Faraday's laws of electrolysis

Result

Faraday's laws of electrolysis

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Statement

Charge passed and moles of product deposited.

Why it matters

galvanic-cell-emf and gibbs-emf-relation established how a spontaneous redox reaction delivers electrical work; Faraday's laws are the same electron bookkeeping run against the current, forcing a non-spontaneous redox reaction to proceed by driving charge through a cell from an external source. This is the quantitative bridge between an electrical measurement — current and time, both trivial to measure precisely — and a chemical one, the moles (and hence mass) of a substance produced or consumed at an electrode. Electroplating, electrorefining of metals, and large-scale industrial processes such as the chlor-alkali process and aluminium extraction all rest on exactly this relationship.

It also gives the Faraday constant itself, \(F\), a direct operational meaning: the charge carried by one mole of electrons, linking the macroscopic world of ammeters and stopwatches to the microscopic count of electrons transferred per ion.

Hypotheses

All charge passed through the cell drives exactly one specified electrode reaction (100% current efficiency).In a real cell, some current can be diverted into a competing electrode reaction (commonly the electrolysis of water itself, or a side reaction at high overpotential) rather than the reaction of interest; Faraday's law as stated below then over-predicts the yield of the intended product unless corrected by a measured efficiency factor. The number of electrons transferred per formula unit, \(n\), is fixed and known from the balanced half-reaction.Faraday's law converts charge to moles of electrons directly; converting further to moles of product requires dividing by the correct stoichiometric \(n\), which must be identified correctly for the actual electrode reaction occurring, not assumed from the ion's formula alone. At high current density, mass-transport limitations (the electrode reaction consuming ions faster than they can diffuse to the electrode) can also depress the effective current efficiency, an issue explored further once this unit reaches electroplating and electrorefining in practice.

Proof

Q = It

For a constant current \(I\) flowing for a time \(t\), the total charge passed is simply their product — the basic definition of electric current as charge per unit time, rearranged. A

n_{e^-} = \frac{Q}{F}, \qquad F = 96485\,\text{C/mol}

The Faraday constant \(F\) is defined as the magnitude of charge carried by one mole of electrons; dividing the total charge passed by \(F\) therefore gives the number of moles of electrons that have been transferred at the electrode. A

n_{\text{substance}} = \frac{n_{e^-}}{n}

The balanced half-reaction fixes a definite stoichiometric ratio between electrons transferred and formula units of product (e.g. \(\text{Cu}^{2+}+2e^-\to\text{Cu}\) has \(n=2\)); dividing the moles of electrons by this \(n\) gives the moles of product actually formed or consumed. A

m = n_{\text{substance}}\times M

Multiplying by the molar mass \(M\) of the substance converts the mole quantity from Step 3 into a directly measurable mass, closing the chain from electrical measurement to chemical yield. A

Result

m = \frac{ItM}{nF}

Reading. The mass of substance produced or consumed at an electrode is directly proportional to the charge passed (current times time) and to the substance's molar mass, and inversely proportional to the number of electrons its formation requires — a single equation combining both of Faraday's original laws of electrolysis.

Scope. Strictly valid only at 100% current efficiency for the specified electrode reaction (Hypotheses); real industrial cells routinely report and correct for efficiencies below this ideal.

Corollaries & converses

For a fixed charge passed, the mass of two different substances deposited at two different electrodes stands in the ratio of their equivalent weights, \(M/n\) — Faraday's original second law, recovered directly from the Result held at constant \(Q\).
galvanic-cell-emf and gibbs-emf-relation describe the reverse process: a spontaneous cell reaction delivering charge outward to do electrical work. Electrolysis is the identical electron bookkeeping run against spontaneity, forcing the reaction backward by supplying charge from an external source; the same \(n\) and \(F\) govern both directions.
electrochemical-series ranks which species is reduced or oxidised preferentially when several are available at an electrode (for instance, in aqueous solution, whether a metal ion or water itself is reduced at the cathode), determining which reaction Faraday's law should actually be applied to.

Fails without

Drop the 100% current-efficiency hypothesis (Hypotheses), letting some charge divert into a competing electrode reaction: the actual mass of the intended product falls below the Result's prediction, and only a separately measured current-efficiency factor, multiplying the Result, can restore agreement with the true yield.
Apply the Result with the wrong stoichiometric \(n\) for the actual electrode reaction occurring: since \(n\) enters the Result as a simple divisor, an incorrect \(n\) scales the predicted mass by the wrong integer factor throughout, even though every other quantity (current, time, molar mass) may be entered correctly.

