Strongest to Weakest Oxidant


by Endorser
Tags: oxidant, strongest, weakest
Endorser
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#1
Aug19-10, 09:58 PM
P: 42
Hello All I have been given a question form my chem teacher asking to place the following; Zn, Fe, Cu, I^- in order from strongest to weakest oxidant.

I have attempted two methods to solve this question

1st using the oxidation reaction table which lists as follows:
Zn>Fe>Cu>I

2nd using reduction half-reaction table which lists as following:
Zn>Cu>2I^->Fe^2+

I am unsure what 1 to use so please help me

Thanks
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Endorser
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#2
Aug19-10, 10:43 PM
P: 42
I have the following elements Zn, I^-, Fe^3+, Cu and i would like to know how to place these in order from strongest to weakest oxidant.

Thanks
Borek
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#3
Aug21-10, 02:18 AM
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Check standard potential tables.

Endorser
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#4
Aug21-10, 06:22 AM
P: 42

Strongest to Weakest Oxidant


Quote Quote by Borek View Post
Check standard potential tables.
So is the correct answer; (strong to weak)

Fe^2+
2I^-
Cu
Zn

just want some confirmation on this

thanks
Redbelly98
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#5
Aug21-10, 11:45 AM
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Moderator's note: two identical threads merged into one.
Borek
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#6
Aug21-10, 04:57 PM
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Please list half reactions in which each given substance behaves as an oxidizer and list standard potentials of these reactions.

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Endorser
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#7
Aug21-10, 09:17 PM
P: 42
[QUOTE=Borek;2848106]Please list half reactions in which each given substance behaves as an oxidizer and list standard potentials of these reactions./QUOTE]

Full

Zn2+ (aq)| Zn(S) || Cu2+(aq) | Cu(S)
I2(aq) | l-(aq) ||Cu2+(aq) | Cu(S)
I2(aq) | l-(aq) || Fe3+(aq) | Fe2+(aq)
Zn2+ (aq)| Zn(S) || I2(aq) | l-(aq)
Zn2+ (aq)| Zn(S) || Fe3+(aq) | Fe2+(aq)
Fe3+(aq) | Fe2+(aq) || Cu2+(aq) | Cu(S)

Half

Oxidation
Zn(S) → Zn2+(aq) + 2e-

Cu(S) → Cu2+(aq) + 2e
2I-(aq) → I2(aq) + 2e-

Zn(S) → Zn2+(aq) + 2e-

Zn(S) → Zn2+(aq) + 2e-
Cu(S) → Cu2+(aq) + 2e

Reduction
Cu2+(aq) + 2e- → Cu(S)
I2(aq) + 2e- →2I-(aq)

Fe3+(aq) +e- →Fe2+(aq)

I2(aq) + 2e- →2I-(aq)

Fe3+(aq) +e- →Fe2+(aq)

Fe3+(aq) +e- →Fe2+(aq)
Borek
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#8
Aug22-10, 02:25 AM
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Seems to me like you have listed everything just in case, as you have no idea what to look for.

I asked for half reactions, so full reactions are ruled out.

When substance acts as an oxidant (or oxidizer), what does happen to it? Does it gain, or loses electrons?

Finally, you have listed reactions, but omitted potentials, which are the most important part when looking for the answer to your question.

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Endorser
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#9
Aug22-10, 02:38 AM
P: 42
When substance acts as an oxidant (or oxidizer), what does happen to it? Does it gain, or loses electrons?

OXI RED

Oxidation is increase Reduction is decrease
Borek
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#10
Aug22-10, 03:08 AM
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Redox reaction means transfer of electrons - that means one substance gains electrons, other loses them. That in turn means one substance gets reduced, while the other gets oxidized.

Now, what happens to the substance that acts as an oxidizer? If you don't know, try to analyze the situation. Imagine there two substances reacting, Ox, and Red.

Substance Ox acts as an oxidizer.

The other substance is oxidized - does it get, or lose electrons?

If so, what happened to the electrons of the Ox substance - did it get more, or lost some?

If so, was it reduced, or oxidized?

And please, try to answer these questions, and not evade the answer but saying something that is basically right but not related to the question, as you did twice already. It won't work with me.

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Endorser
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#11
Aug22-10, 04:13 AM
P: 42
If so, was it reduced, or oxidized?


Zn -> Zn^2+ + 2e^-

thus is has oxidized as Oxidation is gain in electrons
Borek
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#12
Aug22-10, 05:17 AM
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OXI RED - Oxidation is increase Reduction is decrease - in terms of oxidation number.

OIL RIG - Oxidation is Loss Reduction is Gain - in terms of electrons.
Endorser
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#13
Aug23-10, 05:31 AM
P: 42
Quote Quote by Borek View Post
OXI RED - Oxidation is increase Reduction is decrease - in terms of oxidation number.

OIL RIG - Oxidation is Loss Reduction is Gain - in terms of electrons.
ahhhh... i c

so
Zn -> Zn^2+ + 2e^-

would be reduction due to OIL RIG
Borek
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#14
Aug23-10, 06:00 AM
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Yep. Now, go back and do as asked earlier.

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