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Electro-Hydrolysis

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1.
You are given a couple of bags of ferrous iodide, and are instructed that you need to obtain the iron out of it. You need to find the most efficient way, and it should be as economical as possible. You decide to use electrolysis as your method, but you are not sure whether perform it on the aqueous solution or the molten salt. If its possible, electrolysis on the aqueous solution would be the cheapest, but you are not sure whether it is possible to do it that way. To decide what to do, you look up the standard reduction potentials for the components involved and the reactions that likely will take place at the cathode and anode.
Anode:

Cathode:

You look at the reactions and decide on your course of action. What do you decide to do and why?

You decide to do electrolysis on the molten salt because it is the only possible way to do it. You couldn't get any iron metal from doing electrolysis in aqueous solution.

You decide to do the electrolysis in aqueous solution because the reduction potentials are close enough to favor the deposition of iron metal on the cathode instead of the production of hydrogen gas.

You decide to do electrolysis of the molten salt because the reduction potentials for water make the reactions in aqueous solution favor the production of hydrogen and oxygen gases.

You do the electrolysis in the aqueous solution because ferrous iodide is explosive.

2.
From the previous question, calculate the overall reduction potential of the entire reaction , using the given reduction potentials.

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Solution Preview

Please see the attached file.

Here we see that the there is a bigger gap between the reduction potentials of these two reactions:
Anode
Cathode:
Note that the gap for Fe reaction (-0.44) would be smaller.
This means that under hydrolysis conditions, oxygen and hydrogen gases will be favored. This is not what we want. So you want to choose the answer: "You decide ...

Solution Summary

The solution provides detailed explanations on the concept of electro-hydrolysis and includes step-by-step calculations.

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