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Oxidation-Redox Reactions

From: OpenStax College Chemistry book

Earthโ€™s atmosphere contains about 20% molecular oxygen, O2, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O2, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.

Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

2Na(๐‘ )+Cl2(๐‘”)โŸถ2NaCl(๐‘ )2Na(s)+Cl2(g)โŸถ2NaCl(s)

It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called aย half-reaction:

2Na(๐‘ )โŸถ2Na+(๐‘ )+2eโˆ’Cl2(๐‘”)+2eโˆ’โŸถ2Clโˆ’(๐‘ )2Na(s)โŸถ2Na+(s)+2eโˆ’Cl2(g)+2eโˆ’โŸถ2Clโˆ’(s)

These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons, the โ€œsโ€ subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:๐จ๐ฑ๐ข๐๐š๐ญ๐ข๐จ๐ง๐ซ๐ž๐๐ฎ๐œ๐ญ๐ข๐จ๐ง==loss of electronsgain of electronsoxidation=loss of electronsreduction=gain of electrons

In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.๐ซ๐ž๐๐ฎ๐œ๐ข๐ง๐  ๐š๐ ๐ž๐ง๐ญ๐จ๐ฑ๐ข๐๐ข๐ณ๐ข๐ง๐  ๐š๐ ๐ž๐ง๐ญ==species that is oxidizedspecies that is reducedreducing agent=species that is oxidizedoxidizing agent=species that is reduced

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:

H2(๐‘”)+Cl2(๐‘”)โŸถ2HCl(๐‘”)H2(g)+Cl2(g)โŸถ2HCl(g)

The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.

  1. The oxidation number of an atom in an elemental substance is zero.
  2. The oxidation number of a monatomic ion is equal to the ionโ€™s charge.
  3. Oxidation numbers for common nonmetals are usually assigned as follows:
    • Hydrogen: +1 when combined with nonmetals, โˆ’1 when combined with metals
    • Oxygen: โˆ’2 in most compounds, sometimes โˆ’1 (so-called peroxides, O22โˆ’),O22โˆ’), very rarely โˆ’1/2 (so-called superoxides, O2โˆ’),O2โˆ’), positive values when combined with F (values vary)
    • Halogens: โˆ’1 for F always, โˆ’1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
  4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.

EXAMPLE 4.5

Assigning Oxidation NumbersFollow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:

(a) H2S

(b) SO32โˆ’

(c) Na2SO4

Solution(a) According to guideline 1, the oxidation number for H is +1.

Using this oxidation number and the compoundโ€™s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

charge on H<sub>2</sub>S=0=(2ร—+1)+(1ร—x)

x=0โˆ’(2ร—+1)=โˆ’2

(b) Guideline 3 suggests the oxidation number for oxygen is โˆ’2.

Using this oxidation number and the ionโ€™s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

charge on SO32โˆ’ =โˆ’2=(3ร—โˆ’2)+(1ร—๐‘ฅ)

๐‘ฅ=โˆ’2โˆ’(3ร—โˆ’2)=+4

(c) For ionic compounds, itโ€™s convenient to assign oxidation numbers for the cation and anion separately.

According to guideline 2, the oxidation number for sodium is +1.

Assuming the usual oxidation number for oxygen (โˆ’2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:

charge on SO42โˆ’โˆ’=โˆ’2=(4ร—โˆ’2)+(1ร—๐‘ฅ)

๐‘ฅ=โˆ’2โˆ’(4ร—โˆ’2)=+6