Video Transcript
In the electrolysis of aqueous copper sulfate, if carbon electrodes are used, a colorless gas is given off at the anode and the electrolyte changes from a blue to a colorless solution. Which statement explains these observations? (A) The colorless gas is hydrogen, and the solution changes to pure water. (B) The colorless gas is hydrogen, and the solution changes to sulfuric acid. (C) The colorless gas is carbon dioxide, and the solution changes to pure water. (D) The colorless gas is oxygen, and the solution changes to carbonic acid. (E) The colorless gas is oxygen, and the solution changes to sulfuric acid.
This is a question about electrolysis, which is a process where electricity is passed through an electrolyte, causing it to decompose. According to the problem, a colorless gas is given off at the anode when we carry out this electrolysis. The second observation is that at the end of the reaction, the electrolyte becomes a colorless solution. Let’s describe what happens during this entire process, the electrolysis of aqueous copper sulfate, in order to identify these two substances.
At the beginning of this process, we have two electrodes connected to a power supply dipped in a container of aqueous copper sulfate. Aqueous copper sulfate contains copper two ions, sulfate ions, and water molecules. When electricity passes through the substance, ions are attracted to the electrode of opposite charge. So the copper ions and sulfate ions gather at the cathode and anode, respectively. However, something happens to the water molecules as well. The electricity can cause some of the molecules to decompose into a hydrogen ion and a hydroxide ion. Electrons will flow from the anode through the circuit to the cathode. At each electrode, one of those two ions will react to form a new product. The reaction will involve donating electrons or accepting electrons.
To determine which ion will react at the anode, there’s a simple rule we can follow. If a halide is present, halogen gas will be produced. If not, oxygen gas will be produced from the hydroxide ion. The hydroxide ions react to produce oxygen gas alongside water and electrons that flow through the circuit. This is the colorless gas given off at the anode that the question mentions. If the colorless gas is oxygen, the correct the answer must be choice (D) or choice (E).
To determine what the solution might change to, let’s take a look at the reaction that occurs at the cathode. To determine what happens at the cathode, the rule we can follow is that the more stable element will be the one that reacts. In this case, copper is more stable than hydrogen. When the copper ions accept electrons, solid copper is plated on the surface of the electrode. The ions that haven’t reacted, in other words the ones that remain behind in the solution, are hydrogen ions and sulfate ions. Putting these ions in a two-to-one ratio to balance out the charge gives us an H2SO4 solution. H2SO4 is the chemical formula of sulfuric acid, the solution named in answer choice (E), the correct answer.
We started with a blue aqueous copper sulfate solution. But when the cation of the solution changed from copper to hydrogen, the color of the solution changed with it from blue to colorless. So, which statement explains these observations? That’s the statement in choice (E). The colorless gas is oxygen, and the solution changes to sulfuric acid.
