How this instrument works
Electrolysis uses electrical current to drive a chemical reaction that wouldn't happen on its own — most commonly, reducing metal ions in solution into solid metal deposited on an electrode, the basis of electroplating and industrial metal refining. Michael Faraday worked out in 1833 that the amount of substance produced or deposited depends on nothing but the total electric charge passed through the cell and the specific reaction's electron-transfer chemistry — not on the voltage, the electrode material, or how fast versus slow the current was delivered.
The key link is the Faraday constant, F ≈ 96,485 coulombs per mole — the electric charge carried by exactly one mole of electrons. Multiplying current by time gives total charge passed (Q = I·t, in coulombs); dividing that charge by F converts it into moles of electrons delivered; and dividing again by n, the number of electrons each ion needs to be reduced to metal, gives moles of metal deposited. Multiplying by molar mass converts that mole count into the deposited mass in grams — the full chain the formula m = (I·t·M)/(n·F) compresses into one line.
The electron count n depends entirely on the ion's charge: Cu²⁺ needs 2 electrons to become copper metal, so n = 2, while Ag⁺ needs only 1, so n = 1 — meaning that for the identical charge passed, silver deposits twice as many moles as copper would, gram for gram scaled by their different molar masses. Getting n right is the one place this calculation depends on chemistry knowledge rather than pure arithmetic; everything else follows directly from Faraday's constant.
- Enter the electrolysis cell's current into Current (A).
- Enter how long the current ran into Time (s) — convert minutes or hours to seconds first.
- Enter the deposited element's molar mass into Molar mass of deposited element (g/mol).
- Enter how many electrons each ion needs to be reduced into Number of electrons transferred, n — 2 for Cu²⁺, 1 for Ag⁺, 3 for Al³⁺, and so on, matching the ion's charge.
- Read the deposited mass off Mass deposited (g); n must be greater than zero, since dividing by zero electrons has no physical meaning.
Worked example — electroplating copper for one hour
Enter 2 into Current (A), 3600 into Time (s) (one hour), 63.546 into Molar mass of deposited element (g/mol), and 2 into Number of electrons transferred, n — a classic copper-electroplating setup reducing Cu²⁺ to Cu metal. Mass deposited (g) reads 2.3710 g.
By hand: total charge Q = I×t = 2 × 3600 = 7200 coulombs. Dividing by the Faraday constant gives moles of electrons, 7200/96485 ≈ 0.074625 mol; dividing by n = 2 gives moles of copper deposited, ≈ 0.037312 mol; multiplying by copper's molar mass, 63.546 g/mol, gives mass ≈ 2.3710 g — matching the instrument's readout exactly, since it's the same chain of arithmetic compressed into one formula.
Questions
What does the Faraday constant actually represent?
It's the total electric charge carried by exactly one mole of electrons — about 96,485 coulombs. Because chemistry counts reactions in moles but electrical measurements count charge in coulombs, the Faraday constant is the conversion bridge between the two, letting you translate a measured current and time (which give total charge) into a mole count of electrons actually delivered to the electrode.
How do I know what value to use for n, the number of electrons transferred?
It equals the ionic charge of the species being deposited, since that's how many electrons each ion needs to gain to become neutral metal: Cu²⁺ needs 2 (n=2), Ag⁺ needs 1 (n=1), Al³⁺ needs 3 (n=3). Check the half-reaction for your specific electrolysis setup — the coefficient on the electrons in that balanced equation is your n.
Does the electrode material or the electrolyte affect how much mass deposits?
Not according to Faraday's law itself — the deposited mass depends only on total charge passed (current × time), the deposited species' molar mass, and its electron-transfer number n. In real cells, side reactions (like competing hydrogen evolution) can reduce the fraction of current that actually goes toward the intended deposit, an effect called current efficiency, but the ideal Faraday's-law prediction this instrument computes assumes 100% efficiency.
What happens if I pass exactly one Faraday's worth of charge?
Passing exactly F = 96,485 coulombs delivers exactly one mole of electrons by definition, so for a one-electron reduction (n=1) it deposits exactly one mole of that metal — for silver, that's exactly 107.8682 g, silver's molar mass, with no rounding needed. This is a useful sanity check: it's built into the Faraday constant's very definition, independent of trusting any other part of the formula.
How do I convert minutes or hours into the seconds this instrument expects?
Multiply minutes by 60, or hours by 3600, to get seconds — the formula's Faraday constant is defined in coulombs per mole using SI seconds, so current and time must combine to give charge in coulombs (amperes × seconds) for the arithmetic to come out right. A run time of 1 hour, for instance, is entered as 3600 seconds, as in the worked example above.