Faraday-Law Deposition Calculations

Introduction

Electroplating a watch, refining copper for wiring and extracting aluminium from its oxide are all governed by one idea: electrons are a reagent. Each ion that is discharged at an electrode needs a fixed number of electrons, and the number of electrons supplied is fixed by the current and the time. Faraday's laws turn this into a straightforward stoichiometry problem in which the "limiting reagent" is the charge passed.

Core explanation

From current to charge. The charge passed is

Q = I × t

with I in amperes and t in seconds, giving Q in coulombs. Convert minutes or hours to seconds first; this is the most frequent unit slip.

From charge to moles of electrons. One mole of electrons carries 96 485 C (the Faraday constant, F). So

n(e⁻) = Q/F

From electrons to product. Use the half-equation to find the electron-to-product ratio. For Cu²⁺ + 2e⁻ → Cu, two moles of electrons deposit one mole of copper; for Ag⁺ + e⁻ → Ag the ratio is 1 : 1; for Al³⁺ + 3e⁻ → Al it is 3 : 1. In general

n(product) = n(e⁻)/z

and the mass follows from m = n × M. For gases, n(H₂) = n(e⁻)/2 from 2H⁺ + 2e⁻ → H₂, and a volume can be found with the ideal-gas equation.

Combined expression. Putting the steps together, m = ItM/(zF). Using it is fine, but working in stages makes the reasoning visible and helps catch errors.

Reverse problems. To find the time needed for a target mass, run the chain backwards: mass → moles of metal → moles of electrons (multiply by z) → charge (multiply by F) → time (divide by I).

Cells in series. When several electrolytic cells are connected in series, the same current flows through each for the same time, so each receives the same number of moles of electrons. The amounts of product differ only because z and M differ.

Current efficiency. Side reactions, such as hydrogen evolution at a cathode, consume some charge. Current efficiency = (actual product ÷ theoretical product) × 100%. Always calculate the theoretical amount first, then apply the efficiency.

Formulae

Q = It; n(e⁻) = Q/F; n(product) = n(e⁻)/z; m = ItM/(zF).

Step-by-step reasoning

1. Convert time to seconds and calculate Q = It. 2. Find n(e⁻) = Q/96 485. 3. Write the half-equation and read off z. 4. Divide by z to find moles of product. 5. Multiply by molar mass, then apply any current efficiency.

Visual explanation

Picture a conveyor belt delivering electrons to the cathode at a steady rate set by the current. Silver ions each need one electron, so every electron that arrives adds one atom. Copper ions each need two, so the same belt builds copper at half the atom rate. Aluminium needs three electrons per atom, so it grows at one third of the silver atom rate.

Real-world analogy

Electrons are like tickets and each metal ion has an entry price. A silver ion enters for one ticket, a copper ion for two, an aluminium ion for three. The current decides how fast tickets are printed; the entry price decides how many ions get in.

Real-world example

Aluminium smelting uses enormous currents, typically hundreds of kiloamperes per cell, because each aluminium atom needs three electrons and the molar mass is low. Producing one tonne of aluminium requires roughly 10⁷ mol of electrons, which is why smelters are built near cheap electricity supplies.

Why?

Why is the mass deposited proportional to charge? Each electron discharges a fixed fraction of an ion, and the number of electrons is Q/e. Doubling either the current or the time doubles the number of electrons and therefore doubles the number of atoms deposited.

Common misconception

"One mole of electrons deposits one mole of any metal." This ignores the charge on the ion. One mole of electrons deposits one mole of silver but only half a mole of copper and one third of a mole of aluminium.

Worked example

Question: How long must a current of 1.50 A flow to deposit 0.500 g of nickel from Ni²⁺(aq)? (M(Ni) = 58.69 g mol⁻¹)

Reasoning: n(Ni) = 0.500 ÷ 58.69 = 8.52 × 10⁻³ mol. From Ni²⁺ + 2e⁻ → Ni, n(e⁻) = 2 × 8.52 × 10⁻³ = 1.704 × 10⁻² mol. Q = 1.704 × 10⁻² × 96 485 = 1644 C. t = Q/I = 1644 ÷ 1.50 = 1096 s.

Answer: About 1.10 × 10³ s, or 18.3 minutes.

Quick check

1. What mass of copper is deposited from Cu²⁺(aq) by a current of 2.00 A flowing for 30.0 minutes? (M(Cu) = 63.55 g mol⁻¹) Answer: Q = 2.00 × 1800 = 3600 C; n(e⁻) = 0.0373 mol; n(Cu) = 0.0187 mol; mass ≈ 1.19 g.

Exam focus

Examiners reward a clear chain: Q, then n(e⁻), then n(product), then mass. Show the half-equation to justify z, and convert time to seconds. In series-cell questions, state that the moles of electrons are equal in every cell before comparing products.

Advanced insight

Faraday's laws are exact for the charge that actually reduces the target ion; deviations reveal competing reactions. Measuring current efficiency is therefore a diagnostic tool: a copper-plating bath showing 93% efficiency tells an engineer that about 7% of the charge is lost to side reactions, usually hydrogen evolution, which can also affect the quality of the deposit.

Summary

Charge is Q = It, moles of electrons are Q/F, and moles of product are n(e⁻)/z, where z comes from the half-equation. Multiplying by molar mass gives the mass deposited. Reverse problems run the same chain backwards. In series cells the moles of electrons are equal everywhere, and current efficiency scales the theoretical yield.

Practice questions

1. What mass of silver is deposited by 0.500 A in 965 s? (M(Ag) = 107.87 g mol⁻¹) Answer: Q = 482.5 C; n(e⁻) = 5.00 × 10⁻³ mol = n(Ag); mass ≈ 0.540 g. 2. How long does it take to deposit 27.0 g of aluminium at 10.0 A? Answer: n(Al) = 1.00 mol, so n(e⁻) = 3.00 mol and Q = 2.89 × 10⁵ C; t = 2.89 × 10⁴ s ≈ 8.04 h. 3. Cells containing Ag⁺ and Cu²⁺ are in series. If 1.08 g of silver is deposited, what mass of copper is deposited? Answer: n(e⁻) = 1.08/107.87 = 0.0100 mol; n(Cu) = 0.00500 mol; mass ≈ 0.318 g. 4. A plating run should give 1.19 g of copper but gives 1.10 g. Calculate the current efficiency. Answer: Efficiency = 1.10/1.19 × 100 ≈ 92%.