Oxidation and Reduction

One of the most important chapter not only in inorganic but also in organic chemistry and biochemistry!

Oxidation

Fundamentally oxidation means, a compound reacts with Oxygen, O2. Burning a piece of wood or any compound is not else but a reaction with oxygen, O2. However it’s to remember, the air consist’s of about 80 % nitrogen, N2, and only about 20 % oxygen. Nitrogen, often used as a protection gas due to not affecting  unstable compounds, does not at all react. It is Oxygen. Sugar e.g. glucose, C6H12O6 and O2 turns into carbondioxide, CO2 and water, H2O.  In the human organism, when effort is made, sugar is transformed by enzymes (biocatalysts) into CO2 and H2O.

Reduction

Reducing on the opposite originally means removal of oxygen from a compound. From acetic acid, CH3COOH, ethanol, CH3CH2OH ( alcohol ) is produced, which is a reduction. By burning lime, CaCO3 is reduced to calcium oxide CaO. This is also a reduction in alchemical sense .

CaO reacts with carbon, C, to calcium carbide, CaC2 and carbon monoxide, CO, at 2000C °. Calcium carbide and water are converted into ethyne ( acetylene ) C2H2 and calcium hydroxide. 

 

Oxidation and reduction have today an entirely different meaning in modern chemistry:

Review again the chapter, ion, anion and cation and resee electronegativity!  it’s recommended to understand what follows!

• By oxidation an atom gives away one or more electrons.
• By reducing an atom accepts one or more electrons.

Alkali earth metal and transition metals mainly tend to oxidation, non-metals to reduction.

In covalent compounds the more electropositive atom release one or more electrons and provide them to the more electronegative one.

 

The number of electrons an atom gives away or accepts is designated by oxidation numbers, usually with Roman numerals.

 

For each element itself, without any connection, the oxidation number is zero.

 

Example: Elemental sulfur, which can be found in nature as yellow orthorhombic crystals (eg in Wallis, Switzerland and in Sicily) has oxidation number zero, as well O2, Cl2, sodium, Na, magnesium and copper, etc.

 

Otherwise an atom of a compound, which is the more elektropositve one, spends one or more electrons to the more electronegative one. So it accepts the same number of electrons. Let’s have some examples to determine oxidation numbers:

For sodium chloride, NaCl we get Na = + I and Cl = -I. The sodium is now a cation and the chlorine an anion.

H, hydrogen fluoride: H = + I, F = – I.

Let’s control: the sum of wrested and added electrons must equal zero.

H2O, water: H = + I, O = – II, the calculation is as follows: 2 * (+ I) + (- II) = 0

NaH, sodium: Na = + I, H = -I!  Attention! Hydrogen is more electronegative than sodium. Hydrogen can either have oxidation number + I and – I!

 

If a compound has a ionic charge, if it ‘s an anion or a cation, the total of the oxidation number must result in that ionic charge:

Let’s see the permanganate ion, MnO4-. Here the ionic charge is -1. Oxidation number of Mn = + VII and O = -II, numeric results are 7 – 8 = -1 or described in Roman numerals: VII + [4 * (- II)] = – I.

Ta6O19 (8), tantalum oxide ion: ion charge – 8. O has oxidation number -2. 19 * -2 = -38: The following equation has to be solved:

6Ta – 38 = -8

Ta = (-8 + 38) / 6 = 30/6 = 5

Control: 6 * 5 + 19 * (- 2) = 30 – 38 = -8.

metals generally have positive oxidation numbers, however, the oxidation numbers of the following transition metals can also be negative

Co, Re, Ir with -I

Fe, Ru, Os with -II,

V, Cr, and Mn can also be -III.

For this complex [Mn (CO) 4] 3- the oxidation number of Mn is –III !

The rare earths (lanthanides and actinides) only have positive oxidation numbers in compounds.

complex compounds of the following transition metals or central atoms V, Cr, Mn, Fe, Co, Ni, Mo, Ru, Rh, Pd, Pt, Ir, Os, Zn, the oxidation number can also be 0. In the case of [Ni (CN) 4] 4- The oxidation number of nickel is 0.

All semi- and non-metals except boron, B, germanium, Ge, oxygen and fluorine may have both positive and negative oxidation numbers.

