Electrolysis Calculator
Calculate the mass deposited or produced in an electrolysis cell using Faraday's law. Enter the current, time, molar mass, and electrons transferred per formula unit to get charge, moles, and deposited mass.
Electrolysis Deposit from Faraday's Law
Electrolysis Formula
Where:
- \(Q\) — electric charge passed through the cell (coulombs)
- \(I\) — current (amperes)
- \(t\) — time the current flows (seconds)
- \(z\) — electrons transferred per formula unit of product
- \(F\) — Faraday constant, \(96\,485.33\ \text{C/mol}\)
- \(n\) — moles of product formed (mol)
- \(M\) — molar mass of the product (g/mol)
- \(m\) — theoretical deposited mass (g)
These equations express Faraday's laws of electrolysis: the amount of a substance produced at an electrode is directly proportional to the quantity of electric charge passed through the cell.
Electrolysis – Calculation Example
Example: a copper strip is electroplated in a CuSO4 bath. A current of 2 A flows for 30 minutes. Copper (\(M = 63.546\) g/mol) deposits from Cu2+ ions, so \(z = 2\).
\[ Q = I \cdot t = 2\ \text{A} \times 1800\ \text{s} = 3600\ \text{C} \]
\[ n = \frac{Q}{z \cdot F} = \frac{3600}{2 \times 96\,485.33} = 0.018656\ \text{mol} \]
\[ m = n \cdot M = 0.018656 \times 63.546 \approx 1.1855\ \text{g} \]
Result: the charge passed is 3600 C, producing 0.018656 mol of copper, which corresponds to approximately 1.1855 g of copper deposited on the electrode.
How to Use the Electrorolysis Calculator
- Enter the Current flowing through the cell and select the unit (A or mA) from the dropdown.
- Enter the Time the current flows and choose seconds, minutes, or hours.
- Enter the Molar Mass of the deposited or produced substance in g/mol (e.g. 63.546 for Cu, 107.868 for Ag).
- Enter the number of Electrons per Product Unit (\(z\)): 1 for Ag+ → Ag, 2 for Cu2+ → Cu, 3 for Al3+ → Al.
- Read the results: the Electric Charge in coulombs, the Moles of Product formed, and the Theoretical Deposited Mass in grams.
What Is Electrolysis?
Electrolysis is a chemical process that uses direct electric current to drive a non-spontaneous redox reaction. Inside an electrolytic cell, ions migrate toward electrodes: positive ions (cations) gain electrons at the cathode, while negative ions (anions) lose electrons at the anode. This is how metals are electroplated, water is split into hydrogen and oxygen, and pure aluminum and chlorine are industrially produced.
The quantitative relationship between electricity and chemical change is described by Faraday's laws. The first law states that the mass of a substance altered at an electrode is proportional to the quantity of charge passed. The second law states that for the same charge, the amounts of different substances produced are proportional to their equivalent weights (molar mass divided by \(z\)). Together, they give the single equation implemented in this calculator: \(m = \frac{I \cdot t \cdot M}{z \cdot F}\).
The three results build on each other. The charge \(Q\) tells you how much electricity actually flowed. Dividing by \(z \cdot F\) converts that charge into moles of product — one mole of electrons (96,485 C) reduces one mole of singly charged ions, half a mole of doubly charged ions, and so on. Multiplying by the molar mass finally gives the mass you would expect to deposit, in grams.
If you want to compare the calculated theoretical deposit with what you actually measured on an electrode, you can express the outcome as a percentage using the Percent Yield calculator. To understand which species reacts at each electrode, the Balancing Redox Reactions calculator helps you work out the half-reactions and the value of \(z\).
Common Values of z and M
| Half-reaction | z | M (g/mol) |
|---|---|---|
| Ag+ + e− → Ag | 1 | 107.868 |
| Cu2+ + 2e− → Cu | 2 | 63.546 |
| Zn2+ + 2e− → Zn | 2 | 65.38 |
| Ni2+ + 2e− → Ni | 2 | 58.693 |
| Al3+ + 3e− → Al | 3 | 26.982 |
| 2H+ + 2e− → H2 | 2 | 2.016 |
Assumptions and Limitations
- 100% current efficiency: the calculator assumes every electron goes into the target reaction. In real cells, side reactions (such as hydrogen evolution) consume part of the current, so the actual deposit is smaller. Divide the theoretical mass by the efficiency factor if you know it.
