Using oxidation says

Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be useful to review and be familiar with the following concepts:

oxidation and reduction in terms of electron transfer electron-half-equations

To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., (ceV^2+) and (ceV^3+)). The 2+ ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:

< ceV ightarrow V^2+ + 2e^- label1>

The vanadium in the ( ceV^2+) ion has an oxidation state of +2. Removal of another electron gives the (ceV^3+) ion:

< ceV^2+ ightarrow V^3+ + e^- label2>

The vanadium in the (ceV^3+ ) ion has an oxidation state of +3. Removal of another electron forms the ion (ceVO2+):

< ceV^3+ + H_2O ightarrow VO^2+ + 2H^+ + e^- label3>

The vanadium in the (ceVO^2+) is now in an oxidation state of +4.

You are watching: What is the oxidation state of s in so32-


Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations ef1 and ef2), but not for the ion in Equation ef3).


The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth electron to form another the (ceVO_2^+) ion with the vanadium in a +5 oxidation state.

< ceVO^2+ + H_2O ightarrow VO_2^+ + 2H^+ + e^->

Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of zero.

If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is common for nonmetals such as sulfur:

< ceS + 2e^- ightarrow S^2- >

Here the sulfur has an oxidation state of -2.



Determining oxidation states

Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules provide a simpler method.

See more: Where In A Cell Does Glycolysis Occur? Where Does Glycosylation Occur In The Cell





Using oxidation states



Using oxidation states to identify what has been oxidized and what has been reduced

This is the most common function of oxidation states. Remember:

Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state

In each of the following examples, we have to decide whether the reaction is a redox reaction, and if so, which species have been oxidized and which have been reduced.


Example (PageIndex4):

This is the reaction between magnesium and hydrogen chloride:

< ceMg + 2HCl -> MgCl2 +H2 onumber>

Solution

Assign each element its oxidation state to determine if any change states over the course of the reaction:

*
*
api/deki/files/190651/padding.GIF?revision=1" /> -->