Determine analyte mass using Faraday's Law of Electrolysis
μC
mC
C
Enter the integrated area under the current vs. time peak (Q).
Please enter a valid positive number.
Molecular weight of the substance being oxidized (e.g., Cu = 63.55).
Please enter a valid molar mass.
Number of electrons involved in the oxidation reaction per molecule.
Value must be at least 1.
Calculated Mass
0.00
μg
Total Charge (Coulombs):0.0005 C
Moles of Substance:0.00 mol
Mass in Nanograms (ng):0.00 ng
Formula Used: Mass = (Q × M) / (n × F)
Mass vs. Peak Charge
Relationship between measured charge and calculating mass for current Molar Mass/Electron settings.
Estimated sensitivity levels for current substance settings.
Signal (Charge)
Calculated Mass
Concentration Regime
Understanding How to Calculate Weight from Oxidation Peak
In analytical chemistry and electrochemistry, the ability to calculate weight from oxidation peak data is fundamental for quantifying substances. Whether you are performing Cyclic Voltammetry (CV), Stripping Voltammetry, or coulometric titrations, the area under an oxidation peak represents the total charge passed during the reaction. By applying physical constants, this charge can be directly converted into the mass of the analyte.
This calculator utilizes Faraday's Law of Electrolysis to provide precise mass conversions from electrochemical data. It is an essential tool for researchers, students, and lab technicians working with quantitative analysis.
What is "Calculate Weight from Oxidation Peak"?
The phrase refers to the process of converting the electrical signal (current integrated over time) from an electrochemical experiment into a tangible physical quantity: mass. When a substance undergoes oxidation at an electrode, electrons are transferred. The total number of electrons transferred is proportional to the number of molecules reacted.
Key Concept: The "Peak" refers to the signal in a voltammogram. The "Area" under this peak (Current × Time) is the Total Charge (Q).
This calculation is commonly used by:
Electrochemists determining the amount of material deposited or stripped from an electrode.
Analytical Chemists quantifying trace metals in water samples via Anodic Stripping Voltammetry.
Battery Researchers calculating the active material mass utilized during charge/discharge cycles.
Formula and Mathematical Explanation
To calculate weight from oxidation peak, we use Faraday's Law. The derivation is straightforward:
1. Calculate Total Charge (Q):
If your instrument gives you the peak area in Volts×Seconds or Amps×Volts, you must convert it to Coulombs (Amps × Seconds).
Q = ∫ I dt
2. Convert Charge to Moles:
Faraday's constant (F) represents the charge of one mole of electrons.
Moles = Q / (n × F)
3. Convert Moles to Mass:
Using the molar mass (M) of the substance.
Mass = Moles × M
The Combined Formula:
m = (Q × M) / (n × F)
Variable Definitions
Variables used in the Faraday Calculation
Variable
Meaning
Unit
Typical Range
m
Mass of substance
Grams (g)
ng to g
Q
Total Charge (Peak Area)
Coulombs (C)
μC to C
M
Molar Mass
g/mol
1 – 300+ g/mol
n
Electrons Transferred
Unitless
1 to 4
F
Faraday's Constant
C/mol
96,485.33
Practical Examples
Example 1: Analyzing Copper Stripping
A researcher performs Anodic Stripping Voltammetry on a water sample to detect Copper (Cu). The oxidation peak area is measured as 500 μC.
Parameters: Molar Mass of Cu = 63.55 g/mol, Oxidation state transition Cu(0) → Cu(2+) implies n = 2.
Calculation: Q = 500 × 10-6 C
m = (0.0005 × 63.55) / (2 × 96485)
m ≈ 1.64 × 10-7 grams = 0.164 μg
Example 2: Silver Deposition
In a plating experiment, the charge consumed to oxidize a Silver (Ag) layer is 1.5 Coulombs.
Parameters: Molar Mass of Ag = 107.87 g/mol, n = 1 (Ag → Ag+ + e–).
Calculation: m = (1.5 × 107.87) / (1 × 96485)
m ≈ 0.001677 grams = 1.68 mg
How to Use This Calculator
Enter Peak Area: Input the numeric value of the area under your oxidation peak. Ensure you select the correct unit (microCoulombs, milliCoulombs, or Coulombs) from the dropdown.
Input Molar Mass: Enter the atomic or molecular weight of the species responsible for the peak.
Specify Electrons (n): Enter the number of electrons transferred in the oxidation half-reaction. (e.g., Fe2+ → Fe3+ is n=1).
Review Results: The calculator instantly provides the mass in micrograms (μg), milligrams (mg), and nanograms (ng), alongside the molar quantity.
Key Factors That Affect Accuracy
When you calculate weight from oxidation peak, several factors can influence the accuracy of your result:
Baseline Subtraction: The most common error source is improper integration of the peak. Background current (capacitive current) must be subtracted to get the true faradaic charge.
Electrode Area: While mass calculation depends on charge, the peak shape and resolution depend on the electrode surface area.
Diffusion Coefficients: If using peak height (current) instead of area to estimate mass (via Randles-Sevcik), diffusion rates heavily impact the result. Integration (Area) is generally more robust for total mass.
Side Reactions: If the oxidation current includes current from solvent decomposition or other interfering species, the calculated mass will be overestimated.
Variable 'n' Values: In complex organic oxidations, the number of electrons transferred can sometimes be non-integer or pH-dependent, complicating the calculation.
Adsorption: If the analyte adsorbs to the electrode surface, the peak may be sharper, but the total charge should still follow Faraday's law assuming 100% stripping efficiency.
Frequently Asked Questions (FAQ)
Q: Can I use this for reduction peaks?
Yes. The physics are identical. For a reduction peak, simply use the absolute value of the charge area. The formula remains valid.
Q: What if I only have Peak Current (Amps), not Area?
You cannot accurately calculate total mass from peak current alone without knowing the scan rate, diffusion coefficient, and time. However, if the peak is a perfect triangle (approximation), Area ≈ (Peak Current × Base Width) / 2.
Q: Why is my calculated weight lower than expected?
This often happens in stripping voltammetry if the deposition step was not 100% efficient or if the stripping time was insufficient to oxidize all material on the electrode.
Q: Is Faraday's constant always 96485?
For most practical laboratory calculations, 96485 C/mol is sufficient. High-precision physics may use 96485.332, but the difference is negligible for standard analysis.
Q: Does temperature affect this calculation?
Temperature does not change Faraday's Law directly. However, temperature affects diffusion and kinetics, which changes the shape of the peak, but not the theoretical charge-to-mass relationship.
Q: How do I handle polymers?
For polymers, use the molar mass of the monomer unit and the 'n' value per monomer unit to calculate the mass of the polymer film.
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