Oxidation state...

[Temperature101]

What is the definition of oxidation state? For instance if they say some metal is in high oxidation state...Does that mean that metal can be easily reduced.


I come across this question in BR about Galvanic cell:

What BEST describes the initial composition of a galvanic cell?

A. The cathode half-cell contains a cation in a high oxidation state, and the anode half-cell contains a metal with a low ionization energy.

B. The cathode half-cell contains a cation in a high oxidation state, and the anode half-cell contains a metal with a high ionization energy.

C. The cathode half-cell contains a cation in a low oxidation state, and the anode half-cell contains a metal with a low ionization energy.

D. The cathode half-cell contains a cation in a low oxidation state, and the anode half-cell contains a metal with a high ionization energy.

The answer is A.

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[ljc]

Oxidation state usually implies an atom has completely lost or partially lost part of its electron density (or gained, if oxidation state is negative).

Metals don't like to be reduced, they like to be oxidized. So they like being +1,+2,+3, generally. Technically, an "oxidized" metal would be more easily reduced than an unoxidized one, though you don't generally hear about metals being reduced (unless they are acting as some sort of oxidizing agent for another metal, see below)

This is the case in the question you posted. A metal with a high oxidation state will accept electrons from another metal which wants to get rid of its electrons. For instance, Copper likes to be oxidized, but Zinc likes to be oxidized EVEN more. So if you combine oxidized Copper with pure Zinc metal, Zinc will give away it's electrons very easily (low ionization energy), while Copper takes them on. Copper will be reduced, and Zinc oxidized.

Hopefully that helps. Let me know if anything was unclear.

Edit: Mixed up Zinc and Copper.

 

[bluecabinet]

Low ionization energy for anode is good in this case because electrons need to be given off from anode. Low ionization energy makes sure that the process easy to accomplish given the low energy hoop to go over.

High oxidation state in cation is good in this case also because cathode side of the cell needs to be reduced with electron. High oxidation state insures that there is a ROOM for reduction..For example, if cathode cell metal is already fully reduced aka in LOW oxidation state, it can't be reduced because there is simply no more room for electron acceptance..

Hence two combos, low ionization energy and high oxidation state, accomplishes the greatest delta V (potential difference).

Might not be completely right...but that's the thought process I used to get the right answer.

 

[ljc]

Low ionization energy for anode is good in this case because electrons need to be given off from anode. Low ionization energy makes sure that the process easy to accomplish given the low energy hoop to go over.

High oxidation state in cation is good in this case also because cathode side of the cell needs to be reduced with electron. High oxidation state insures that there is a ROOM for reduction..For example, if cathode cell metal is already fully reduced aka in LOW oxidation state, it can't be reduced because there is simply no more room for electron acceptance..

Hence two combos, low ionization energy and high oxidation state, accomplishes the greatest delta V (potential difference).

Might not be completely right...but that's the thought process I used to get the right answer.

This is completely right. One important thing that I forgot earlier: you need to recognize that the cathode is where reduction is occurring, and the anode is where oxidation is occurring.