"The oxidized species of the couple is very oxidizing". Does this make sense?

[zenterix]

Homework Statement

To summarize my question, consider the reaction that occurs in an electrode

$$\mathrm{F_2(g)+2e^-\rightarrow 2F^-}\ \ \ E^\circ=2.87\text{V}$$

Relevant Equations

The notes I am reading say that "A large positive $E^\circ$ for $\mathrm{F_2/F^-}$ means the oxidized species of the couple is very oxidizing."

I don't understand this quote because as far as I can tell, the oxidized species is $\mathrm{F^-}$ and it is a reducing agent.

The notes from MIT OCW I am reading contain the following

A large positive $E^\circ$ means the element or compound is easy to reduce.

$E^\circ$ denotes a standard electrode (aka redox) potential.

Therefore, the quote above says that a positive $E^\circ$ means that relative to a standard hydrogen electrode, the electrode in question operates as a cathode, and the more positive the value is the more reduction happens for the redox pair at the cathode.

$\Delta E^\circ_{cell}=E^\circ(\text{cathode})-E^\circ(\text{anode})$ is simply the cell potential, aka, the electromotive force generated by an electrochemical cell, when products and reactants are in their standard states.

Here is an example

For the electrochemical cell $\mathrm{Zn(s)\ |\ Zn^{2+}(aq)\ ||\ Cu^{2+}(aq)\ |\ Cu(s)}$ we have the reactions

$\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^-}$ with $E^\circ=-0.76\text{V}$ at the anode. Atoms in a bar of solid zinc are oxidized.

$\mathrm{Cu^{2+}(aq)+2e^-\rightarrow Cu(s)}$ with $E^\circ=0.34\text{V}$ at the cathode. Copper ions in solution are reduced.

The cell potential is $\Delta E^\circ_{cell}=E^\circ(\text{cathode})-E^\circ(\text{anode})=1.10\text{V}$.

Okay, so then there is the following in the notes

$$\mathrm{F_2(g)+2e^-\rightarrow 2F^-}\ \ \ E^\circ=2.87\text{V}$$

$\mathrm{F_2(g)}$ is being reduced, as far as I can tell.

Is $\mathrm{F_2}$ a good oxidizing agent?

Yes, because it is easy to reduce.

Okay, but I do not understand the following quote

A large positive $E^\circ$ for $\mathrm{F_2/F^-}$ means the oxidized species of the couple is very oxidizing.

The oxidized species is $\mathrm{F^-}$. Is it not the reducing agent? Is it not then "very reducing" rather than "very oxidizing"?

 

[zenterix]

Note that there is even a table

 

[Borek]

The oxidized species is $\mathrm{F^-}$.

Is it?

 

[zenterix]

Is it?

In the 2nd post you can see it says "it is easy to add electrons to $\mathrm{F_2}$", which as far as I can tell becomes $\mathrm{2F^-}$, which would need to lose the electrons to become $\mathrm{F_2}$.

Doesn't oxidation mean that a species loses electrons?

 

[zenterix]

I think what is happening here is possibly not a chemistry issue but a semantics issue.

Apparently, according to my new interpretation of what the notes are saying

$\mathrm{F^-}$ is the reduced species (since $\mathrm{F_2}$ is reduced to $\mathrm{F^-}$) and

$\mathrm{F_2}$ is the oxidized species (since $\mathrm{F^-}$ is oxidized to $\mathrm{F_2}$.

Since the $E^\circ$ for $\mathrm{F_2(g)+2e^-\rightarrow 2F^-}$ is positive then it means this reaction is spontaneous, and $\mathrm{F_2}$ is a good oxidizing agent.

$\mathrm{F_2}$ is the oxidized species and is very oxidizing.

Semantically, "oxidized species" seems to mean "the species that is the result of an oxidation" and not "the species that is oxidized".

That is my understanding right now, and if it is correct, then I find the term "oxidized species" to be ambiguous and they should have written

"A large positive $E^\circ$ for $\mathrm{F_2/F^-}$ means the species that undergoes a reduction is very oxidizing."

"A large positive $E^\circ$ for $\mathrm{F_2/F^-}$ means the species that is the result of oxidation is very oxidizing."

 

[zenterix]

A species that undergoes reduction is an oxidizing agent (because it oxidizes another species).

One could say

A reduced species is an oxidizing agent.

But "reduced species" might mean "a species that is reduced" or "a species that has been reduced".