How To Make Acidified Sodium Dichromate Molecular

Oxidation numbers and redox reactions

1) Oxidation numbers:

With every oxidation and reduction electrons are shifted. You move from one element to another. The change is noticeable in the elements concerned through a change in their electrical charge. We have such redox systems in a very large number of simple chemical processes. Above all, however, such shifts take place in the more complicated reactions - and one can benefit from this when setting up the equations. Since electrons cannot be lost any more than atoms or molecules can, they always have to be found somewhere in the equation, which makes finding the right coefficients a lot easier.

In real ion reactions, the oxidation numbers can easily be determined. They correspond to the charges of the individual ions. In magnesium chloride MgCl2, the magnesium is positively charged twice, it has the oxidation number +2; the chlorine is simply negatively charged, it has an oxidation number of -1.

But not all chemical reactions are really ionic reactions. Many of them follow different rules - but this doesn’t matter in this context: You can pretend that all molecules in equations consist of ions.

This does not represent the chemical reality correctly, but it makes the writing of equations much easier.

For example, the water molecule is not ionized, but it can still be broken down into two positively charged hydrogen ions and a doubly negatively charged oxygen ion:

2 H+ and O2-

The first rule can be derived from this:

When writing equations with the help of oxidation numbers, one always assumes that hydrogen in compounds has the oxidation number +1, oxygen in compounds has the oxidation number -2.

This immediately results in a number of different oxidation numbers, if you consider that a molecule must appear neutral to the outside world.

The formula of hydrochloric acid is HCl. Since H with the mental decomposition of the molecule according to the above rule H+ has to be written, only Cl remains for chlorine-. Accordingly, chlorine is always simply negatively charged in chlorides. The same applies to all elements of the halogens (fluorine, chlorine, bromine, iodine, astatine).

Accordingly, one can also determine the oxidation number of iron in iron chloride FeCl2 determine. Two negative chloride ions are opposed to two positive charges on iron, i.e. Fe2+. Iron therefore has the oxidation number +2 in the iron chloride present.

There is, however, another ferric chloride, which has the formula FeCl3 owns. According to the example above, the iron in this compound must have an oxidation number of +3.

All of the compounds mentioned so far can be understood as true ionic compounds. In order not to conflict with these "real" ions, the oxidation numbers are not written at the top right after the symbol, as is usual with real ion charges, but rather vertically above the symbol.

Different simple molecules can be assigned their oxidation numbers according to the same principle.

Elements that are not in connection, but rather “elementary” (pure ferrous metal, gold, etc.) are given the oxidation number 0 by definition. The elementary gases that occur diatomically (H2, O2, N2, etc.), are assigned the oxidation number 0.

The first equations can be solved according to this system.

When chlorine gas is introduced into iron (II) chloride solution, iron (III) chloride is formed.

It can be clearly seen that the two elementary chlorine atoms with the oxidation number o change into two negatively charged chloride ions with the oxidation number -1. So you have each picked up an electron.

Every chlorine atom has an electron from one fetched because the iron shows up on the right hand side of the equation as on, so this one has lost an electron because it has become one oxidation state more positive.

Since the chlorine absorbed a total of two electrons, two electrons were also released from the iron; but since each Fe only provides one electron, 2 Fe must have been involved in the reaction.

So the equation must be:

The whole interplay boils down to a redox reaction:

(Submission of 2 e-: Oxidation of iron)

(Image of 2 e-: Reduction of chlorine)

Training with oxidation numbers

1) Tin (II) chloride is oxidized to tin (IV) chloride by oxygen and hydrochloric acid, with water being formed at the same time.

The chlorine in this equation is always in combination on the right and left and therefore has the oxidation number -1. It is the same with hydrogen H, which here always has the oxidation number +1.

Oxygen, on the other hand, appears elementary on the left with the oxidation number 0 and on the right in connection with the oxidation number -2, so it has been reduced.

(Uptake of 2 * 2 e- = 4 e-; oxygen has been reduced)

Tin is the element that was oxidized in the reaction. In tin (II) chloride, tin has the oxidation number +2, as it is linked to 2 Cl with the oxidation number -1 and must be neutral to the outside. In tin (IV) chloride it has the oxidation number +4, because it is now connected to 4 chlorine atoms.

(Release of 2 e-; tin has been oxidized)

Since oxygen absorbs 4 electrons, but tin only supplies 2 electrons per atom, 2 tin atoms must be brought into the reaction. Molecular oxygen provides 2 oxygen atoms, therefore 2 oxygen atoms must also appear on the right side. This can be achieved by 2 H2O.

