Redox Reactions-Understanding, Formulas and Examples
Friday, March 13, 2020
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Kang Gus Tri-This reading activity is the most disliked activity of most students. What's more to read the textbook is a thing that is not preferred by the present, moreover, the material that it reads is long added more lazy students to read the book.
For that path you take is right that has visited my blog, because I will summarize the subject matter for you all to not be too tired to read a book that is so thick, and very rambling.
For that please see carefully and carefully the material we will convey this, namely about the "redox reaction ".
Redox reaction
A. Redox reaction concept
Redox reaction is a reduction and oxidation reaction. Oxidation and reduction differences will be explained in the table below.
B. Oxidation numbers
Oxidation number provisions:
1. Free element has oxidation number = 0. Examples of free elements such as O2, Na, N2, and C;
2. Metal group IA, IIA, and IIIA have oxidation number in accordance with its classification. Example KOH oxidation number K = + 1;
3. Oxygen has oxidation number =-2, except:
4. Hydrogen has oxidation number = + 1, except in hydride compounds (hydrogen compounds with metals) have oxidation number =-1, for example in NaH, KH, and CaH2;
5. Total oxidation number of a compound = 0;
6. Number of oxidation of an ion = its load;
7. The Halogen element as Halia has oxidation number =-1.
C. Redox
There are 2 ways of redox reaction, namely:
1. Half Reaction method
The steps are:
* Write reduction reactions and oxidation separately;
* As the number of atoms that are changing the number of oxidation;
* As the number of atoms O with regard to acidic or alkaline conditions;
* Acidic conditions: The less O section is added H2O, the less H section plus H+;
* Alkaline Condition: Excess section O added H2O and Setarakan Atom H by adding OH-;
* As a second the amount of charge on the section that overloads positively;
* Put the amount of electrons both reactions by multipressing them;
* Sum both reactions.
2. Oxidation method Number
The steps include:
* Calculate the oxidation number of elements that have changed the oxidation number;
* Setarakan the number of elemental atoms that change the number of oxidation;
* Increase the number of oxidation changes in the increment and its decline by multiplying by the X Factor;
* Make the X Factor as an element multiplier or compound in the reaction;
* Setarakan charge by adding ion H+ (acid) and OH-ion (base);
* As the sum of H atom by adding H2O.
Enough of our material about the "redox reaction" on this occasion, here's a summary or summary of the redox reaction that I have summarized for all of us, may this my blog can help you in doing your job. And may you be the one who is more fond of reading, because with many reading we can open new horizons of this world.
Hopefully it can be useful for all of us, thank you.
For that path you take is right that has visited my blog, because I will summarize the subject matter for you all to not be too tired to read a book that is so thick, and very rambling.
For that please see carefully and carefully the material we will convey this, namely about the "redox reaction ".
Redox reaction
 |
A. Redox reaction concept
Redox reaction is a reduction and oxidation reaction. Oxidation and reduction differences will be explained in the table below.
Oxidation reaction
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Reduction reaction
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1. Oxygen binding reaction
Example:
CH4 (g) + 2O2 (g) → CO2 (g)
+ 2H2O(g)
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1. Oxygen release reaction
Example:
Cr2O2 (s) + 2Al(s) → Al2O2 (s) + 2Cr(s)
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2. Electron discharge reaction
Example: Na → Na+ + e-
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2. Electron binding reaction
Example: S2+ + 2e-→ S
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3. Increased oxidation reaction
Example: 4FeO + O2 → 2Fe2O3
The oxidation number of Fe
in 4FeO is + 2, whereas in 2Fe2O3 is + 3. Since the Fe element had an
increase in the number of oxidation, i.e. from + 2 to + 3, 4FeO experienced
an oxidation reaction.
The reductors are 4FeO and the substance of the oxidation result is
2Fe2O3.
Substances that experience
oxidation are called reductors.
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3. Decreased oxidation number reaction
Example: 2SO4 → 2SO2 + O4
The number of oxidation S
in 2SO4 is + 6, whereas in 2SO2 is + 4. Because the S element has an increase
in the number of oxidation, i.e. from + 6 to + 4, 2SO4 is subjected to
reduction reactions.
The Oxygenator is 2SO2 and the reduction substance is 2SO4.
Substances that are
subjected to reduction called oxidizing.
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B. Oxidation numbers
Oxidation number provisions:
1. Free element has oxidation number = 0. Examples of free elements such as O2, Na, N2, and C;
2. Metal group IA, IIA, and IIIA have oxidation number in accordance with its classification. Example KOH oxidation number K = + 1;
3. Oxygen has oxidation number =-2, except:
- Perioxide compound, oxidation number O =-1, example in H2O2, Na2O2
- Superoxide compounds, oxidation number O =-1/2, example in KO2;
4. Hydrogen has oxidation number = + 1, except in hydride compounds (hydrogen compounds with metals) have oxidation number =-1, for example in NaH, KH, and CaH2;
5. Total oxidation number of a compound = 0;
6. Number of oxidation of an ion = its load;
7. The Halogen element as Halia has oxidation number =-1.
C. Redox
There are 2 ways of redox reaction, namely:
1. Half Reaction method
The steps are:
* Write reduction reactions and oxidation separately;
* As the number of atoms that are changing the number of oxidation;
* As the number of atoms O with regard to acidic or alkaline conditions;
* Acidic conditions: The less O section is added H2O, the less H section plus H+;
* Alkaline Condition: Excess section O added H2O and Setarakan Atom H by adding OH-;
* As a second the amount of charge on the section that overloads positively;
* Put the amount of electrons both reactions by multipressing them;
* Sum both reactions.
2. Oxidation method Number
The steps include:
* Calculate the oxidation number of elements that have changed the oxidation number;
* Setarakan the number of elemental atoms that change the number of oxidation;
* Increase the number of oxidation changes in the increment and its decline by multiplying by the X Factor;
* Make the X Factor as an element multiplier or compound in the reaction;
* Setarakan charge by adding ion H+ (acid) and OH-ion (base);
* As the sum of H atom by adding H2O.
Enough of our material about the "redox reaction" on this occasion, here's a summary or summary of the redox reaction that I have summarized for all of us, may this my blog can help you in doing your job. And may you be the one who is more fond of reading, because with many reading we can open new horizons of this world.
Hopefully it can be useful for all of us, thank you.
