Predicting Deposited Metal Amount

Introduction

A charge meter does not directly report grams of metal. To predict an ideal deposit, convert charge to electron moles, use the electrode half-reaction to find metal moles, and multiply by metal molar mass. The ion's electron requirement is the key chemical step.

Core explanation

For Mᶻ⁺ + z e⁻ → M, each metal atom needs z electrons. If charge Q passes entirely through this reduction, electron amount is Q/F and theoretical metal amount is Q/(zF). Theoretical mass is m = QMᵣ/(zF), where Mᵣ is metal molar mass. This compact formula follows from the conversion chain and should not be used without checking the actual oxidation state and half-reaction.

For copper, Cu²⁺ + 2e⁻ → Cu, so z = 2. Passing 19,297 C gives 0.200 mol electrons and 0.100 mol Cu ideally. At 63.5 g mol⁻¹, that is 6.35 g Cu. For silver(I), Ag⁺ + e⁻ → Ag, the same charge gives 0.200 mol Ag, with a different mass because silver molar mass is about 107.9 g mol⁻¹.

Current and time can supply Q: a 2.00 A current for 1000 s gives 2000 C. Then n(e⁻) ≈ 2000/96,485 = 0.0207 mol. If only Cu²⁺ reduction occurs, n(Cu) ≈ 0.0104 mol and mass ≈ 0.658 g. Keeping extra digits through intermediate steps avoids unnecessary rounding.

Theoretical amount is an upper result under the specified single-reaction model. Hydrogen evolution or another cathode reaction can use some current, reducing metal deposition. Metal may also peel off or be lost during weighing. If efficiency is given, multiply theoretical amount by the fraction of charge used for the target reaction; do not subtract an arbitrary percentage from current before converting unless that fraction is explicitly defined.

Deposit thickness needs further information. If the metal mass is known, volume is mass divided by metal density; average thickness is deposited volume divided by coated area, assuming a uniform layer. Total mass alone does not show whether the coating is even across a complex object.

Step-by-step reasoning

1. Calculate Q = It with seconds if current is provided. 2. Divide Q by F to obtain electron moles. 3. Write the metal-ion reduction and read z. 4. Divide electron moles by z for deposited metal moles. 5. Multiply by molar mass and apply supplied efficiency if relevant.

Visual explanation

Draw four boxes: 2000 C → 0.0207 mol e⁻ → 0.0104 mol Cu → 0.658 g Cu. Under the electron-to-copper arrow, write Cu²⁺ + 2e⁻ → Cu.

Real-world analogy

If each finished device requires two identical parts, twenty parts can make ten devices. A divalent metal ion similarly needs two electrons for one deposited atom.

Real-world example

A plating shop can compare predicted mass from logged charge with the mass increase of a part. A shortfall may reflect side reactions, coating loss or measurement errors, rather than an incorrect Faraday constant.

Why?

Why does Cu²⁺ need two electrons? Its +2 charge must be neutralized to form Cu metal in the written half-reaction, so electron and charge balance require two.

Common misconception

“Charge divided by F always equals moles of deposited metal.” It equals moles of electrons. Divide by the number of electrons per deposited atom.

Worked example

A 1.50 A current runs for 20.0 minutes in an ideal Ni²⁺ plating cell. Time is 1200 s; charge is 1800 C. Electron amount is 1800/96,485 ≈ 0.0187 mol. Nickel amount is half, 0.00933 mol; with molar mass 58.7 g mol⁻¹, theoretical deposit is about 0.548 g. A different current efficiency would change actual recovered mass.

Quick check

1. How many moles Al can be deposited ideally from 0.300 mol electrons if Al³⁺ + 3e⁻ → Al? Answer: Divide by three electrons per aluminum atom: 0.300/3 = 0.100 mol Al.

Exam focus

Show the electron-mole intermediate and metal half-reaction. Keep current-time conversion, Faraday constant and z factor distinct.

Advanced insight

The mass formula assumes all current is faradaic and assigned to the chosen metal. Double-layer charging and competing processes can complicate a short or nonsteady experiment.

Summary

Theoretical metal mass follows charge → electron amount → metal amount → grams. The electron coefficient in the balanced reduction determines how many metal moles form per Faraday of charge. Actual deposit can be lower.

Practice questions

1. What metal moles form from 0.200 mol electrons for Cu²⁺ reduction? Answer: Two electrons per Cu atom give 0.100 mol Cu. 2. Find charge from 2.00 A over 5.00 minutes. Answer: Time is 300 s, so Q = 2.00 × 300 = 600 C. 3. Why might measured copper mass be less than the theoretical charge-based value? Answer: Some charge may drive side reactions or deposited copper may be lost during collection and weighing.