Redox Reactions (ORR): Examples

What is iad? Examples of such reactions can be found not only in inorganic, but also in organic chemistry. In the article we will give definitions of the main terms used in the analysis of such interactions. In addition, we will present some of the ERIs, examples and solutions of chemical equations that will help to understand the algorithm of actions.

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Basic definitions

But first, let's recall the basic definitions that will help to understand the process:

  • An oxidizer is an atom or an ion capable of accepting electrons in the process of interaction. In the form of serious oxidizing agents are mineral acids, potassium permanganate.
  • A reducing agent is an ion or an atom that donates valence electrons to other elements.
  • The process of attaching free electrons is called oxidation, while recoil is called reduction.

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Action algorithm

How to parse the equation of iad? Examples offered to graduates of schools suggest the distribution of coefficients by electronic balance.Here is the procedure:

  1. First, it is necessary to set the values ​​of oxidation states for all elements in simple and complex substances involved in the proposed chemical transformation.
  2. Next, select those elements that have changed the digital value.
  3. The signs "+" and "-" indicate the received and given electrons, their number.
  4. Further, between them determine the smallest common multiple, determine the coefficients.
  5. The resulting numbers put in the reaction equation.

First example

How to complete the task associated with IAD? The examples offered at the final exams in grade 9 do not imply the addition of substance formulas. The children, as a rule, need to determine the coefficients and substances that have changed the valence values.

Consider those IAD (reactions), examples of which are offered to graduates of 11th grade. Schoolchildren should independently supplement the equation with substances and only after that by electronic balance arrange the coefficients:

H2O2+ H2SO4+ KMnO4= Mn SO4+ O2+ …+…

To begin with, we will arrange the oxidation states in each compound. So, in hydrogen peroxide at the first element, it corresponds to+1have oxygen-1. In sulfuric acid the following indicators:+1, +6, -2(in sum, we get zero).Oxygen is a simple substance, so it has a zero oxidation rate.

In potassium permanganate, as well as in manganese sulfate (2), we obtain the following values:

K+Mn+7O4-2, Mn+2S+6O4-2

Putting the values ​​of the elements proposed in the task, you need to finish the IAD. Examples of such interactions are similar, therefore, when solving, it is necessary to identify atoms (ions) that exhibit oxidizing and reducing properties.

So, as one of the missing products of the reaction will be the salt of potassium, namely sulfate. The second substance is water, since the process involves sulfuric acid with hygroscopic properties.

The next step is to draw up an electronic balance of this process:

  • 2O–gives 2 electrons =O205(reducing agent);
  • Mn+7accepts 5 electrons =Mn+22(oxidizer).

In the process of placing the coefficients, we necessarily sum up the sulfur atoms, we get the ready-made equation of the process:

5H2O2+ 3H2SO4+ 2KMnO4= 2Mn SO4+ 5O2+ 8H2O + K2SO4

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Difficult moments

What difficulties do schoolchildren have when analyzing IAD? The examples offered in the final tests in chemistry, the guys have to finish on their own, which causes them difficulties.

Suppose the proposed scheme:FeCl2+ HCl + K2Cr2O7= FeCl3+ CrCl3+ …+…

It is necessary to add the missing substances and arrange the necessary stereochemical coefficients in the equation. In the proposed assignment, the degree of oxidation changes iron: with+2on+3therefore exhibits restorative properties. Potassium bichromate acts as an oxidizing agent, lowering the value of the degree of oxidation with+6before+3.

The missing products of the reaction will be water, as well as potassium chloride. They do not participate in the electronic balance, since the elements included in their composition do not show a change in their numerical value. The electronic balance for this process will be as follows:

  • Fe+2gives 1 electron =Fe+36(reducing agent);
  • 2cr+6takes 6 e =2cr+31(oxidizer).

When placing the coefficients in this scheme, we summarize the chlorine atoms:

6 FeCl2+ 14HCl + K2Cr2O7= 6 FeCl3+ 2CrCl3+ 2KCl + 7H2O

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Another example of interaction

We will continue the conversation on how to properly disassemble the IAD. Chemistry (examples of such reactions in it are common) not only explains the algorithm of actions, but also characterizes the essence of the processes taking place. Consider another example of the interaction, accompanied by the formation of new chemicals:

KMnO4+ H2SO4+ KI = MnSO4+ I2+…+…

In this example, there are two elements that change the degree of oxidation: iodine and manganese. Find out what substances will be formed as products of this chemical reaction.

Since sulfuric acid is involved in the process, water will be one of the substances formed. In the right side there is no compound in potassium, therefore the second product will be the sulfate of this alkali metal.

The electronic balance for this interaction is as follows:

  • Mn+7takes 5 e =Mn+22, is an oxidizing agent;
  • 2I-gives 2e =I205, acts as a reducing agent.

At the final stage of this task, we place the coefficients in the finished scheme and get:

2KMnO4+ 8H2SO4+ 10KI = 2MnSO4+ 5I2+ 6K2SO4+ 8H2O

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Conclusion

These processes have found serious application in chemical analysis. With their help, you can open and separate different ions, carry out the method of oximetry.

A variety of physical and chemical methods of analysis are based on IAD. The theory of acid and basic interactions explains the kinetics of the processes that occur, and allows us to carry out quantitative calculations using equations.

In order for schoolchildren who have chosen chemistry to pass at the final exam successfully pass these tests, it is necessary to work out an IHB equalization algorithm based on electronic balance.Teachers work with their students on the method of arranging the coefficients, using various examples from inorganic and organic chemistry.

Assignments related to the determination of oxidation states of chemical elements in simple and complex substances, as well as the drawing up of a balance between adopted and given electrons, are an indispensable element of examination tests at the basic, general level of training. Only in the case of successful completion of such tasks can we talk about the effective mastering of the school course of inorganic chemistry, and also expect to receive a high mark on the OGE, EGE.

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