The redox process is now complete and the equation only needs to be balanced. On the right side there are 8 Cl, which can be removed by adding 4 HCl (of course plus the 2 SnCl2) on the left side.

2) Sulphurous acid oxidizes to sulfuric acid in the air.

Oxygen is in the elementary state on the left and thus has the oxidation number 0. On the right (and left side) it is in combination and has the oxidation number -2.

(Uptake of 2 * 2 e- = 4 e-; oxygen has been reduced)

Sulfur S is in the H2SO3 with the oxidation number +4. The molecule is electrically neutral to the outside. The oxidation numbers of O (-2) and H (+1) are known. Oxygen O occurs 3 times, i.e. 3 * -2 = -6 and hydrogen H 2 times, i.e. 2 * +1 = +2. The sum of the occurring H and O atoms results in -4 ([-6] + [+ 2] = -4). The sulfur has to balance this charge -4, which is why it is assigned the oxidation number +4 so that the molecule appears neutral to the outside world.

The oxidation number of sulfur in the H is calculated using the same method2SO4 and get the oxidation number +6 for sulfur in this compound.

During the transition from sulphurous acid to sulfuric acid, each sulfur atom has to give up 2 electrons.

At the transition from O2 into the oxygen of the compound H2SO3 4 electrons were absorbed. To deliver 4 electrons we need 2 sulfur atoms. Hence the equation reads:

3) By reducing diboron trioxide with metallic magnesium, boron is obtained.

What is the oxidation number of boron in the diboron trioxide (B.2O3)?

Boron is linked to 3 oxygen atoms. By convention, each of these O atoms has the oxidation number -2, together that is -6.

This -6 must balance the two boron atoms, since the charge of the molecule is zero, both boron atoms have the oxidation number +3 (2 * +3 = +6). Now the two charges equalize to zero ([oxygen] -6 + [boron] +6 = 0 [B2O3]).

On the right-hand side of the equation, boron is in the elemental state (B), its oxidation number is 0.

During the reaction, boron takes 2 * 3 = 6 electrons:

Boron has thus been reduced.

Magnesium oxidation number:

Magnesium is in its elemental state on the left. It therefore has the oxidation number 0.

On the right side, Mg is linked to an oxygen atom (MgO). This molecule is uncharged and oxygen always has an oxidation number of -2. In order to balance this charge on the O atom, magnesium has an oxidation number of +2 in this compound.

In order to be able to provide the 6 electrons for the boron, 3 Mg is required for the reaction.

So the equation is:

Mnemonics:

The oxidation number is understood to be the charges that an atom in a molecule would have if the molecule were made up of just ions.

You can easily calculate the oxidation number of an element if you remember:

- H has the oxidation number +1 (exception: HLi)

- O has the oxidation number -2,

- and the entire molecule must be neutral.

- By definition, free elements have the oxidation number 0.

- In many Compounds have halogens (7th HG) the oxidation number -1.

- Alkali metals (1st HG) always have the oxidation number +1.

- Alkaline earth metals (2nd HG) always have the oxidation number +2.

4) What is the oxidation number of the chromium in the potassium dichromate?

Potassium dichromate has the formula K.2Cr2O7.

Potassium (K) is in the 1st main group and therefore has the oxidation number +1. As always, oxygen (O) is assigned the oxidation number -2.

The subtotal for this molecule is:

2 * +1 (from K) + 7 * -2 (from O) = -12

The two remaining Cr atoms have to balance this value of -12, so each Cr has the oxidation number +6 (2 * +6 = +12).

The assigned oxidation numbers are thus:

5) What is the oxidation number of nitrogen (N) in nitric acid HNO3?

H has the oxidation number +1, oxygen -2.

The subtotal for this molecule is:

1 * +1 (from H) + 3 * -2 (from O) = -5

This -5 has to balance the nitrogen atom and is assigned the oxidation number +5.

6) What is the oxidation number of nitrogen in ammonium chloride (NH4Cl)?

H has the oxidation number +1, chlorine has the oxidation number -1 (7th HG !!!).

Subtotal:

4 * +1 (from H) + 1 * -1 (from Cl) = +3

So nitrogen (N) in ammonium chloride has the oxidation number -3

7) What is the oxidation number of manganese (Mn) in potassium permanganate (KMnO4)?

K has the oxidation number +1 (1st HG), oxygen -2.

