electrolysis and Faraday laws

Electrolysis is the process in which an electric current drives a non-spontaneous chemical reaction in an electrolyte, depositing or dissolving material at the electrodes.

The prerequisites — current, voltage, resistance, and Ohm's law — determine how much charge passes: the current through the cell is set by the applied voltage and the cell resistance, $I = U/R$, and the total charge is $Q = I t$.

Faraday's first law states that the mass $m$ of substance transformed at an electrode is proportional to the charge: $m = kQ = kIt,$ where $k$ is the electrochemical equivalent.

Faraday's second law expresses $k$ through molar mass $M$ and valence $n$: $k = \frac{M}{nF},$ so $m = \frac{M I t}{n F}.$

The constant $F \approx 96485\ \text{C/mol}$ is the charge of one mole of electrons.

Thus measuring current and time lets you predict the mass, and measuring mass lets you find valence or molar mass.

A copper sulfate solution is electrolyzed with a steady current $I = 2.0\ \text{A}$ for $t = 30\ \text{min}$. Copper has molar mass $M = 63.5\ \text{g/mol}$ and valence $n = 2$. Find the mass of copper deposited at the cathode.

A cell containing silver nitrate solution has resistance $R = 5.0\ \Omega$ and is connected to a $U = 10\ \text{V}$ source. Silver has $M = 107.9\ \text{g/mol}$ and $n = 1$. Find the mass of silver deposited in $t = 1.0\ \text{h}$.

Two electrolytic cells are connected in series: one contains copper sulfate ($M_{\text{Cu}} = 63.5\ \text{g/mol}$, $n_{\text{Cu}} = 2$) and the other aluminium chloride ($M_{\text{Al}} = 27.0\ \text{g/mol}$, $n_{\text{Al}} = 3$). The same current $I = 3.0\ \text{A}$ flows for $t = 20\ \text{min}$. Find the masses of copper and aluminium deposited and the ratio $m_{\text{Al}}/m_{\text{Cu}}$.

Use an interactive electrolysis simulation (for example, a virtual lab where you can set current, time, ion charge, and molar mass). Do not perform real electrolysis with reactive salts at home; work only in a simulation or under a teacher's supervision.

1. Set a fixed current, e.g. $I = 1.0\ \text{A}$, and record the deposited mass for several times: $t = 60, 120, 180\ \text{s}$.

2. Plot $m$ versus $Q = I t$; fit a straight line and find its slope.

3. Compare the slope with $M/(nF)$ for the chosen ion; explain any difference.

4. Repeat with a different ion of the same valence and check how the slope changes.

Show that deposited mass is directly proportional to charge and that the slope of the line equals $M/(nF)$, the electrochemical equivalent.

A graph with charge $Q$ on the horizontal axis and mass $m$ on the vertical axis, containing a straight line through the origin with slope $M/(nF)$. Beside it, a schematic of an electrolytic cell with a voltage source, anode, cathode, and arrows showing positive and negative ion migration.

In a glass jar of silver nitrate solution $\mathrm{AgNO_3}$ hang a silver anode and a steel spoon — the cathode, with an area of about $30\ \mathrm{cm^2}$. The supply drives a current $I = 2.0\ \mathrm{A}$ for $t = 5\ \mathrm{min} = 300\ \mathrm{s}$. The molar mass of silver is $M = 107.87\ \mathrm{g/mol}$, its density is $\rho = 10.5\ \mathrm{g/cm^3}$, and the Faraday constant is $F = 96\,485\ \mathrm{C/mol}$.

The cathode reaction is $\mathrm{Ag^+} + e^- \to \mathrm{Ag}$, so for silver $z = 1$. Faraday's first law $m = \frac{MIt}{zF}$ gives $m = \frac{107.87 \cdot 2.0 \cdot 300}{1 \cdot 96\,485} \approx 0.67\ \mathrm{g}$ of silver, that is a layer of thickness $m/(\rho S) \approx 0.021\ \mathrm{mm}$ — the same order as decorative plating on jewellery.

In an electrolyser of acidified water the electrodes carry a current $I = 1.5\ \mathrm{A}$ for exactly $t = 20\ \mathrm{min} = 1200\ \mathrm{s}$, so the charge passed through the solution is $Q = It = 1800\ \mathrm{C}$. The released gases are collected in two tubes above the electrodes at standard conditions, where the molar volume of a gas is $22.4\ \mathrm{L/mol}$.

The number of moles of electrons is $n_e = Q/F = 0.01866\ \mathrm{mol}$. Hydrogen forms by $2\mathrm{H^+} + 2e^- \to \mathrm{H_2}$, where $z = 2$, so $n(\mathrm{H_2}) = \frac{Q}{2F} = 0.00933\ \mathrm{mol}$ and $V = 0.00933 \cdot 22.4 \approx 0.209\ \mathrm{L} = 209\ \mathrm{mL}$. Oxygen forms by $2\mathrm{H_2O} \to \mathrm{O_2} + 4\mathrm{H^+} + 4e^-$, where $z = 4$, so twice fewer moles appear: $104\ \mathrm{mL}$. Faraday's second law predicts this fixed $2:1$ volume ratio, and it does not depend on which electrolyte is used.

In a refinery tank of copper sulphate solution $\mathrm{CuSO_4}$ a current $I = 250\ \mathrm{A}$ flows for $t = 24\ \mathrm{h} = 86\,400\ \mathrm{s}$, with the cell voltage near $U = 0.3\ \mathrm{V}$. For copper $M = 63.55\ \mathrm{g/mol}$, and the cathode reaction is $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$. The metal produced goes into wires and cables, including the one that charges a phone.

The charge is $Q = It = 2.16 \times 10^7\ \mathrm{C}$. Faraday's first law gives $m = \frac{MIt}{zF} = \frac{63.55 \cdot 250 \cdot 86\,400}{2 \cdot 96\,485} \approx 7.1\ \mathrm{kg}$ of $99.99\%$ copper per day from one tank. It is exactly $z = 2$ from the $\mathrm{Cu^{2+}}$ ion that fixes the electrochemical equivalent $M/(zF) = 0.329\ \mathrm{mg/C}$: every coulomb passed deposits a strictly determined mass — this is what the second law states for any element.