Electrolysis Calculator
Free Electrolysis Calculator for electrochemistry. Enter variables to compute results with formulas and detailed steps.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Electrolysis Calculator
Calculator
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Formula: m = (I * t * M) / (n * F)
Worked example โ Mass deposited = 7.114 g of copper
Formula
m = (I * t * M) / (n * F)
The mass deposited (m) equals current (I) times time (t) times molar mass (M), divided by the number of electrons per ion (n) times Faraday's constant (F = 96,485 C/mol). This is the quantitative form of Faraday's first law of electrolysis.
Worked Examples
Example 1: Copper Electroplating
Problem:How much copper is deposited by passing 3 amperes for 2 hours through CuSO4 solution? Cu2+ requires 2 electrons, M = 63.55 g/mol.
Solution:Q = I * t = 3 * 7200 = 21,600 C Moles of electrons = 21600 / 96485 = 0.2239 mol Moles of Cu = 0.2239 / 2 = 0.1119 mol Mass = 0.1119 * 63.55 = 7.114 g
Result:Mass deposited = 7.114 g of copper
Example 2: Aluminum Production
Problem:Calculate the mass of aluminum produced by 100 A for 24 hours. Al3+ requires 3 electrons, M = 26.98 g/mol.
Solution:Q = 100 * 86400 = 8,640,000 C Moles of electrons = 8640000 / 96485 = 89.55 mol Moles of Al = 89.55 / 3 = 29.85 mol Mass = 29.85 * 26.98 = 805.5 g
Result:Mass produced = 805.5 g (0.806 kg) of aluminum
Frequently Asked Questions
What is electrolysis?
Electrolysis is a chemical process that uses electrical energy to drive a non-spontaneous chemical reaction. It involves passing a direct current through an electrolyte (a solution or molten compound containing ions) using two electrodes. At the cathode (negative electrode), cations gain electrons (reduction), and at the anode (positive electrode), anions lose electrons (oxidation). Common applications include electroplating metals, refining copper, producing aluminum from bauxite, generating hydrogen and oxygen from water, and manufacturing chemicals like chlorine and sodium hydroxide. The process is governed by Faraday's laws of electrolysis.
What are Faraday's laws of electrolysis?
Faraday's first law states that the mass of substance deposited or dissolved at an electrode is directly proportional to the total electric charge passed through the electrolyte. Mathematically, m = (Q * M) / (n * F), where m is mass, Q is charge (current times time), M is molar mass, n is the number of electrons transferred per ion, and F is Faraday's constant (96,485 C/mol). Faraday's second law states that when the same quantity of charge passes through different electrolytes, the masses of substances deposited are proportional to their equivalent weights (molar mass divided by the charge number). These laws are fundamental to quantitative electrochemistry.
What is Faraday's constant and why is it important?
Faraday's constant (F) equals 96,485 coulombs per mole and represents the total electric charge carried by one mole of electrons (approximately 6.022 times 10 to the 23rd electrons). It connects the macroscopic world of measured currents and times to the molecular world of moles and atoms. To deposit one mole of a monovalent metal (like silver), you need to pass exactly 96,485 coulombs of charge. For a divalent metal (like copper from Cu2+), you need twice that amount (192,970 C) per mole. Faraday's constant is one of the most precisely known physical constants and is essential for all quantitative electrochemistry calculations.
What factors affect electrolysis efficiency?
Several factors influence electrolysis efficiency. Current efficiency refers to the fraction of total charge that produces the desired product; side reactions like hydrogen evolution or oxygen evolution can reduce this below 100 percent. Overpotential is the extra voltage needed beyond the thermodynamic minimum to drive the reaction at a practical rate, and it varies with electrode material, surface area, and current density. Temperature affects both ion mobility and reaction kinetics, with higher temperatures generally improving efficiency. Electrode material and geometry influence the overpotential and current distribution. Solution concentration and agitation ensure adequate ion supply to the electrodes and prevent concentration polarization.
References
Background & Theory
History
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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