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

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.

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.