Reducing

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chiddler

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Which is the strongest reducing agent?

F-
Cl-
Hg(2+)
Au(3+)












Answer is Cl-. Why isn't it F-?

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Ok yea, F is more electronegative. It won't ever give up its electrons. The chart you linked to is showing oxidation potentials. More negative potential = less likely to happen. So, since 2F- has a more negative oxidation half reaction than 2Cl-, 2F- is less likely to get oxidized. So it is less likely to act as a reducing agent.

On the other hand, F2 has a reduction potential of +2.87, whereas Cl2 has a reduction potential of only +1.36. F2 is more likely to act as an oxidizing agent (gets reduced) than Cl2 because even though Cl2 will gladly take the electrons, F2 won't let anyone have electrons if he has space for them.
 
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F is more electronegative. It will never ever ever give up its electrons.

How am I supposed to use the numbers provided to me, then? What does the -2.87 V tell me?

Come to think of it, shouldn't it be Au?
 
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How am I supposed to use the numbers provided to me, then? What does the -2.87 V tell me?

Come to think of it, shouldn't it be Au?

You have to look at which side of the reaction the electrons are on. If they are on the product side, then the reactant is getting oxidized (losing electrons) and is therefore acting as the reducing agent.

If the electrons are on the reactant side, then the reactant is getting reduced (gaining electrons) and is therefore acting as the oxidizing agent.

Regardless of which half potential you are looking at, if the Ecell number is negative, the reaction is less likely to happen. So since 2F- ----> F2 + 2e- has a negative Ecell number, 2F- is probably not going to get oxidized, and therefore 2F- will not act as a reducing agent.

For Au and Hg they switched the reactions around on you (look at where the electrons are!). Both Au3+ and Hg2+ have positive reduction half potentials, so they are readily reduced and will act as oxidizing agents (Au3+ more-so than Hg2+ since the Ecell is more positive for Au3+).
 
You have to look at which side of the reaction the electrons are on. If they are on the product side, then the reactant is getting oxidized (losing electrons) and is therefore acting as the reducing agent.

If the electrons are on the reactant side, then the reactant is getting reduced (gaining electrons) and is therefore acting as the oxidizing agent.

Regardless of which half potential you are looking at, if the Ecell number is negative, the reaction is less likely to happen. So since 2F- ----> F2 + 2e- has a negative Ecell number, 2F- is probably not going to get oxidized, and therefore 2F- will not act as a reducing agent.

For Au and Hg they switched the reactions around on you (look at where the electrons are!). Both Au3+ and Hg2+ have positive reduction half potentials, so they are readily reduced and will act as oxidizing agents (Au3+ more-so than Hg2+ since the Ecell is more positive for Au3+).

sorry to have made you type all that.
 
How am I supposed to use the numbers provided to me, then? What does the -2.87 V tell me?

Come to think of it, shouldn't it be Au?

ohhh! tricky chart. the top two are oxidations and bottom two are reductions!

Yea. I don't think the MCAT will actually give you different half reactions like that, but they might ask you about oxidizing potential and give you reduction potential values.

Without even consulting a chart, you can rule out C and D on the original question because an anion is always going to be a better reducing agent than a cation. Cations are already electron deficient and won't want to lose even more electrons.
 
sorry to have made you type all that.

No don't be. Electrochemistry is actually one of the three things I'm spending today going over. It showed up on one of the AAMCs I took and the passage took me way too long because I wasn't solid on reading these charts, among other things.
 
You are reading it as:

Which is the most likely to be reduced?

when the question is: "which is the most likely to be oxidized."

A reducing agent is oxidized.
 
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