Subtotal:

1 * +1 (from K) + 4 * -2 (from O) = -7

Manganese has an oxidation number of +7 in potassium permanganate.

8) What is the oxidation number of platinum (Pt) in potassium hexachloroplatinate (K2[PtCl])?

K has the oxidation number +1 (1st HG), chlorine -1 (7th HG).

Subtotal:

2 * +1 (from K) + 6 * -1 (from Cl) = -4

In potassium hexachloroplatinate, platinum has an oxidation number of +4.

9) What is the oxidation number of tungsten in metatungstic acid (H.8W.12O40)?

Hydrogen has the oxidation number +1, oxygen -2.

Subtotal:

8 * +1 (from H) + 40 * -2 (from O) = -72

To neutralize the molecule, 72 positive charges are necessary. These share the 12 tungsten atoms, each W receives 6 positive charges (+72: 12 = +6). The oxidation number of tungsten is +6.

10) What is the oxidation number of nitrogen in hydrazine (H.2N-NH2)?

Elements that bind “their own kind” get the oxidation number 0 (Cl-Cl, H-H, O-O, etc.). For the N-N bond in hydrazine, the oxidation number 0 is assigned to both nitrogen atoms. Only the two H's on each nitrogen count in this connection.

2) redox reactions

The oxidation numbers are very helpful for solving redox equations.

However, in order to be able to solve equations at all, the starting materials and end products must be known.

Such a reaction equation, the masses of which are not balanced, is called a “skeleton”.

1) Nitric acid oxidizes iodine to iodic acid, whereby nitrogen oxide and water are also formed.

First you have to determine which atoms will change their oxidation number.

Right side (starting materials, starting materials):

Left side (products):

Nitrogen in ENT3: +5

Iodine in its elemental state: 0

Iodine in HJO3: +5

Nitrogen in NO: +2

All oxygen and hydrogen atoms have retained their oxidation number.

To make things easier, an equation is set up that only includes the atoms that also change their oxidation number:

Every nitrogen (N) changes from +5 to +2, so it takes 3 electrons and is reduced.

Every iodine (I) changes from 0 to +5, so it gives off 5 electrons and is oxidized.

So that as many electrons can be absorbed as emitted, 5 N and 3 J have to be included in the equation.

N takes 3 electrons per atom: 5 * 3 e- = 15 e-

J releases 5 electrons per atom: 3 * 5 e- = 15 e-

This gives the coefficients of the atoms that change their oxidation number.

The hydrogen and oxygen atoms must now be balanced between right and left. On the right-hand side, and 10 are too few. By adding a water molecule (H.2O) this deficit can be eliminated.

2) Sulfur is oxidized to sulfuric acid by concentrated nitric acid, producing nitrogen.

"Skeleton" equation:

Again, we first consider only the atoms that change their oxidation number:

Educts

Products

Elemental sulfur: 0

Nitrogen in ENT3: +5

Sulfur in H2SO4: +6

Nitrogen in NO: +2

Every sulfur goes from 0 to +6, so it gives off 6 electrons and is oxidized.

Every nitrogen changes from +5 to +2, takes 3 electrons, is reduced.

If you coordinate the absorption and release of electrons again, you can see that 2 N have to be introduced into the equation in order to accommodate the 6 electrons released by the sulfur.

S gives 6 e-: 1 * 6 e- = 6 e-

N takes 3 e-: 2 * 3 e- = 6 e-

From this follows the equation:

If you do the math, you can see that there are corresponding quantities of atoms on the right and left in the equation. It is not necessary to add water molecules or the like on either side.

3) Phosphorus is oxidized to phosphoric acid by concentrated nitric acid, producing nitric oxide.

Framework:

Oxidation numbers:

Electron shift:

P emits 5 e-, it was oxidized: 3 * 5 e- = 15 e-

N takes 3 e-, it was reduced: 5 * 3 e- = 15 e-

To balance the 5 e of phosphorus and the 3 e of nitrogen, the corresponding coefficients are used:

With this the essential thing has already been achieved; there is only an equalization of the H and O between left and right.

On the right, in the current state of the equation, and 2 O are too much. These are in the form of 2 H2O added on the left.

Exercises:

Solutions:

3) disproportionation:

So far, only redox equations have been dealt with, in which the atoms involved changed from one particular oxidation number to another.

However, it is not uncommon for an atom to change from a certain oxidation number to a higher one and splits a lower component. This phenomenon is called Disproportionation.