Common errors

Omitting the stoichiometric \(n\) entirely, or reading it off the ionic charge without checking the actual balanced half-reaction (e.g. treating \(\text{Al}^{3+}\to\text{Al}\) as if \(n=1\) rather than \(3\)).
Confusing the Faraday constant \(F=96485\,\text{C/mol}\) with Avogadro's number \(N_A=6.022\times10^{23}\,\text{mol}^{-1}\); \(F=N_Ae\), the product of the two, not either alone.
Forgetting to convert time into seconds before substituting into \(Q=It\), since \(I\) is conventionally in amperes (coulombs per second).
Applying the Result to a cell with multiple competing electrode reactions without first identifying, or correcting for, the actual current efficiency of the reaction of interest (Hypotheses).

Discussion

Michael Faraday established these quantitative laws of electrolysis in 1833, well before the electron itself was identified. At the time, the striking proportionality between charge passed and the amount of substance liberated was itself among the strongest available pieces of evidence that electricity, like matter, might be fundamentally discrete rather than a continuous fluid — a suspicion only fully vindicated decades later once the electron's charge \(e\) was measured directly and the relation \(F=N_Ae\) could be verified.

Industrially, the Result underlies electroplating (depositing a thin, controlled-thickness protective or decorative metal layer), electrorefining (purifying a crude metal by dissolving it at an anode and redepositing it purely at a cathode), and bulk electrolytic production of reactive elements that cannot be obtained by chemical reduction alone, such as aluminium (Hall–Héroult process) and chlorine and sodium hydroxide (chlor-alkali process).

Common misconception: that the current alone determines the amount of substance produced, independent of which reaction is occurring. The Result makes explicit that the same charge produces different amounts of different substances, scaled inversely by each reaction's own \(n\) — charge measures electrons transferred, not moles of product directly.

Worked examples

\text{Cu}^{2+}+2e^-\to\text{Cu}, \quad I=2.00\,\text{A},\ t=3600\,\text{s} \ (1\,\text{hour})

\(Q=It=2.00\times3600=7200\,\text{C}\); \(n_{e^-}=7200/96485=0.0746\,\text{mol}\); with \(n=2\) electrons per copper ion, \(n_{\text{Cu}}=0.0373\,\text{mol}\). A

m_{\text{Cu}} = n_{\text{Cu}}\times M_{\text{Cu}} = 0.0373\times63.55 = 2.37\,\text{g}

Applying Step 4 of the Proof directly with copper's molar mass gives the mass of copper deposited at the cathode over the one-hour electrolysis. A

m_{\text{Cu}} \approx 2.37\,\text{g}

Reading. A precisely measurable electrical quantity (2.00 A for one hour) predicts an equally precise chemical yield, assuming ideal current efficiency.

Scope. The identical calculation, with \(n=1\) in place of \(n=2\), would give roughly double the molar (though not mass) yield for silver deposition from \(\text{Ag}^+\) under the same charge — illustrating Faraday's second law directly.

Problems

A current of 5.00 A is passed through molten \(\text{Al}_2\text{O}_3\) for 4.00 hours, depositing aluminium at the cathode (\(\text{Al}^{3+}+3e^-\to\text{Al}\), \(M_{\text{Al}}=27.0\,\text{g/mol}\)). Find the mass of aluminium deposited.
Solution
\(Q=It=5.00\times(4.00\times3600)=72000\,\text{C}\). \(n_{e^-}=72000/96485=0.746\,\text{mol}\). With \(n=3\), \(n_{\text{Al}}=0.746/3=0.2486\,\text{mol}\). \(m=0.2486\times27.0=6.71\,\text{g}\).
What current, held constant for exactly 30.0 minutes, is required to deposit 1.00 g of silver from \(\text{Ag}^+\) (\(n=1\), \(M_{\text{Ag}}=107.9\,\text{g/mol}\))?
Solution
\(n_{\text{Ag}}=1.00/107.9=9.27\times10^{-3}\,\text{mol}\). \(n_{e^-}=n_{\text{Ag}}\times1=9.27\times10^{-3}\,\text{mol}\). \(Q=n_{e^-}F=9.27\times10^{-3}\times96485=894.6\,\text{C}\). \(t=30.0\,\text{min}=1800\,\text{s}\), so \(I=Q/t=894.6/1800=0.497\,\text{A}\).
The same charge, 10000 C, is passed through two separate cells in series, one depositing copper (\(n=2\)) and one depositing silver (\(n=1\)). Find the ratio of the masses of copper to silver deposited, and confirm it equals the ratio of their equivalent weights \(M/n\).
Solution
Since the same \(Q\) and \(F\) apply to both, \(m_{\text{Cu}}/m_{\text{Ag}} = (M_{\text{Cu}}/n_{\text{Cu}})/(M_{\text{Ag}}/n_{\text{Ag}}) = (63.55/2)/(107.9/1) = 31.78/107.9 = 0.295\). This matches the Corollaries' statement that equal charge deposits masses in the ratio of equivalent weights directly, independent of computing each mass separately.