Quantitative electrolysis and Faraday's laws: WACE Year 12 Chemistry

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:

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:

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.

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. A useful sanity check at the anode and cathode of the same cell: the same number of electrons passes through both, so the moles of product at each electrode are in the ratio of the inverse of their electron requirements. Hydrogen ($2\text{e}^-$ per mole) and oxygen ($4\text{e}^-$ per mole) are therefore produced in a $2:1$ mole ratio, exactly as the electrolysis of water predicts.

Cells in series and current efficiency

When two electrolytic cells are connected in series, the same charge passes through both, so the moles of electrons are identical in each cell. This is the basis of the classic "silver coulometer" question: measure the silver deposited in one cell to find the charge, then apply that charge to a second cell containing a different metal. The number of electrons is the link.

Real cells are never perfectly efficient. Some current is wasted in side reactions (such as the reduction of water instead of the target ion), so the actual mass deposited is less than the theoretical Faraday prediction. Current efficiency is the ratio

If a question gives an efficiency below $100\%$, calculate the theoretical mass with Faraday's laws first, then multiply by the efficiency to get the real yield.

Why the relationships hold

Faraday's first law (mass proportional to charge) and second law (for the same charge, mass proportional to molar mass divided by the number of electrons, the chemical equivalent) both fall out of the single idea that each ion is discharged by a fixed whole number of electrons. The Faraday constant is just Avogadro's number multiplied by the charge on one electron, $F = N_A \times e = (6.022 \times 10^{23})(1.602 \times 10^{-19}) = 96\,485\ \text{C mol}^{-1}$, which is why $96\,500\ \text{C mol}^{-1}$ is used in WACE calculations.

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.