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I understand the principles of electrolysis of salts in aqueous solutions, but there are two points on which i am unsure.

  • At the positive electrode, how can you work out which ions will preferentially donate electrons?

  • If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?

    • If so, what decides which ions are formed?
  • What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

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    $\begingroup$ I would love to answer this question if you could make it neater and easier to read. Please bullet what you don't understand and why. Thanks. $\endgroup$ – Asker123 May 17 '15 at 19:39
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    $\begingroup$ I edited in some legibility, but the questions might be a bit broad. $\endgroup$ – tschoppi May 17 '15 at 19:41
  • $\begingroup$ just changed it for clarity $\endgroup$ – ziggy May 17 '15 at 19:44
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    $\begingroup$ Alright, I will begin. $\endgroup$ – Asker123 May 17 '15 at 19:58
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For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

Don’t be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of $2.2\ \mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.

$$\begin{alignat}{2} \ce{O2_{(g)} + 4H+ + 4e- &-> H2O_{(l)}}\quad&&1.23\ \mathrm{V}\\ \ce{F2_{(g)} + 2e- &-> 2F-}\quad&&2.87\ \mathrm{V}\\ \ce{2H2O{(l)} + 2e- &-> H2_{(g)} + 2OH-}\quad&&{-}0.83\ \mathrm{V}\\ \ce{Ni^2+ + 2e- &-> Ni_{(s)}}\quad&&{-}0.25\ \mathrm{V} \end{alignat}$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^2+}$ and $\ce{F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since fluorine already exists reduced, and it does not like being oxidized. Then you have nickel and water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:

Equation 1 (this one is flipped) and 4. $\ce{Ni^2+}$ is being reduced while $\ce{H2O}$ is being oxidized.

In general …

  1. Eliminate which reaction cannot take place.
  2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

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At the positive electrode, how can you work out which ions will preferentially donate electrons?

In an electrolytic cell, the positive electrode is called the anode, and the oxidation half of the reaction occurs here. The oxidation reactions which have less negative standard oxidation potential are more likely to occur, though most resources give the standard reduction potentials. Wikipedia is one such resource.

If you wish to know which species are more likely to donate electrons, these will be the half reactions that have the less negative standard reduction potentials.

If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?

I wrote a response to this exact question here.

What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?

In an electrolytic cell, the negative electrode is called the cathode, and the reduction half of the reaction occurs here. When an species is reduced, typically it changes phases (goes from aqueous ion to being a gas, a liquid, or a solid). Most likely if an ion is reduced to another ion here, this ion will then also be reduced.

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