Boron has oxidation number + III, germanium+ IV, oxygen -I and -II and fluorine only -I.

Possible oxidation numbers of important elements of organic chemistry:

Carbon C: -4. 2. 4

Nitrogen N, -3, 2, 3, 4, 5

Phosphorus P: -3, 3, 5

Sulfur S: -2, 2, 4, 6

Chlorine and bromine Br Cl per: -1, 1, 3, 5, 7

Iodine I: -1, 1, 5, 7

 

Reduction-oxidation reactions, Redox reactions

 

Thus oxidation may be carried out, a reduction must take place at the same time. Oxidation without reduction is impossible and vice versa. An electron that is an atom is removed is added to another atom.Reduction-oxidation reactions are therefore called short redox reactions.

The compound or element that reactant removes one or more electrons, is called oxidant. In return, the substance, which oxidizes itself and provides electrons, is designated as a reducing agent.

Have a look at salt as an example:

2Na + Cl2 ===> 2 NaCl

Elemental sodium reacts with chlorine. Both elements have oxidation number Zero. After the reaction, the product, NaCl, has the following oxidation numbers Na (+I), Cl (-I). Sodium has given away one electron it is oxidized. Chlorine has added an electron, it is reduced. Sodium has given an electron to the reactant chlorine. Sodium is a reducing agent Chlorine has accepted an electron from sodium. Chlorine is the oxidant.

 

Determination of the reaction equation in redox reactions

The above example with NaCl is very simple. Normally redox reactions take place in an aqueous environment.

In redox reactions, the reaction equation is defined so that the number of emitted electrons from the reducing agent corresponds to the number of recorded by the oxidant. In the example, NaCl, this rule may be understood. The number of given electrons by sodium matches with the number of accepted ones by Chlorine. sodium oxidates 2 and chlorine reduces 2 electrons.

Not always may the situation be that easy.

In following examples redox stoichiometry is some more difficult to formulate:

This Redox equation is wrong, although the stoichiometry matches according to the numbers of atoms: (in brackets the oxidation numbers in roman figures)

Fe3+(+III)   +   S2-(-II)  ===>  Fe2+(+II)  +  S(0)

The Fe-cation takes 1 electron, while the sulfur emits to 2 electrons. How is it to be solved this problem?

We decompose the redox equation in reduction and oxidation steps.

Reduction of iron, Fe3+(III) takes on an electron, it is reduced by 1 electron:

e-  +  Fe3+  ===>  Fe2+

The sulfur or sulphide anion S2-(-II) is oxidized to 2 electrons:

S2  ===>  S  +  2e-

We look now for a balance. This is done by multiplying the iron equation by 2.

2e-  +  2Fe3+  ===>  2Fe2+

We add now the two partial equations and we get. (Electrons disappear)

2Fe3+(+III)  +  S2-(-II)  ===>  2Fe2+(+II)  +  S(0)

Control: iron absorbs a total of 2 electrons, while the sulfur is poorer by 2 electrons. .

In this example, the ion charges coincide,

left 6 – 2 = 4+

right 2*2+ = 4+.

 

In the following example however this is not the case:

The following redox equation also takes place in an aqueous environment. As you may already know, or in the next chapter to learn, in one liter of neutral water there are each 10exp-7 moles of hydroxide (OH-)ions  and hydrogenium (H3O+) ions. The pH is 7, the water is neutral, or neither acidic nor alkaline.

Cr2O7(2-)  +  6Cl-  ===>  2CR3+  +  3Cl2

This equation is also stoichiometric incomplete, although the number of 6 electrons reduzed by chrome and oxidized by chlorine matches. But in addition to the numbers of oxygens missing on the right, the ionic charge doesn’t match:

Left (2-) + (6-) = 8-

Right = 6+

The compensation can bring the balance, with positively charged cations, here H+ (protons), or negatively charged anions, here OH- (hydroxide ions) and the corresponding equivalent H2O

Supplement with protons:

On the right side the ion charge is 6+ on the left side 8-.

How many H+ or OH- and H2O do we need.

We calculate ionic charge products ./. Ion charge reactants

(6+)  –  (-8) = 6 + 8 = 14

And we can now add the redoxequation with 14 H+ ions on the left side and 7 H2O on the right side or 14 OH- ions on the right side and 7 H20 on the left side.

