WAChemistrySyllabus dot point

How can we calculate the mass of substance produced at an electrode from the current and time of electrolysis?

Apply Faraday's laws to relate charge, current, time and the amount of substance produced or consumed at an electrode

A focused answer to the WACE Year 12 Chemistry dot point on quantitative electrolysis, using the relationships Q = It and Q = nF with the Faraday constant to calculate the amount of substance deposited or liberated at electrodes, with a worked example and common exam mistakes.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Electrolysis is quantitative: a measured amount of electricity deposits or liberates a predictable amount of substance. This relationship is summarised by Faraday's laws of electrolysis.

The two key relationships

First, the charge passed depends on current and time:

Q = It

where $Q$ is charge in coulombs (C), $I$ is current in amperes (A), and $t$ is time in seconds (s).

Second, the charge relates to moles of electrons through the Faraday constant:

Q = nF

where $n$ is the moles of electrons. Combining these gives a complete chain from current and time to moles of electrons.

Using the half-equation ratio

The half-equation tells you how many electrons are needed per mole of product. For copper, $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ , so 2 mol of electrons deposits 1 mol of copper. For silver, $\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}$ , only 1 mol of electrons is needed per mole. This ratio is the step most often missed.

Mass of copper deposited

Question

A current of $2.00$ A is passed through copper(II) sulfate solution for $30.0$ minutes. Calculate the mass of copper deposited at the cathode. ( $F = 96\,500$ C mol $^{-1}$ , molar mass of Cu $= 63.5$ g mol $^{-1}$ .)

Solution

Step 1, charge: $t = 30.0 \times 60 = 1800$ s, so $Q = It = 2.00 \times 1800 = 3600$ C.

Step 2, moles of electrons: $n(\text{e}^-) = Q / F = 3600 / 96\,500 = 0.0373$ mol.

Step 3, half-equation ratio: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ , so moles of Cu $= 0.0373 / 2 = 0.01865$ mol.

Step 4, mass: $m = n \times M = 0.01865 \times 63.5 = 1.18$ g.

So about $1.18$ g of copper is deposited.

Calculating gas volumes

For a gaseous product such as oxygen or hydrogen, convert moles of electrons to moles of gas using the half-equation, then to volume using the molar gas volume at the stated conditions (for example $24.79$ L mol $^{-1}$ at SLC, or via $pV = nRT$ ). For example, $2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-$ means 2 mol of electrons gives 1 mol of hydrogen gas.

Why this matters

Faraday's laws let industry control electroplating thickness, predict copper refining output, and size the power supply for aluminium smelting. In the examination this is a reliable calculation question that draws together half-equations, the mole concept and electrical quantities.

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