Chemistry - Stability of Compounds

I don't understand, in my CGP book it states: " Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+. Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions. "

How does the reducing agent relate to stability, and why are Iodide ions used for iron iodide, and Chloride ions used for iron chloride?

You can consider the standard electrode potentials.

The $\mathrm{E^{\circ}(V)}$ of $\mathrm{Cl_2}$ reduction is 1.36
For $\mathrm{Fe^{3+}}$ to be reduced, the $\mathrm{E^{\circ}(V)}$ value would have to be greater than 1.36. It isn't.

What does EV mean and I don't understand, how does that relate to the stability of a compound?

Woah, never been quoted before.
The E part means 'Electrode Potential'
The V part means 'Voltage'


Essentially, they're both the same thing.

Electrode Potential is the potential of an element to accept electrons. The more positive this number is, the more it's likely to accept electrons.

So going back to what Kvone the Arcane said
The of reduction is 1.36

This is pretty high. It's likely to accept electrons and be reduced to Chloride Ions.
i.e. $\mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}$

Now the of $\mathrm{Fe^{3+}}$ is +0.77.

Because it's lower than our E of Cl₂, it's going to remain oxidised. It's not going to get converted into Fe^2+.

If this isn't making sense, let's look at that statement you said:
"Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions"

We have two reactions.

$\mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}$
$\mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}$

If we want to reduce the Fe³⁺ ions to Fe²⁺ ions, the bottom reaction has to go forward, agreed?
In order to do that, we need to get electrons.
To get electrons, we can oxidise Cl^-, i.e. the top reaction has to go backwards.

But, the higher the E, the more it's likely to accept electrons.
In other words, the higher the E, the more forward the reaction is.
Our top reaction, the Chlorine one has the higher E, so it's going to be forward.
Thus, the bottom reaction is backward.
As a result, we end up with Cl^- and Fe³⁺. That's why it's stable, those two ions remain.

That's the reason for that other sentence
"Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+"

Again our two reactions are:
$\mathrm{I_2} + \mathrm{2e^-} \longrightarrow \mathrm{2I^-}$
$\mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}$

However, the E of I₂ is +0.54.
The E of Fe³⁺/Fe²⁺ is +0.77.

So this time, the bottom reaction is going to go forward.
This means Fe³⁺ is going to be reduced to Fe²⁺


That's why if you try and create a solution of Fe(III) and I^-, you can't get the two ions to live happily. I^- is a strong enough reducing agent which causes the solution to be really unstable, because these ions are going to react. Stability depends upon the difference in E (electrode Potential) and which reactions need to go forward and backward.

Have you not done Electrochemistry?

Also, please ask away, I feel I've scared you with this long post

Woah, never been quoted before.
The E part means 'Electrode Potential'
The V part means 'Voltage'


Essentially, they're both the same thing.

Electrode Potential is the potential of an element to accept electrons. The more positive this number is, the more it's likely to accept electrons.

So going back to what Kvone the Arcane said
The of reduction is 1.36

This is pretty high. It's likely to accept electrons and be reduced to Chloride Ions.
i.e. $\mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}$

Now the of $\mathrm{Fe^{3+}}$ is +0.77.

Because it's lower than our E of Cl₂, it's going to remain oxidised. It's not going to get converted into Fe^2+.

If this isn't making sense, let's look at that statement you said:
"Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions"

We have two reactions.

$\mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}$
$\mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}$

If we want to reduce the Fe³⁺ ions to Fe²⁺ ions, the bottom reaction has to go forward, agreed?
In order to do that, we need to get electrons.
To get electrons, we can oxidise Cl^-, i.e. the top reaction has to go backwards.

But, the higher the E, the more it's likely to accept electrons.
In other words, the higher the E, the more forward the reaction is.
Our top reaction, the Chlorine one has the higher E, so it's going to be forward.
Thus, the bottom reaction is backward.
As a result, we end up with Cl^- and Fe³⁺. That's why it's stable, those two ions remain.

That's the reason for that other sentence
"Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+"

Again our two reactions are:
$\mathrm{I_2} + \mathrm{2e^-} \longrightarrow \mathrm{2I^-}$
$\mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}$

However, the E of I₂ is +0.54.
The E of Fe³⁺/Fe²⁺ is +0.77.

So this time, the bottom reaction is going to go forward.
This means Fe³⁺ is going to be reduced to Fe²⁺


That's why if you try and create a solution of Fe(III) and I^-, you can't get the two ions to live happily. I^- is a strong enough reducing agent which causes the solution to be really unstable, because these ions are going to react. Stability depends upon the difference in E (electrode Potential) and which reactions need to go forward and backward.

Have you not done Electrochemistry?

Also, please ask away, I feel I've scared you with this long post

Ahah, to be honest when I first saw the post it really did scare me lol, but if I'm going to be honest when I read into it and understood it carefully it truly makes sense! So thank you so much you're amazing. I haven't actually learnt electrochemistry no as it doesn't come in AS, but yes I was just confused about the concept. Just one thing - so can you say that an unstable compound is one that is more likely to react - and does it always involve reduction reactions or would a compound being oxidised also be considered unstable?

No problem!
Ah, I should have realised that you're in AS, yeah this is not an easy topic to learn at AS, let alone at A2, and condensing it into that post may still be insufficient.

Generally speaking yes, unstable compounds are more reactive because they're in a high energy state which means the particles have more potential chemical energy to transferred into other forms.

What I mean by that is, say somehow I add some heat energy to a box. If I add a tiny bit, there's not that much heat loss from the box, but if I add a shed load of heat energy, there's a good chance heat will escape from the box. The box is really unstable with all this energy, it can't contain it. Likewise, this unstable compound has all this potential energy that it needs to transfer.


It doesn't always involve reduction reactions, an unstable compound can be oxidised to something more stable.

For example, going back to your example on the Iron Iodide solution, we said Fe³⁺ is unstable relative to I^-, so it tends to be reduced. And we can argue it the other way, I^- is in it's unstable form when present with Fe³⁺, so it tends to be oxidised to I_2.
Any reaction with reduction must involved oxidation, so you can think of stability from both sides.

When you get to A2, you're actually going to cover this idea of stability more and see what it means for a compound to be stable, and why unstable species tend to react.