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CHEMSIM v1.0
⚡ Electrolysis & Faraday's Laws
#23

§1 Interactive Simulation

Aqueous CuSO₄
Molten NaCl
Water Electrolysis
3D Cell View
Electroplating
⚡ THREE.JS 3D ELECTROLYSIS
Mass vs Charge
Mass vs Time
Equivalents Comparison
Current Efficiency
Energy Cost
V (applied)
—
Current I
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Charge Q
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Mass dep.
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Mol dep.
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Energy
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e⁻ flow
Ion migration
Show products
Bubbles (gas)

§2 The Idea, Step by Step

⚡ From a stubborn reaction to grams of metal

Some chemical changes simply refuse to happen on their own. A lump of copper will never crawl out of a blue copper-sulfate solution by itself — that would be running the reaction uphill. Electrolysis is the trick of pushing it uphill anyway, by forcing electric current through the liquid. Hook up a battery: at one metal plate (the cathode) the electrons you supply grab onto positive ions and turn them into solid metal; at the other plate (the anode) something is forced to give those electrons up. The more electricity you push, the more metal you make. That is the whole idea — now let's make it exact.

Counting the electricity

"Amount of electricity" means electric charge, $Q$, measured in coulombs. A steady current $I$ (amperes) flowing for a time $t$ (seconds) delivers $Q = I\,t$. In the 1830s Michael Faraday found that the mass deposited is simply proportional to that charge. Each ion needs a fixed number of electrons to become a neutral atom — call it $n$ ($n=2$ for $\text{Cu}^{2+}$, $n=1$ for $\text{Ag}^+$) — and one mole of electrons always carries the same charge, Faraday's constant $F = 96{,}485$ C. Stitch these together and you get the one equation that runs this whole page:

$$m = \frac{I\,t\,M}{n\,F}$$

where $M$ is the molar mass of the deposited element.

One worked number. Run $I = 2$ A for $t = 600$ s through copper sulfate ($M = 63.55$ g/mol, $n = 2$). Then $Q = It = 1200$ C, and $m = \dfrac{(1200)(63.55)}{(2)(96485)} = 0.395$ g of copper plates onto the cathode. Watch the "Mass dep." readout climb to about this value.

The two ideas hiding inside

First, $n$ is "how many electrons per ion," so a triply-charged ion like $\text{Al}^{3+}$ swallows three electrons per atom — one reason aluminium smelting is such an electricity hog. Second, no real cell is perfect: side reactions (often reducing water instead of the metal) steal a slice of the current, so the honest formula carries a current efficiency $\eta_I$ a little below 1:

$$m = \eta_I\,\frac{I\,t\,M}{n\,F}$$

Voltage plays a separate role. The applied voltage must beat the reaction's own reverse EMF plus extra "overpotential," $V_{\text{applied}} = |E^\circ| + \eta_{\text{over}} + IR$. Voltage decides whether the reaction runs at all; charge decides how much metal you get. On the panel, current $I$ and time $t$ multiply into the charge $Q$, the preset sets $M$ and $n$, the efficiency slider trims the mass, and the voltage slider only feeds the energy-cost readout.

Try this in the sim above. (1) On "Mass vs Charge," change $I$ or $t$ — the dot slides along the same straight line, because mass depends only on $Q = It$, not on how fast you delivered it. (2) Switch the preset from $\text{Cu}^{2+}$ ($n=2$) to $\text{Al}^{3+}$ ($n=3$) and compare the moles deposited at the same charge — more electrons per atom means fewer atoms. (3) Open "Equivalents Comparison" to see why silver ($n=1$) out-deposits aluminium gram-for-coulomb.

