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What Is The Oxidation State Of S In Feso4 ? How Can I Calculate Oxidation Number For Feso4

Using oxidation states

This page explains what oxidation states (oxidation numbers) are and how to calculate and use them.

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Oxidation states (oxidation numbers)

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. Vanadium forms a number of different ions – for example, V2+ and V3+. The 2+ ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:

< V ightarrow V^{2+} + 2e^->

The vanadium is now said to be in an oxidation state of +2. Removal of another electron gives the V3+ ion:

< V^{2+} ightarrow V^{3+} + e^->

The vanadium now has an oxidation state of +3. Removal of another electron forms the ion VO2+.

< V^{3+} + H_2O ightarrow VO^{2+} + 2H^+ + e^->

The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn”t always the same as the charge on the ion (that was true for the first two cases but not for the third). 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 ion. The oxidation state of the vanadium is now +5.

< VO^{2+} + H_2O ightarrow VO_2^{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:

< S + 2e^- ightarrow S^{2-} >

Here the sulfur has an oxidation state of -2.

Summary

The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.

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

Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations.

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:

The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl2, S8, and large structures of carbon or silicon each have an oxidation state of zero. The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion. The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left. Some elements almost always have the same oxidation states in their compounds:

 Element Usual oxidation state Exceptions Group 1 metals Always +1 Group 2 metals Always +2 Oxygen Usually -2 Peroxides and F2O (see below) Hydrogen Usually +1 Metal hydrides (-1) (see below) Fluorine Always -1 Chlorine usually -1 Compounds with O or F (see below)

The reasons for the exceptions

Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.

Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).

Oxygen in peroxides: Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.

Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.

Oxygen in F2O: The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.

Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference. An example of this situation is given below.

Example 1: Chromium

What is the oxidation state of chromium in Cr2+?

SOLUTION

For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion)

What is the oxidation state of chromium in CrCl3?

This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let n equal the oxidation state of chromium:

n + 3(-1) = 0

n = +3

The oxidation state of chromium is +3.

Example 2: Chromium

What is the oxidation state of chromium in Cr(H2O)63+?

SOLUTION

This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.

The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3.

What is the oxidation state of chromium in the dichromate ion, Cr2O72-?

The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don”t forget that there are 2 chromium atoms present.

2n + 7(-2) = -2

n = +6

Example 3: Copper

What is the oxidation state of copper in CuSO4?

SOLUTION

Unfortunately, it isn”t always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulfur) with variable oxidation states.

In cases like these, some nadechworld.comical intuition is useful. Here are two ways of approaching this problem:

Recognize CuSO4 as an ionic compound containing a copper ion and a sulfate ion, SO42-. To form an electrically neutral compound, the copper must be present as a Cu2+ ion. The oxidation state is therefore +2.

Recognize the formula as being copper(II) sulfate (the (II) designation indicates that copper is in a +2 oxidation state, as discussed below).

In naming compounds

You will have come across names like iron(II) sulfate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.

This can also be extended to negative ions. Iron(II) sulfate is FeSO4. The sulfate ion is SO42-. The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion.

The sulfite ion is SO32-. The oxidation state of the sulfur is +4. This ion is more properly named the sulfate(IV) ion. The ate ending indicates that the sulfur is in a negative ion.

FeSO4 is properly named iron(II) sulfate(VI), and FeSO3 is iron(II) sulfate(IV). Because of the potential for confusion in these names, the older names of sulfate and sulfite are more commonly used in introductory nadechworld.comistry courses.

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 4

This is the reaction between magnesium and hydrogen chloride:

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

Example 6

The reaction between chlorine and cold dilute sodium hydroxide solution is given below:

< 2NaOH + Cl_2 ightarrow NaCl + NaClO + H_2O>

It is probable that the elemental chlorine has changed oxidation state because it has formed two ionic compounds. Checking all the oxidation states verifies this: