After writing of any reaction in it it is necessary to place coefficients. Sometimes it is possible to make it by usual mathematical selection. In other cases it is necessary to use special methods: method of electronic balance or method of semi-reaction.
Instruction
1. If reaction is not oxidation-reduction, i.e. takes place without change of oxidation levels, then selection of coefficients comes down to simple mathematical calculations. The amount of the substances received as a result of reaction has to be equal to amount of the substances entering it. For example: BaCl2 + K2SO4 = BaSO4 + KCl. We consider amounts of substances. Ba: 2 in the left member of equation - 2 in right. Cl: 2 in left – 1 in right. We equalize, we put coefficient 2 before KCl. We receive: BaCl2 + K2SO4 = BaSO4 + 2KCl. We count amounts of other substances, all of them coincide.
2. In oxidation-reduction reaction, i.e. the reaction which is taking place with change of oxidation levels, coefficients place either by method of electronic balance, or by a semi-reaction method. The method of electronic balance consists in equalizing of number of the electrons given by reducer and numbers of the electrons accepted by oxidizer. It should be noted that reducer – atom, a molecule or an ion which gives electrons, and oxidizer – atom, a molecule or the ion attaching electrons. Let's sort on an example: H2S + KMnO4 + H2SO4 = S + MnSO4 + K2SO4 + H2O. At first we define what substances changed oxidation level. It is Mn(от +7 до +2), S(от-2 до 0). We show process of return and accession of electrons by means of the electronic equations. Coefficients we find by the rule of the smallest multiple multiplier.Mn (+7) +5e = Mn (+2)/2S(-2) – 2e = we substitute S (0)/5dalee in the equation of reaction the received coefficients: 5H2S + 2KMnO4 + H2SO4 = 5S + 2MnSO4 + K2SO4 + H2O. But equalizing on it comes to an end very seldom, it is necessary to count still amounts of other substances and to balance them as we did it in reactions without change of oxidation levels. After equalizing we will receive: 5H2S + 2KMnO4 + 3H2SO4 = 5S + 2MnSO4 + K2SO4 + 8H2O.
3. The following method consists in drawing up semi-reactions, i.e. the ions real-life in solution undertake already (for example, not Mn(+7), but MnO4(-1)). Then semi-reactions are summarized in the general equation and with its help coefficients are placed. For an example we will take the same reaction: H2S + KMnO4 + H2SO4 = S + MnSO4 + K2SO4 + N2O.Sostavlyaem semi-reaction.MnO4 (-1) – Mn (+2). We look at the reaction environment, in this case it sour, because of presence of sulfuric acid. Means it is equalized by hydrogen protons, we do not forget to fill missing oxygen with water. We receive: MnO4 (-1) + 8H (+1) + 5e = Mn (+2) +4N2O.Drugaya semi-reaction looks so: H2S – 2e = S + 2H (+1). We put both semi-reactions, previously having balanced number of the given and received electrons, using the rule of the smallest multiple multiplier: H2S – 2e = S + 2H (+1)/5MnO4 (-1) + 8H (+1) + 5e = Mn (+2) + 4H2O/25H2S + 2MnO4 (-1) + 16H(+1) = 5S + 10H (+1) + 2Mn (+2) +8N2OSokrativ hydrogen protons, we receive: 5H2S + 2MnO4 (-1) + 6H(+1) = 5S + 2Mn (+2) +8N2O.Perenosim coefficients in the equation in a molecular form: 5H2S + 2KMnO4 + 3H2SO4 = 5S + 2MnSO4 + K2SO4 + 8N2O.Kak you see result same, as well as at application of a method of electronic balance. In the presence of the alkaline environment of semi-reaction are equalized with the help hydroxide ions (OH(-1))