Let’s do that with protons first

14H+ + Cr2O7(2-) + 6Cl-  ===>  2CR3+ + 3Cl2 + 7H2O

And now with OH-

Cr2O7(2-) + 6Cl- + 7H2O  ===>  14OH- + 2 Cr3+ + 3Cl2

Control ionic charges:

first equation with H+

Left: (14+)  + (2-) + (6-)   = 14 – 2  – 6  = 6+

Right: = 6+

Second equation

Left: (2-) + (6-) = -2 – 6  =   -8

Right: (14-) + 2*3+ = -14 + 6 = -8

 

Example Redoxequation in an acidic aqueous solution:

H2SeO3 + H2S + ===> Se HSO4-

Determination of oxidation numbers (in brackets)

H2(+I)Se(+IV)O3(-II)  +  H2(+I)S(-II)  ===>  Se(0) + H(+I)S(+VI)O4-(-II)

Selenium, Se, reduced 4 electrons and sulfur oxidized 8 electrons.

Finding the balance is obtained by multiplying selenide and Se by 8, hydrogen sulfide and hydrogen sulfate by 4.

8H2SeO3 + 4H2S ===> 8Se + 4HSO4-

Now you can simplify this equation even by shortening

2H2SeO3 + H2S + ===> 2Se + HSO4-

Selenium has now received (reduced) a total of 8 electrons, sulfur has given away (oxidized) 8 electrons

On the right there is still a ionic charge.

The equation can now be supplemented with one proton and 2 equivalents of water, so that the figure of H and O correspond. Because here there is an acidic environment, this is done merely with protons, H+.

2H2SeO3 + H2S ===> 2Se HSO4- + H+ + 2H2O

In a basic aqueous solution of the ionic charge redoxequation is balanced with corresponding hydroxide ions, OH-, and H2O.

 

Redox reactions in organic chemistry

 

In order not to press ahead to far in organic chemistry, only some basic and practical examples are mentioned presenting some important oxidizing and reducing agents.

oxidation

A compound containing one or more groups OH (hydroxide) is an alcohol. It is designated R ending OH, R-OH. R is the alkyl part of any organic compound, OH is the hydroxy group OH. The compound can have several OH-groups, especially sugars do have. For example glucose has 5 OH-groups.

Any alcohol, R-COH, can be oxidized into aldehyde, RCHO. This aldehyde can get a further oxidation step to carboxylic acid, RCOOH.

Example:

Ethanol, CH3CHOH (ingredrient of spirits, wine and beer) is produced by the first oxidizing step to acetaldehyde, CH3CHO and by the 2nd oxidation to acetic acid, CH3COOH.

CH3CH2OH ===> CH3CHO ===> CH3COOH

There is a third oxidation to peracetic acid, CH3COOOH, however this compound is rather unstable.

The reduction proceeds in the reverse direction of acetic acid to the organic compound, which is totally free of oxygen.

CH3COOH ===> CH3CHO ===> CH3CH2OH ===> CH3CH3 (Ethan)

Practical example.

In earlier times Doctors detected sugar in urine by a solution of a silversalt.

The so-called Tollens detection was applied:

Glucose is an aldehyde, RCHO, and is excreted in urine.

The sugar detection is carried out by reduction of silver, and oxidation of glucose.

elemental silver is produced, which is reflected on a mirror.

RCHO + Ag+ ===> Ag (silver mirror) + RCOOH

Silver acts as an oxidizing agent.

 

Important oxidizing agents in organic chemistry

Dichromate, K2Cr2O7 or Na2Cr2O7,

Chromium trioxide, CrO3

Potassium permanganate, KMnO4

Bromine, Br2

 

Important reducing agents in organic chemistry

Natriunborhydrid, NaBH4

Lithium aluminum hydride, LiAlH4

All these reagents are also available commercially .

Little Exercise: What are the oxidation numbers of LiAlH4 ?

Answer : Li (+I) , Al (+III), insignificant, but H (-I ) !! It is the hydride ion which has reducing function.

 

Small Introduction to electrochemistry

The salt battery

The most simple example of a battery. In a saturated salt solution consisting of NaCl , which means solution of Na+ and Cl-.

At the anode oxidation and at the cathode reduction is taking place.

More on this topic in physics or higher lessons.

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