- Constant current: the formula applies to steady direct current. For varying current, integrate \(I\) over time to find \(Q\) first, then use \(n = Q/(zF)\).
- One product per electrode: if several ions compete at the same electrode, only the favored reaction proceeds appreciably, and you should apply the calculation to that reaction only.
- Gas products: for gases such as H2 or O2, the same equations give moles; convert to volume with the ideal gas law or molar volume at your conditions (see the Combined Gas Law calculator).
- Sign convention: enter the magnitude of the current. The direction of the reaction (deposition or dissolution) depends on which electrode you consider.
Practical Tips
- For quick mental checks, remember that 1 mole of electrons ≈ 96,500 C, so 1 A for about 26.8 hours delivers one faraday of charge.
- In electroplating shops, mass is often converted to coating thickness: divide the deposited mass by the plated area and the metal's density.
- When a solution concentration is needed instead of a deposit mass — for example, after dissolving the product — the Molarity calculator or the Normality calculator can finish the analysis.
Frequently Asked Questions
How do I calculate the deposited mass in electrolysis by hand?
Multiply current by time in seconds to get charge: \(Q = I \cdot t\). Divide by the product of electrons transferred \(z\) and Faraday's constant (96,485 C/mol) to get moles: \(n = Q/(zF)\). Finally multiply by molar mass \(M\): \(m = n \cdot M\). Example: 2 A for 30 min with Cu²⁺ gives \(3600/(2 \times 96485) \times 63.546 \approx 1.19\) g of copper.
What is Faraday's constant and why is it 96,485 C/mol?
Faraday's constant \(F\) is the total charge of one mole of electrons. Since one electron carries 1.602 × 10⁻¹⁹ C and one mole contains 6.022 × 10²³ particles, their product is about 96,485 C/mol. This value links electricity to chemistry: passing one faraday of charge (roughly 1 A for 26.8 hours) transfers exactly one mole of electrons through the cell.
How do I find the value of z (electrons transferred) for my reaction?
Balance the half-reaction and count the electrons on one side. Ag⁺ + e⁻ → Ag has \(z = 1\); Cu²⁺ + 2e⁻ → Cu has \(z = 2\); Al³⁺ + 3e⁻ → Al has \(z = 3\). For O₂ from water (2H₂O → O₂ + 4H⁺ + 4e⁻), \(z = 4\) per molecule. The charge on the ion is a good shortcut for metal deposition, but always check the balanced half-reaction for complex species like permanganate or dichromate.
How do I convert the deposited mass into coating thickness?
Divide the mass by the plated area to get mass per unit area, then divide by the metal's density. For example, 1.1855 g of copper spread over 50 cm² gives 0.02371 g/cm²; dividing by copper's density (8.96 g/cm³) yields about 26.5 µm of coating. This works only if the deposit is uniform, which in practice requires good agitation and electrode positioning.
Why is my actual deposited mass lower than the calculated value?
The calculator assumes 100% current efficiency, but real cells lose current to side reactions — most commonly hydrogen evolution competing with metal deposition. If your actual mass is 85% of the theoretical value, the current efficiency is 85%. You can estimate it as (actual mass ÷ theoretical mass) × 100%, and account for it by dividing the theoretical mass by the efficiency in future predictions.
Can I use this calculator for gases like hydrogen or oxygen?
Yes. The same equations give moles of gas: use \(z = 2\) for H₂ and \(z = 4\) for O₂, with the appropriate molar mass (2.016 or 32.00 g/mol). To get volume, multiply moles by the molar volume at your conditions — 22.4 L/mol at STP (0 °C, 1 atm) or 24.5 L/mol at 25 °C — or apply the ideal gas law \(PV = nRT\) at other pressures and temperatures.
What happens if the current changes over time?
The formula \(Q = I \cdot t\) only applies to constant current. For varying current, first find the total charge by integrating current over time, \(Q = \int I\,dt\). For simple cases, split the run into segments of roughly constant current, compute \(I \cdot t\) for each, and sum them. Then divide the total \(Q\) by \(zF\) and multiply by \(M\) as usual.