Before giving examples of redox reactions with a solution, we single out the basic definitions associated with these transformations.
Those atoms or ions that during the course of the interaction change the degree of oxidation with decreasing (accepting electrons) are called oxidizing agents. Among the substances possessing such properties, one can note strong inorganic acids: sulfuric, hydrochloric, nitric.
Oxidizing agent
Strong oxidizing agents include permanganates and alkali metal chromates.
The oxidizing agent accepts the number of electrons in the course of the reaction that it needs before the completion of the energy level (establishment of the completed configuration).
Reducing agent
Any redox reaction scheme involves the identification of a reducing agent. It includes ions or neutral atoms that can increase the oxidation state during the course of the interaction (they give electrons to other atoms).
Typical reducing agents include metal atoms.
Processes in OVR
What else are characterized by OVR? Redox reactions are characterized by a change in the oxidation state of the starting materials.
Oxidation involves the process of recoil of negative particles. Recovery involves accepting them from other atoms (ions).
Parsing algorithm
Examples of redox reactions with a solution are offered in various reference materials designed to prepare high school students for final tests in chemistry.
In order to successfully cope with the tasks proposed in the Unified State Examination and the Unified State Examination, it is important to have an algorithm for compiling and analyzing redox processes.
- First of all, charge quantities are put down for all elements in substances proposed in the scheme.
- Atoms (ions) are written out from the left side of the reaction, which, during the interaction, changed indicators.
- With an increase in the degree of oxidation, the sign “-” is used, and with a decrease in “+”.
- Between the given and received electrons is determined by the lowest common multiple (the number by which they are divided without a remainder).
- When dividing the NOC into electrons, we obtain stereochemical coefficients.
- We place them before the formulas in the equation.
The first example from the OGE
In the ninth grade, not all students know how to solve redox reactions. That is why they make a lot of mistakes, do not get high scores for OGE. The algorithm of actions is given above, now we will try to work it out on concrete examples.
The peculiarity of the tasks relating to the arrangement of the coefficients in the proposed reaction, issued to graduates of the main stage of training, is that both the left and right sides of the equation are given.
This greatly simplifies the task, since it is not necessary to independently invent interaction products, to select the missing starting materials.
For example, it is proposed using electronic balance to identify the coefficients in the reaction:
CuO + Fe = FeO + Cu
At first glance, stereochemical coefficients are not required in this reaction. But, in order to confirm their point of view, charge elements are necessary for all elements.
In binary compounds, which include copper oxide (2) and iron oxide (2), the sum of oxidation states is zero, given that in oxygen it is -2, in copper and iron this indicator is +2. Simple substances do not give (do not accept) electrons; therefore, they are characterized by a zero oxidation state.
We will compose an electronic balance, showing with a “+” and “-” sign the number of electrons received and given during the interaction.
Cu 2+ + 2e = Cu0;
Fe 0 -2e = Fe 2+ .
Since the number of electrons received and given during the interaction is the same, it makes no sense to find the least common multiple, determine stereochemical coefficients, put them in the proposed interaction scheme.
In order to get the maximum score for the task, it is necessary not only to write down the examples of redox reactions with the solution, but also to write out separately the formula of the oxidizing agent (CuO) and the reducing agent (Fe).
The second example with OGE
Here are some more examples of redox reactions with a solution that ninth-graders may choose to choose as the final exam.
Suppose, it is proposed to arrange the coefficients in the equation:
Na + HCl = NaCl + H 2 .
In order to cope with the task, it is first important to determine the degree of oxidation in each simple and complex substance. In sodium and hydrogen, they will be equal to zero, since they are simple substances.
In hydrochloric acid, hydrogen has a positive and chlorine a negative oxidation state. After arranging the coefficients, we obtain a reaction with the coefficients.
The first sample of the assignment from the exam
How to supplement redox reactions? Examples of the solution encountered in the exam (grade 11) involve the addition of omissions, as well as the arrangement of the coefficients.
For example, you need to electronically supplement the reaction:
H 2 S + HMnO 4 = S + MnO 2 + ...
Define the reducing agent and oxidizing agent in the proposed scheme.
How to learn how to make redox reactions? The sample involves the use of a specific algorithm.
First, in all substances given by the condition of the problem, it is necessary to set the oxidation state.
Next, you need to analyze which substance may become an unknown product in this process. Since there is an oxidizing agent (manganese in its role), a reducing agent (sulfur is it), the oxidation state does not change in the desired product, therefore, it is water.
Talking about how to solve redox reactions, we note that the next step will be the preparation of the electronic ratio:
Mn +7 takes 3 e = Mn +4 ;
S -2 gives 2e = S 0 .
Manganese cation is a reducing agent, and sulfur anion is a typical oxidizing agent. Since the smallest multiple between the received and given electrons will be 6, we obtain the coefficients: 2, 3.
The final step will be to put the coefficients in the original equation.
3H 2 S + 2HMnO 4 = 3S + 2MnO 2 + 4H 2 O.
The second sample OVR in the exam
How to make redox reactions? Examples with the solution will help to work out the algorithm of actions.
It is proposed by the electronic balance method to fill in the gaps in the reaction:
PH 3 + HMnO 4 = MnO 2 + ... + ...
We set the oxidation state of all elements. In this process, the oxidizing properties are manifested by manganese, which is part of manganese acid, and the reducing agent must be phosphorus, changing its oxidation state to positive in phosphoric acid.
According to the assumption made, we obtain a reaction scheme, then we compose the electronic balance equation.
P -3 gives 8 e and turns into P +5 ;
Mn +7 takes 3e, turning into Mn +4 .
The NOC will be 24, so phosphorus must have a stereometric coefficient of 3, and manganese -8.
We put the coefficients in the resulting process, we get:
3 PH 3 + 8 HMnO 4 = 8 MnO 2 + 4H 2 O + 3 H 3 PO 4 .
The third example from the exam
By electron-ion balance, you need to make a reaction, indicate a reducing agent and an oxidizing agent.
KMnO 4 + MnSO 4 + ... = MnO 2 + ... + H2SO 4 .
According to the algorithm, we place the oxidation state of each element. Next, we determine those substances that are missing in the right and left parts of the process. Here, a reducing agent and an oxidizing agent are given, therefore, the oxidation state in the missing compounds does not change. The lost product will be water, and the starting compound is potassium sulfate. We get a reaction scheme for which we compose an electronic balance.
Mn +2 -2 e = Mn +4 3 reducing agent;
Mn +7 + 3e = Mn +4 2 oxidizing agent.
We write the coefficients in the equation by summing the manganese atoms on the right side of the process, since it relates to the disproportionation process.
2KMnO 4 + 3MnSO 4 + 2H 2 O = 5MnO 2 + K 2 SO 4 + 2H 2 SO 4 .
Conclusion
Redox reactions are of particular importance for the functioning of living organisms. Examples of OVR are the processes of decay, fermentation, nervous activity, respiration, and metabolism.
Oxidation and reduction are relevant for the metallurgical and chemical industries, thanks to such processes it is possible to recover metals from their compounds, protect them from chemical corrosion, and process them.
To compose the redox process in organic or inorganic chemistry, it is necessary to use a certain algorithm of actions. First, in the proposed scheme, the degrees of oxidation are arranged, then those elements that increase (lower) the indicator are determined, and the electronic balance is recorded.
Next, between the received and given electrons, it is necessary to determine the smallest multiple, and calculate the coefficients mathematically.
If you follow the sequence of actions proposed above, you can easily cope with the tasks offered in the tests.
In addition to the electronic balance method, the arrangement of the coefficients is also possible by drawing up half-reactions.