§3 Equation Derivation

⚡ Faraday's Laws of Electrolysis

$$\boxed{\;m = \frac{Q\,M}{n\,F} = \frac{I\,t\,M}{n\,F}\;}\qquad\text{(combined Faraday's First and Second laws)}$$

Symbol Definitions

SymbolMeaningSI Unit
$m$Mass of substance deposited or liberatedg
$Q$Total electric charge passedC (coulomb)
$I$Constant currentA (ampere)
$t$Time current flowss
$M$Molar mass of substanceg/mol
$n$Charge number of ion (electrons per ion)—
$F$Faraday's constant96 485 C/mol e⁻
$M/n$Equivalent weight (electrochemical equivalent × F)g/equiv

Step-by-Step Derivation

1Faraday's First Law (1834). The mass of substance produced at an electrode is directly proportional to the charge passed: $$m \propto Q\;\Longrightarrow\;m = Z\,Q$$ where $Z$ is the electrochemical equivalent — mass deposited per coulomb (g/C).
2Faraday's Second Law. When the same charge is passed through different electrolytes, the masses deposited are proportional to their equivalent weights ($M/n$): $$\frac{m_1}{m_2} = \frac{(M/n)_1}{(M/n)_2}$$ This implies $Z = (M/n)/F$, where $F$ is a universal constant.
3Combine the two laws. Substituting $Z = M/(nF)$ into $m = ZQ$: $$\boxed{m = \frac{QM}{nF} = \frac{ItM}{nF}}$$ using $Q = It$ for constant current.
4Moles deposited. Dividing by molar mass $M$: $$n_{\text{mol}} = \frac{m}{M} = \frac{Q}{nF} = \frac{It}{nF}$$ For 1 Faraday (96 485 C), exactly 1 mole of electrons passes — depositing 1/n moles of metal of charge n+.
5Energy and minimum voltage. Electrical energy consumed: $$E_{\text{elec}} = QV = ItV\;\text{(joules)}$$ The minimum theoretical voltage to drive electrolysis equals the magnitude of the cell's reverse $E^\circ$. Real cells need an additional overpotential $\eta$ to overcome kinetic barriers: $$V_{\text{applied}} = |E^\circ_{\text{rxn}}| + \eta_{\text{anode}} + \eta_{\text{cathode}} + IR_{\text{cell}}$$
6Current efficiency. In practice, side reactions (water reduction, etc.) consume some current. If $\eta_I < 100\%$ is the current efficiency: $$m_{\text{actual}} = \eta_I \cdot \frac{ItM}{nF}$$ Industrial electrolysis (Cu refining, Al smelting) typically runs at $\eta_I \approx 90$–$98\%$.

Worked Example — Copper Refining

Problem: Calculate the mass of Cu deposited from CuSO₄(aq) when a current of 5.00 A flows for 30.0 minutes.
Half-reaction: Cu²⁺ + 2e⁻ → Cu, so $n = 2$, $M_{\text{Cu}} = 63.55$ g/mol.
$$Q = It = (5.00)(1800) = 9000\text{ C}$$ $$m = \frac{QM}{nF} = \frac{(9000)(63.55)}{(2)(96485)} = 2.964\text{ g}$$ $$n_{\text{mol}} = \frac{m}{M} = \frac{2.964}{63.55} = 0.04664\text{ mol Cu} = 0.09328\text{ mol e}^-$$ Check: $0.09328\times 96485 = 9000$ C — exactly the charge passed.

📚 Atkins & de Paula — Physical Chemistry, 11th Ed., §6F: "Electrolysis" | Skoog et al. — Fundamentals of Analytical Chemistry, 9th Ed., Ch. 22 | Bockris & Reddy — Modern Electrochemistry, Vol. 2A

§4 Frequently Asked Questions

📚 LibreTexts — "Electrolysis" | Khan Academy — Faraday's laws | MIT OCW 5.111

§5 Common Misconceptions

📚 Sanger & Greenbowe — J. Chem. Educ. 74, 819 (1997) | Garnett & Treagust — J. Chem. Educ. 69, 121 (1992) | Taber — Chemical Misconceptions (RSC, 2002)