Calculation of the Time Required to Deposit a Given Mass in Electrolysis

Understanding the Time Calculation for Depositing Mass in Electrolysis

Electrolysis time calculation determines how long it takes to deposit a specific mass. This article explains the formulas, variables, and practical examples.

Discover detailed tables of common values, step-by-step calculations, and real-world applications for precise electrolysis timing.

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  • Calculate the time to deposit 10 grams of copper at 2 amperes current.
  • Determine the electrolysis duration for 5 grams of silver with 3 A current.
  • Find the time required to deposit 20 grams of zinc using 4 A current.
  • Compute the electrolysis time for 15 grams of nickel at 1.5 amperes.

Comprehensive Tables of Common Electrolysis Parameters

ElementSymbolAtomic/Molar Mass (g/mol)Valency (n)Faraday Constant (F) (C/mol e⁻)Equivalent Weight (g/equiv)Typical Current (A)Typical Deposition Rate (g/s at 1 A)
CopperCu63.5529648531.7750.5 – 50.00033
SilverAg107.87196485107.870.1 – 30.00112
ZincZn65.3829648532.690.5 – 40.00034
NickelNi58.6929648529.3450.5 – 30.00030
GoldAu196.9739648565.660.1 – 20.00068
LeadPb207.2296485103.60.5 – 30.00107
IronFe55.8539648518.620.5 – 40.00019
ChromiumCr51.99639648517.330.5 – 30.00018
AluminumAl26.983964858.990.5 – 50.00009

Fundamental Formulas for Calculating Electrolysis Time

Electrolysis involves the deposition of a substance on an electrode by passing electric current through an electrolyte. The time required to deposit a given mass depends on the current, the substance’s equivalent weight, and Faraday’s laws of electrolysis.

Primary Formula for Time Calculation

The time t required to deposit a mass m is given by:

t = (m Ɨ F) / (I Ɨ M Ɨ n)
  • t = time in seconds (s)
  • m = mass to be deposited (grams, g)
  • F = Faraday constant ā‰ˆ 96485 coulombs per mole of electrons (C/mol e⁻)
  • I = current applied (amperes, A)
  • M = molar mass of the substance (grams per mole, g/mol)
  • n = number of electrons transferred per ion (valency)

This formula is derived from Faraday’s first law, which states that the mass deposited is proportional to the quantity of electricity passed.

Derivation and Explanation

The total charge Q passed through the electrolyte is:

Q = I Ɨ t

From Faraday’s second law, the amount of substance deposited is proportional to the charge divided by the product of valency and Faraday constant:

m = (M / (n Ɨ F)) Ɨ Q

Rearranging for time t:

t = (m Ɨ F Ɨ n) / (I Ɨ M Ɨ n) = (m Ɨ F) / (I Ɨ M Ɨ n)

Equivalent Weight and Its Role

The equivalent weight E is defined as:

E = M / n

Using equivalent weight, the time formula can be simplified:

t = (m Ɨ F) / (I Ɨ E)

This form is often more convenient when equivalent weights are tabulated.

Additional Considerations

  • Current Efficiency (Ī·): Real electrolysis processes rarely achieve 100% efficiency. The formula adjusts to:
t = (m Ɨ F) / (I Ɨ E Ɨ Ī·)
  • Where Ī· is the current efficiency (decimal form, e.g., 0.95 for 95%).
  • Temperature and Concentration: Affect ion mobility and deposition rate but are not directly in the formula.
  • Overpotential and Electrode Surface Area: Influence practical current density and deposition uniformity.

Real-World Examples of Electrolysis Time Calculation

Example 1: Copper Deposition in Electroplating

A manufacturer wants to deposit 10 grams of copper on a metal part using a 2 A current. Calculate the time required assuming 100% current efficiency.

  • Given:
    • m = 10 g
    • I = 2 A
    • M (Cu) = 63.55 g/mol
    • n = 2 (Cu²⁺ ions)
    • F = 96485 C/mol
    • Ī· = 1 (100% efficiency)

Calculate equivalent weight:

E = M / n = 63.55 / 2 = 31.775 g/equiv

Calculate time:

t = (m Ɨ F) / (I Ɨ E Ɨ Ī·) = (10 Ɨ 96485) / (2 Ɨ 31.775 Ɨ 1) ā‰ˆ 15185 seconds

Convert to hours:

15185 s Ć· 3600 ā‰ˆ 4.22 hours

Result: It will take approximately 4.22 hours to deposit 10 grams of copper at 2 A current.

Example 2: Silver Recovery from Wastewater

In a wastewater treatment plant, silver ions are recovered by electrolysis. The goal is to deposit 5 grams of silver using a 3 A current with 90% current efficiency. Calculate the time required.

  • Given:
    • m = 5 g
    • I = 3 A
    • M (Ag) = 107.87 g/mol
    • n = 1 (Ag⁺ ions)
    • F = 96485 C/mol
    • Ī· = 0.9 (90% efficiency)

Calculate equivalent weight:

E = M / n = 107.87 / 1 = 107.87 g/equiv

Calculate time:

t = (m Ɨ F) / (I Ɨ E Ɨ Ī·) = (5 Ɨ 96485) / (3 Ɨ 107.87 Ɨ 0.9) ā‰ˆ 165.5 seconds

Convert to minutes:

165.5 s Ć· 60 ā‰ˆ 2.76 minutes

Result: Approximately 2.76 minutes are needed to deposit 5 grams of silver at 3 A with 90% efficiency.

Additional Insights and Practical Considerations

While the formulas provide a theoretical time, practical electrolysis involves several factors that can affect the actual deposition time:

  • Current Efficiency Variations: Side reactions such as hydrogen evolution reduce efficiency.
  • Electrolyte Concentration: Low ion concentration can limit deposition rate.
  • Temperature Effects: Higher temperatures generally increase ion mobility but may cause unwanted reactions.
  • Electrode Surface Area: Larger surface areas allow higher current densities without excessive overpotential.
  • Mass Transport Limitations: Diffusion rates can limit ion availability at the electrode surface.

Optimizing these parameters is essential for industrial electrolysis processes to achieve desired deposition rates efficiently.

Summary of Key Variables and Their Typical Ranges

VariableDescriptionTypical Range / ValueUnits
mMass to be deposited0.1 – 1000grams (g)
ICurrent applied0.1 – 100amperes (A)
MMolar mass of element1 – 210grams per mole (g/mol)
nNumber of electrons transferred (valency)1 – 3unitless
FFaraday constant96485coulombs per mole (C/mol e⁻)
Ī·Current efficiency0.7 – 1.0decimal (unitless)

Final Technical Notes

Accurate calculation of electrolysis time is critical in industries such as metal refining, electroplating, and wastewater treatment. Understanding the interplay of current, mass, and electrochemical constants enables engineers to design efficient processes.

Always consider practical factors like current efficiency and electrolyte conditions to refine theoretical calculations. Employing the formulas and data tables provided here will enhance precision and operational control in electrochemical deposition tasks.