's Law and Electrolysis

MCAT trap: Swaps anode/cathode polarity between galvanic and electrolytic cells. In an electrolytic cell, the external power source forces the anode to be positive (oxidation) and the cathode to be negative (reduction), opposite to the intuition from galvanic cells.

Faraday's law connects electricity to chemistry: the amount of substance deposited or dissolved at an electrode is directly proportional to the total charge passed through the cell. The core equation is q = n × F × moles, where q is charge in coulombs, n is electrons transferred per ion, and F is Faraday's constant (96,485 C/mol, but use 96,500 on the MCAT). Electrolysis is the forced, non-spontaneous version of electrochemistry — you're pushing current in to drive a reaction that wouldn't happen on its own.

The MCAT tests this in a few distinct ways. Pure recall questions ask you to identify electrode reactions or state the relationship between charge and moles. Calculation questions give you a current, a time, and an ion, then ask for mass deposited — this is where most students drop points because they either forget to multiply current by time to get charge, or they forget to divide by n for multi-electron species. Passage-based questions embed electrolysis in industrial contexts like aluminum smelting (Hall-Héroult process) or electroplating, and ask you to interpret or predict outcomes from data.

What makes this topic genuinely tricky is that students carry over intuitions from galvanic cells and apply them wrong. In a galvanic cell, the anode is the negative terminal; in an electrolytic cell, the external power source flips this — the anode is positive and the cathode is negative. This single swap trips up a surprising number of students. Aqueous electrolysis adds another layer: water itself can compete with dissolved ions at both electrodes, and which species actually reacts depends on comparing reduction potentials. The MCAT rewards students who can reason through electrode product prediction, not just memorize that 'reduction happens at the cathode.'

Common misconceptions

Common mistake

Wrong: In an electrolytic cell, the anode is negative and the cathode is positive (same logic as a galvanic cell).

Right: In an electrolytic cell, the external power source forces the anode to be positive (oxidation) and the cathode to be negative (reduction), opposite to the intuition from galvanic cells.

In a galvanic cell, the anode is negative because it's where spontaneous oxidation releases electrons — that's what makes it the source of current. In an electrolytic cell, an external battery forces the reaction, so the battery's positive terminal pulls electrons away from the anode, making it positive. The chemical definitions stay the same (anode = oxidation, cathode = reduction), but the polarity signs flip. Don't memorize signs — derive them from the direction of electron flow.

Common mistake

Wrong: The number of moles deposited equals the total charge in coulombs divided by Faraday's constant, regardless of the ion's charge.

Right: Moles deposited = total charge / (n × F), where n is the number of electrons transferred per ion; ignoring n gives the wrong answer for multi-electron species like Cu²⁺ or Al³⁺.

Faraday's constant tells you moles of electrons per coulomb, not moles of atoms per coulomb. If you're depositing Cu²⁺, each copper atom requires 2 electrons, so one mole of charge (96,500 C) only deposits half a mole of copper. The formula is moles of substance = q / (n × F). Skipping n works only for singly-charged ions like Ag⁺; for Cu²⁺ (n=2) or Al³⁺ (n=3), forgetting n gives you an answer that's 2x or 3x too high.

Common mistake

Gap: Unaware that water competes with dissolved ions at each electrode during aqueous electrolysis

In aqueous electrolysis, water can be oxidized at the anode (producing O₂) or reduced at the cathode (producing H₂) when the dissolved ions have less favorable electrode potentials, so the actual product depends on comparing reduction potentials of the ion versus water.

Water isn't inert in aqueous electrolysis — it has its own reduction potential (0.00 V for H⁺/H₂ at standard conditions, about -0.83 V for neutral water) and oxidation potential. When a dissolved ion has a less favorable electrode potential than water, water reacts instead. For example, at the cathode in a NaCl solution, Na⁺ has a very negative reduction potential, so water gets reduced to H₂ instead of Na metal depositing. Always ask: which species at this electrode has the more favorable potential?

Common mistake

Wrong: Current (amperes) alone determines how much material is deposited, independent of how long the current flows.

Right: Total charge q = I × t (coulombs = amperes × seconds), and it is total charge—not current alone—that determines moles deposited via Faraday's law.

Current is a rate — coulombs per second. Running 2 A for 10 seconds gives you 20 C of charge; running 2 A for 100 seconds gives you 200 C. The total charge (q = I × t) is what drives the chemistry, not the current alone. This is step one of any electrolysis calculation, and skipping it is the most common arithmetic error. Always convert to total coulombs before applying Faraday's law.

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What the exam tests

Know Faraday's law cold: moles of substance deposited or dissolved equals total charge divided by (n × F), where n is the number of electrons per ion — and be able to rearrange this equation for any unknown.
Given a current (in amperes) and a time (in seconds), calculate the mass of material deposited or volume of gas evolved at an electrode for a specific ionic species.
Apply Faraday's law to real-world industrial processes like electroplating, aluminum production, or electrolysis of water — usually presented in a passage with data you need to interpret rather than just plug in.
Predict what actually forms at each electrode during aqueous electrolysis by comparing the reduction potentials of the dissolved ion versus water, recognizing that the more favorable half-reaction wins.

Can you avoid these mistakes?

A 2.00 A current is passed through a CuSO₄ solution for 30 minutes. How many grams of copper (atomic mass 63.5 g/mol) are deposited at the cathode? (Cu²⁺ + 2e⁻ → Cu)

In the electrolysis of aqueous NaCl, what product forms at the cathode and why — is it Na metal or H₂ gas? What about at the anode — is it Cl₂ or O₂, and what factor determines this?

An electrolytic cell has its positive terminal connected to an electrode submerged in a metal salt solution. Is this electrode the anode or cathode? What reaction occurs there?

You need to deposit 1.08 g of silver (Ag⁺, atomic mass 108 g/mol) using a 0.50 A current. How many seconds must the current run? (Hint: Ag⁺ + e⁻ → Ag, so n = 1)

Faraday's Law and Electrolysis

Common misconceptions

What the exam tests

Can you avoid these mistakes?

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