While so far only electron shifts took place between atoms of different elements, in the following examples the atoms of one and the same element push their electrons towards each other.

1) Nitrous acid (ENT2) tends to do so when heated in nitric acid (ENT3) and nitric oxide (NO).

If one calculates the oxidation numbers for N in these compounds, one obtains:

The oxidation number of the N in HNO2 (+3) stands between the other two values ​​ENT3 (+5) and NO (+2). Part of the molecules of nitrous acid (ENT2) thus oxidizes when heated (from +3 to +5; release of electrons), another part is reduced (from +3 to +5; absorption of electrons)

Electron shift:

To compensate for the emission of 2 e- with an N (lower arrow), 2 times 1 e-, i.e. 2 N, are necessary (upper arrow). A total of 3 N are involved in the reaction:

This compensates for the electron shift. An H is now missing on the right side of the equation2O:

2) When potassium chlorate (KClO) is heated to 400 ° C, potassium perchlorate (KCLO4) and potassium chloride (KCl).

Oxidation numbers:

Electron shift:

So that a single Cl can take up six electrons (upper arrow), three Cls must give off two electrons each (lower) arrow. A total of four Cl are then involved in the reaction.

3) Hypobromous acid (HBrO), disproportionate when heated to form hydrogen bromide (HBr) and bromic acid (HBrO3).

Oxidation numbers:

Electron shift:

In order for a single Br to be able to donate four electrons (lower arrow), two Br must accept two electrons each; In total, three Br are involved in the process:

4) When heating with water (H.2O) white phosphorus disproportionates partly into hydrogen phosphide (PH3) and hypophosphorous acid (H3PO2).

Oxidation numbers:

Electron shift:

So that a single P can take up three electrons (upper arrow), three P must each give off one electron (lower arrow); In total, four P's are involved in the process:

The number of H and O still needs to be balanced on both sides. Left are and 1 O; on the right there are and 6 O, so and 5 O on the left side are not enough. Additional 5 H2O balance this.

Equations that contain disproportionations are not uncommon. At least as often, however, are cases in which an element changes into two different oxidation states without being able to speak of a higher and a lower one.

5) During the oxidation of sulfur (S) by saltpetre (KNO3) sulfur dioxide (SO2), elemental nitrogen (N2) and potassium sulfate (K2SO4):

Oxidation numbers:

So while the nitrogen is completely reduced from the oxidation number +5 to the oxidation number 0, the sulfur atoms on the right-hand side appear partly as +4, partly as +6. Since because of the molecule N2 If at least two nitrogen atoms have to be present on the left side, at least 2 times 5 = 10 electrons are required for the reduction of the nitrogen.

These 10 electrons are made available by the transition of an S with the oxidation number 0 into the state of an S with the oxidation number +4 and an S (0) into the state S (+6).

Electron shift:

So two sulfur atoms are needed to restore the electron balance:

6) potassium dichromate (K2Cr2O7) reacts with hydrogen sulfide (H.2S) with the formation of potassium chromate (K2CrO4), Chromium hydroxide (Cr (OH)3) and sulfur.

Oxidation numbers:

Potassium chromate (K2CrO4) does not change its oxidation number and can be kept out of the considerations.

Electron shift:

So that the two Cr of potassium dichromate (+6) can be reduced to the level of chromium hydroxide, three S must be oxidized from -2 to 0:

In terms of charge, the equation is thus balanced. Now the atoms that have not changed their charge still have to be balanced,

On the right is a molecule of K2CrO4 with an excess Cr present. Since we are on the left because of the occurrence of two coupled Cr in the K2Cr2O7 cannot add a single Cr, we add a molecule on each side (left: K2Cr2O7; right: K2CrO4):

7)

Oxidation numbers and electron shift:

Charge balancing:

Because of the three times 2 nitrate groups (NO3) on the right side we also have to provide 6 additional nitrate groups on the left side; that makes 8 ENT3 and thus also 4 H2O.

8)

Charge balancing:

The charges are thus balanced, but there is 6 Cl too much on the right-hand side, which can only be covered by adding a further 6 HCl on the left-hand side. This results in 16 HCl on the left. In this way we get the right amount of H (16) to oxidize the 2 times 4 = 8 oxygen atoms of the potassium permanganate:

9)

Charge balancing:

Compensation:

Source: Vogt, "Chemical equations - Quite simply", Aulis Verlag Deubner & Co KG Cologne. ISBN 3-7614-1793-4