18,000
C
18,000
C
0.186557
F
9.327875e-2
mol
5.927491
g
5.9275
g/h
18,000
C
18,000
C
0.186557
F
9.327875e-2
mol
5.927491
g
5.9275
g/h
The Electrolysis Calculator applies Faraday's laws of electrolysis to determine the mass of substance deposited or dissolved at an electrode during electrolysis. By entering the current, time, molar mass, and number of electrons transferred, the calculator computes the total charge passed, the mass deposited, the number of moles, and the Faradays of charge. Electrolysis is the process of using electrical energy to drive non-spontaneous chemical reactions at electrodes. This fundamental principle underlies electroplating, metal refining, anodizing, chlorine production, water splitting, and electrowinning. Accurate mass calculations ensure process control, quality assurance, and cost optimization in both laboratory and industrial electrochemistry.
Faraday's combined law of electrolysis states:
$$m = \frac{M \cdot I \cdot t}{n \cdot F}$$
where m is the mass deposited (g), M is the molar mass (g/mol), I is the current (A), t is the time (s), n is the number of electrons transferred per ion, and F = 96,485 C/mol is the Faraday constant.
The total charge passed through the cell is:
$$Q = I \times t$$
The number of moles deposited is:
$$\text{moles} = \frac{Q}{n \cdot F} = \frac{I \cdot t}{n \cdot F}$$
The number of Faradays (moles of electrons) is:
$$\text{Faradays} = \frac{Q}{F} = \frac{I \cdot t}{F}$$
These equations assume 100% current efficiency — all charge goes to the desired reaction. In practice, side reactions (like hydrogen evolution) may reduce the actual yield.
The mass deposited is directly proportional to the current and time (charge passed). Doubling the current or time doubles the mass. A substance with a higher molar mass deposits more mass per Faraday, while a higher valence (n) reduces the mass per Faraday. The Faradays of charge tell you how many moles of electrons were transferred. One Faraday (96,485 C) deposits one mole of a monovalent ion (like Ag⁺), half a mole of a divalent ion (like Cu²⁺), or one-third of a mole of a trivalent ion (like Al³⁺).
Inputs
Results
Q = 5 × 3600 = 18,000 C. m = (63.546 × 18000)/(2 × 96485) = 5.93 g. About 5.93 g of copper deposits in one hour at 5 A. This represents 0.0933 mol Cu²⁺ ions reduced, consuming 0.1866 Faradays of charge.
Inputs
Results
Q = 2 × 1800 = 3600 C. m = (107.868 × 3600)/(1 × 96485) = 4.02 g. Silver being monovalent (n = 1) deposits more mass per coulomb than divalent metals. 4.02 g of silver in 30 minutes at 2 A.
Electrolysis is the process of passing an electric current through an electrolyte to cause a non-spontaneous chemical reaction. Positive ions migrate to the cathode (reduction) and negative ions to the anode (oxidation).
The mass of substance deposited at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte: m ∝ Q = It. More charge means more substance deposited.
When the same quantity of electricity passes through different electrolytes, the masses deposited are proportional to their electrochemical equivalents (M/n). Heavier atoms with lower valence deposit more mass per coulomb.
Different ions require different numbers of electrons for deposition. Cu²⁺ needs 2 electrons, Ag⁺ needs 1, Al³⁺ needs 3. The value of n comes from the oxidation state of the ion being deposited and directly affects the mass calculation.
Multiply the theoretical mass by the current efficiency (as a fraction). For example, if efficiency is 85%, actual mass = 0.85 × theoretical mass. Side reactions like hydrogen evolution or oxygen evolution consume part of the charge.
Major applications include aluminum production (Hall-Héroult process), copper refining, chlor-alkali industry (Cl₂ and NaOH), electroplating (Cr, Ni, Zn, Au), anodizing, and hydrogen production by water splitting.
The Faraday constant F = 96,485 C/mol is the total charge carried by one mole of electrons (Avogadro's number × elementary charge). It connects macroscopic electrical measurements to the number of atoms deposited.
Yes. Molten salt electrolysis is used when aqueous electrolysis is not feasible — for example, producing sodium (Na⁺ has very negative E°) or aluminum (Al³⁺). The Downs process electrolyzes molten NaCl to produce Na and Cl₂.
Higher temperature reduces electrolyte resistance (lower IR drop), increases reaction kinetics (lower overpotential), and can shift competing reaction equilibria. However, it may also accelerate electrode corrosion and side reactions.
The ion with the least negative (or most positive) reduction potential is deposited first. In aqueous CuSO₄, Cu²⁺ (+0.34 V) deposits before H⁺ (0.00 V). This discharge potential also depends on concentration and overpotential.
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