Question

So far I've only encountered formulas to solve for solutions of single molecules while learning about electrolysis, i.e. H_2O or FeCl_3. This generally has involved determining the resultants, identifying whether they are attracted to the anode or cathode, and performing some multiplications.

What approach is used for analyzing complex solutions? For example, an everyday soda solution may contain water (H_20), fructose (C6H_21O6), phosphoric acid (H_3PO_4), and carbon dioxide (CO_2). I'm not sure how to approach the resultants. It would seem the decomposition potential should for each molecule should first be considered to identify which molecules are likely to decompose, but given so many elements how can we determine what resultants will remain? It would seem C6H12O6 alone could result in 6(C) + 6(H_2O) or in 6(C) + 6(H_2) + 3(O_2), for example.

Answer 1

During electrolysis of complex solutions, decomposition potentials guide which substances will decompose, with water often being the most readily electrolyzed component yielding H2 and O2. Other complex organic molecules may not decompose as readily due to higher stability and more complex reaction pathways.

Step-by-step explanation:

When dealing with complex solutions during electrolysis, it's important to consider the decomposition potentials of the individual components and their tendencies to undergo electrolysis.

In the case of an everyday soda solution containing multiple substances such as water (H2O), fructose (C6H12O6), phosphoric acid (H3PO4), and carbon dioxide (CO2), priority is given to the substances with the lowest decomposition potential first. Typically, this would be water since it readily undergoes electrolysis to form H2 and O2 gases at the cathode and anode, respectively.

For instance, when electrolyzing water, you may use the following balanced chemical equation to determine the resultant gases:

2 H2O(l) → 2 H2(g) + O2(g)

This indicates that decomposition of water will proceed at standard electrolysis conditions to yield hydrogen and oxygen gases. Consequently, identifying the products of the electrolysis of more complex organic compounds like fructose is less straightforward and not typically subject to simple aqueous electrolytic decomposition due to higher stability and complex multiple reaction pathways.