Faraday's laws of electrolysis

Faraday's First Law. The mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.

where: $w$ is the mass deposited or liberated (g), $Q = it$ is the charge passed (C), $i$ is the current (A), $t$ is the time (s), and $Z$ is the electrochemical equivalent — the mass deposited per coulomb of charge, a constant specific to each substance.

Faraday's Second Law. When the same quantity of charge is passed through different electrolytes connected in series, the masses of the different substances liberated at their respective electrodes are directly proportional to their chemical equivalent weights.

where: $w_1, w_2$ are the masses deposited in the two cells and $E_1, E_2$ are the equivalent weights (equivalent mass) of the two substances, $E = \dfrac{\text{molar mass}}{\text{number of electrons transferred per ion}}$.

The Faraday constant. One mole of electrons carries a charge of one Faraday, $F = 96{,}500\ \text{C mol}^{-1}$ (from the more precise $96{,}485\ \text{C mol}^{-1}$). Depositing one gram-equivalent of any substance — whatever it is — always requires exactly $1\ F$ of charge. This is the bridge between the two laws: the electrochemical equivalent and the equivalent weight are related by

so $Z$ is not a random per-substance number but simply $E$ scaled by the universal constant $F$. That single relation is why the first law (mass proportional to charge, for one substance) and the second law (mass ratios equal equivalent-weight ratios, across substances) are really the same physics seen two ways.

Linking $F$ to atomic charge. Since a mole of electrons ($N_A = 6.022\times10^{23}$) carries the Faraday, and each electron carries charge $e = 1.602\times10^{-19}\ \text{C}$:

Check: $6.022\times10^{23} \times 1.602\times10^{-19} = 96{,}472\ \text{C mol}^{-1} \approx 96{,}500\ \text{C mol}^{-1}$. This relation lets you move between the macroscopic Faraday constant and the microscopic electronic charge without needing the details of how $e$ was first measured (the Millikan oil-drop experiment is not required at this level — only the relation $F = N_Ae$ itself).

Problem. A current of $2\ \text{A}$ is passed through molten $\text{CuSO}_4$ for $30$ minutes. Calculate the mass of copper deposited at the cathode, and then find the mass of aluminium that the same charge would deposit from molten $\text{AlCl}_3$ connected in series. (Molar mass of Cu $= 63.5\ \text{g mol}^{-1}$, $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$; molar mass of Al $= 27\ \text{g mol}^{-1}$, $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$.)

Step 1. Charge passed: $Q = it = 2 \times (30 \times 60) = 2 \times 1800 = 3600\ \text{C}$. Time was converted to seconds first, since $Q = it$ only gives coulombs when $t$ is in seconds.

Step 2. Equivalent weight of copper: two electrons are transferred per $\text{Cu}^{2+}$ ion, so $E_{\text{Cu}} = 63.5/2 = 31.75\ \text{g eq}^{-1}$.

Step 3. Mass of copper: $w_{\text{Cu}} = \dfrac{E_{\text{Cu}} \, Q}{F} = \dfrac{31.75 \times 3600}{96{,}500} = \dfrac{114{,}300}{96{,}500} = 1.184\ \text{g}$.

Step 4. Equivalent weight of aluminium: three electrons are transferred per $\text{Al}^{3+}$ ion, so $E_{\text{Al}} = 27/3 = 9.0\ \text{g eq}^{-1}$.

Step 5. Apply Faraday's Second Law directly, since the same $3600\ \text{C}$ passes through both cells in series: $\dfrac{w_{\text{Cu}}}{w_{\text{Al}}} = \dfrac{E_{\text{Cu}}}{E_{\text{Al}}} \implies w_{\text{Al}} = w_{\text{Cu}} \times \dfrac{E_{\text{Al}}}{E_{\text{Cu}}} = 1.184 \times \dfrac{9.0}{31.75} = 1.184 \times 0.2835 = 0.336\ \text{g}$.

Both answers check by direct calculation too: $w_{\text{Al}} = E_{\text{Al}}Q/F = (9.0 \times 3600)/96{,}500 = 0.336\ \text{g}$. Final answer: $w_{\text{Cu}} \approx 1.18\ \text{g}$ and $w_{\text{Al}} \approx 0.336\ \text{g}$ — aluminium deposits far less mass for the same charge because each mole of Al needs three electrons, spreading the same charge over a smaller equivalent weight.

• Using molar mass instead of equivalent weight in $w = ZQ$ or $w = Eit/F$ — always divide molar mass by the number of electrons transferred per ion to get $E$.
• Forgetting to convert time to seconds before computing $Q = it$ — minutes or hours left unconverted give a charge, and hence a mass, off by a large factor.
• Applying Faraday's Second Law to cells that are not connected in series — the equal-charge condition needed for $w_1/w_2 = E_1/E_2$ only holds automatically in a series circuit.
• Treating $Z$ (electrochemical equivalent) as one fixed number for all substances — it equals $E/F$ and is different for every ion, so it must be recalculated each time.

Ask the tutor: Want to try a Faraday's-laws numerical with a gas liberated at an electrode instead of a metal deposited, or see how the equivalent weight changes for an ion with a different charge?