Oxidation-reduction reactions are reactions with change of oxidation levels. Often happens so that initial substances are given and it is necessary to write products of their interaction. The same substance can sometimes give various final products in different environments.
Instruction
1. In dependence not only on the reaction course environment and also on oxidation level substance behaves differently. Substance in the highest oxidation level always is oxidizer, in the lowest - reducer. To make acidic environment usually use sulfuric acid (H2SO4), is more rare - nitric (HNO3) and salt (HCl). If necessary to create an alkaline environment we apply sodium hydroxide (NaOH) and potassium hydroxide (KOH). Further we will review some examples of substances.
2. Ion MnO4(-1). In acidic environment turns into Mn(+2), colourless solution. If Wednesday neutral, then is formed by MnO2, the brown deposit drops out. In the alkaline environment we receive MnO4(+2), solution of green color.
3. Hydrogen peroxide (H2O2). If it is oxidizer, i.e. accepts electrons, then in neutral and alkaline environments turns according to the scheme: H2O2 + 2e = 2OH(-1). In acidic environment we will receive: H2O2 + 2H (+1) + 2e = 2N2O.Pri a condition that hydrogen peroxide reducer, i.e. gives electrons, in acidic environment is formed by O2, in alkaline - O2 + H2O. If H2O2 gets on Wednesday with strong oxidizer, it will be reducer.
4. The ion of Cr2O7 is oxidizer, in acidic environment it turns in 2Cr (+3) which have green color. From an ion of Cr (+3) at presence hydroxide ions, i.e. in the alkaline environment is formed by CrO4 (-2) of yellow color.
5. Let's give an example of drawing up reaction.KI + KMnO4 + H2SO4 - In this reaction of Mn is in the highest oxidation level, the t.e is oxidizer, accepting electrons. Acidic environment, on it shows us sulfuric acid (H2SO4). Reducer is I (-1) here, it gives electrons, increasing at the same time the oxidation level. We write down reaction products: KI + KMnO4 + H2SO4 - MnSO4 + I2 + K2SO4 + H2O. We place coefficients by method of electronic balance or by method of semi-reaction, we receive: 10KI + 2KMnO4 + 8H2SO4 = 2MnSO4 + 5I2 + 6K2SO4 + 8H2O.