Electrolysis: Splitting Compounds with Electricity
Watch an electric current split a compound apart. Two electrodes dip into a liquid and are joined to a DC supply: positive ions drift to the negative electrode (the cathode) and gain electrons, negative ions drift to the positive electrode (the anode) and lose electrons, and the products appear as bubbles or as a growing layer of metal. Choose molten lead bromide, copper sulfate solution with inert or copper electrodes, concentrated sodium chloride solution or dilute sulfuric acid, then set the current and how long it flows. The half-equation for each electrode is shown, with the charge passed and the amount of each product worked out from Faraday's laws, for instance 1 A for 20 minutes deposits 0.395 g of copper. With copper electrodes the anode loses exactly the mass the cathode gains, and dilute sulfuric acid gives hydrogen and oxygen in a 2 to 1 volume ratio.
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What you can adjust
- Electrolyte
- 0 – 4
- Current
- 0.1 – 3 A
- Time
- 1 – 60 min
The liquid that carries the current. Each one gives its own products: lead and bromine from molten lead bromide, copper and oxygen from copper sulfate with inert electrodes, copper moving from one copper electrode to the other, hydrogen and chlorine from concentrated salt solution, and hydrogen and oxygen from dilute sulfuric acid.
The steady current from the supply. The charge passed is the current times the time, and the amount of each product is proportional to that charge.
How long the current flows. The animation squeezes this time into a few seconds, and a readout shows the minutes passing.
About this simulation
What it shows
A beaker with a cathode (negative) and an anode (positive) joined to a DC supply. Positive ions drift to the cathode, negative ions to the anode, and the products appear as bubbles or a growing metal layer. A card for each electrode shows its half-equation and the amount formed so far.
The key idea
The current splits the compound: positive ions gain electrons at the cathode and negative ions lose electrons at the anode. The amount formed depends only on the charge passed, since one mole of electrons carries 96 485 coulombs, so doubling the current or the time doubles every product.
Try this
Run the defaults: 1200 C deposits 0.395 g of copper. Set Electrolyte to Dilute sulfuric acid for 149 cm³ of hydrogen and 74.6 cm³ of oxygen, 2 to 1. Then try copper electrodes.
A simple model: a steady current, with no energy lost and every electron used for the products shown. Gas volumes assume room conditions, where one mole of gas fills 24 dm³, and ignore gas that dissolves (which matters for chlorine). Molten lead bromide has to be molten, about 370 °C, so that its ions can move. Concentrated sodium chloride solution is shown, which gives chlorine; a very dilute solution would give oxygen instead. The ions, bubbles and metal layers are drawn larger than true scale, the layer grows faster at first than the mass does so that a thin layer can be seen, and the time is speeded up so that the whole run plays in a few seconds. Only the ions that matter are drawn.