DAT General Chemistry: Faraday's Law and Electrolysis Calculations

DAT High-Yield Concept

Electric Charge Formula: Charge is calculated as $Q = I \times t$ , where current $I$ is in amperes ( $\pu{A} = \pu{C/s}$ ) and time $t$ must be converted to seconds ( $\pu{s}$ ).
The Faraday Constant: One mole of electrons carries a charge of $F \approx \pu{96500 C/mol e-}$ (precisely $\pu{96485 C/mol e-}$ ).
Half-Reaction Stoichiometry: The stoichiometric coefficient $n$ of electrons in a reduction half-reaction bridges moles of electrons to moles of metal deposited or gas evolved.
Electrolytic Driven Reaction: Electrolysis uses external electrical energy to force non-spontaneous redox reactions ( $\Delta G > 0$ ).

Fundamentals of Charge and the Faraday Constant

In an electrochemical cell, an external power supply drives a non-spontaneous oxidation-reduction reaction. Quantifying the amount of chemical reaction requires measuring the total electrical charge passed through the solution.

Electric charge ( $Q$ ) in coulombs ( $\pu{C}$ ) is directly proportional to the electric current ( $I$ ) in amperes ( $\pu{A}$ ) and the duration of current flow ( $t$ ) in seconds ( $\pu{s}$ ):

Q = I \times t

The Faraday Constant ( $F$ ): Represents the magnitude of electric charge per mole of electrons.

F = e \times N_A = (\pu{1.602 x 10^{-19} C/e-}) \times (\pu{6.022 x 10^{23} e-/mol e-}) \approx \pu{96500 C/mol e-}

Using the Faraday constant, total charge $Q$ converts directly to moles of electrons ( $n_{e^-}$ ):

n_{e^-} = \frac{Q}{F} = \frac{I \times t}{F}

The Electrolysis Conversion Chain: Moles of Electrons to Mass

Calculating the mass of a substance deposited or liberated at an electrode requires dimensional analysis following a clear conversion pathway: time and current yield charge, charge yields moles of electrons, electron stoichiometry yields moles of analyte, and molar mass yields mass deposited.

Faraday's law as a conversion chain: amps and seconds to grams — Every electrolysis calculation is the same chain of conversions. Current multiplied by time gives the total charge in coulombs. Dividing that charge by the Faraday constant, which is the charge carried by one mole of electrons, gives moles of electrons. Dividing by the number of electrons each ion needs gives moles of the substance produced at that electrode, and multiplying by the molar mass gives the mass. Only the third step depends on the identity of the ion, so the same quantity of charge deposits half as many moles of a two-plus ion as of a one-plus ion. Watch the units: time must be in seconds, because an ampere is a coulomb per second.

Combining these conversion factors produces the master quantitative electrodeposition formula:

m = \frac{I \times t \times M}{n \times F}

where $m$ is mass deposited ( $\pu{g}$ ), $I$ is current ( $\pu{A}$ ), $t$ is time ( $\pu{s}$ ), $M$ is molar mass ( $\pu{g/mol}$ ), $n$ is moles of electrons required per mole of product reduced or oxidized, and $F$ is the Faraday constant ( $\pu{96500 C/mol e-}$ ).

Quantitative Worked Examples & DAT Traps

Worked Example 1: Calculating Mass Deposited at an Electrode

Problem: A current of $\pu{5.00 A}$ is passed through an aqueous solution of $\ce{CuSO4}$ for $\pu{32.0 min}$ . What mass of solid copper ( $\ce{Cu}$ , $M = \pu{63.55 g/mol}$ ) is deposited at the cathode?

Reduction half-reaction: $\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$

Step-by-step Solution:

Convert time to seconds: $t = \pu{32.0 min} \times \frac{\pu{60 s}}{\pu{1 min}} = \pu{1920 s}$
Calculate total charge: $Q = I \times t = \pu{5.00 A} \times \pu{1920 s} = \pu{9600 C}$
Calculate moles of electrons: $n_{e^-} = \frac{\pu{9600 C}}{\pu{96500 C/mol e-}} = \pu{0.09948 mol e-}$
Apply reduction stoichiometry ( $\pu{2 mol e-}$ per $\pu{1 mol Cu}$ ): $n_{\ce{Cu}} = \pu{0.09948 mol e-} \times \frac{\pu{1 mol Cu}}{\pu{2 mol e-}} = \pu{0.04974 mol Cu}$
Calculate mass deposited: $m_{\ce{Cu}} = \pu{0.04974 mol Cu} \times \pu{63.55 g/mol} = \pu{3.16 g}$

DAT Exam Trap: Students frequently forget to convert minutes or hours into seconds, or mistake the ion valence

n

(for example, assuming

n = 1

for

\ce{Cu^{2+}}

). Always write out the balanced half-reaction first to identify

n

! Applied processes like electroplating (covered in depth in its own lesson) follow this identical mathematical relation.

Worked Example 2: Calculating Required Operation Time

Problem: How many minutes are required to deposit $\pu{1.08 g}$ of silver ( $\ce{Ag}$ , $M = \pu{107.87 g/mol}$ ) from an aqueous solution of $\ce{AgNO3}$ using a steady current of $\pu{2.00 A}$ ?

Reduction half-reaction: $\ce{Ag^+(aq) + e^- -> Ag(s)}$

Step-by-step Solution:

Calculate moles of silver: $n_{\ce{Ag}} = \frac{\pu{1.08 g}}{\pu{107.87 g/mol}} = \pu{0.01001 mol Ag}$
Determine moles of electrons required ( $1:1$ ratio): $n_{e^-} = \pu{0.01001 mol e-}$
Calculate required charge: $Q = \pu{0.01001 mol e-} \times \pu{96500 C/mol e-} = \pu{966 C}$
Calculate time in seconds and convert to minutes: $t = \frac{Q}{I} = \frac{\pu{966 C}}{\pu{2.00 A}} = \pu{483 s}$ $t = \frac{\pu{483 s}}{\pu{60 s/min}} = \pu{8.05 min}$

Summary Comparison Table

Parameter	Symbol	Standard Unit	Key Relationship / Conversion
Electric Current	$I$	Amperes ( $\pu{A}$ )	$\pu{1 A} = \pu{1 C/s}$
Time	$t$	Seconds ( $\pu{s}$ )	$\pu{1 min} = \pu{60 s}$ ; $\pu{1 h} = \pu{3600 s}$
Electric Charge	$Q$	Coulombs ( $\pu{C}$ )	$Q = I \times t$
Faraday Constant	$F$	$\pu{C/mol e-}$	$F \approx \pu{96500 C/mol e-}$
Electron Valence	$n$	$\pu{mol e- / mol product}$	From balanced half-reaction coefficient

DAT Critical Thinking Questions

If two electrolytic cells containing solutions of $\ce{Ag^+(aq)}$ and $\ce{Cu^{2+}(aq)}$ are connected in series, which cathode deposits a greater number of moles of metal during the same operational duration? Explain using Faraday's law.
During the industrial extraction of aluminum from molten $\ce{Al2O3}$ , how many moles of electrons are needed to produce $\pu{1 mol}$ of elemental $\ce{Al(s)}$ ?
If an experimental ammeter systematically reads a current higher than the actual current supplied to an electrolysis cell, how will the calculated theoretical mass compare to the actual mass collected?

References

Sources returned by Google Search grounding during lesson verification.

Faraday's Law - Chemistry LibreTexts — chem.libretexts.org
Basic electrolysis calculations - Chemguide — chemguide.co.uk

Faraday's Law and Electrolysis Calculations

Summary

Practice Questions

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