Redox reactions
A redox reaction is a reaction involving a reduction reaction and an oxidation reaction. The meaning of the oxidation reaction and the reduction reaction develops in accordance with the development of chemistry. Reduction reactions and oxidation reactions occur in everyday life, such as combustion reactions, vinegar making from alcohol, glucose breaking events in the body, iron filings, and so on.
Understanding Redox Reactions
Initially the concept of reduction and oxidation (redox) is limited to reactions involving the release and binding of oxygen. The oxidation reaction is the reaction of oxygen binding by a substance.
A redox reaction is a reaction involving a reduction reaction and an oxidation reaction. The meaning of the oxidation reaction and the reduction reaction develops in accordance with the development of chemistry. Reduction reactions and oxidation reactions occur in everyday life, such as combustion reactions, vinegar making from alcohol, glucose breaking events in the body, iron filings, and so on.
Understanding Redox Reactions
Initially the concept of reduction and oxidation (redox) is limited to reactions involving the release and binding of oxygen. The oxidation reaction is the reaction of oxygen binding by a substance.
Oxidation and Redox Reactions
The more universal concept of redox reactions to explain reactions involving covalent compounds is the concept of redox reactions based on changes in oxidation numbers.
The redox reactions that are difficult to explain with the concept of oxygen and the concept of electrons can be easily explained using the concept of oxidant numbers.
Oxidation number
The oxidation or oxidation state of an element is a positive or negative integer given to an element in forming a compound. The oxidation number of an element is determined by considering the following.
A) The ionic compound
The oxidation number of elements in monoatomic ions is the real charge of the compound ions.
Example:
The NaCl compound, formed from Na + and Cl- ions, then the oxidation number of Na atoms in NaCl is +1, and the oxidation number Cl is -1.
B) Covalent compounds
The thing to note in determining the oxidation number in the covalent compound is the price of the electronegativity scale of each of the constituent atoms.
The atomic elements having a higher electronegative scale value indicate that the attraction of the atoms to the bonding electron pair is stronger. Because it is stronger to attract an electron pair, it becomes negative, and therefore the oxidation number is given a negative number. Atoms that have lower electronegativity prices are given positive oxidation numbers.
Example:
HCl compounds are formed from hydrogen atoms (electronegativity H = 2.0) and chlorine atoms (electronegativity Cl = 3.0) by using a common electron pair. These joint electrons are more attracted to Cl atoms, then the chlorine atom is oxidized -1, whereas the hydrogen atom is given the +1 oxidation number.
Determination of oxidation number
To determine the oxidation number of suatau atoms in a compound can be used some of the following provisions.
1. The oxidation number of free element (not compound) is 0 (zero).
2. The number of algebraic oxidation numbers of all atoms in a compound is 0 (zero).
3. The number of algebraic oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.
4. Certain elements in forming compounds have certain oxidation numbers, for example:
Group IA atoms (Li, Na, K, Rb, Cs, and Fr) in the compound have +1 oxidation states.
Group IIA atoms (Be, Mg, Ca, Sr, and Ba) in the compound have a +2 oxidation number.
Class IIIA (B, Al, and Ga) atoms in the compound have +3 oxidation numbers.
The hydrogen atom (H) in the compound generally has a +1 oxidation number, except in the metal hydride. Metal hydrides are compounds formed from metal and hydrogen elements. In metal hydrides, such as LiH, NaH, CaH2, MgH2, and AlH3, the hydrogen atom is given a -1 oxidation number.
The oxygen atom (O) in the compound generally has an oxidation number of -2, except in the peroxide compound and OF2.
In peroxides, such as H2O2, Na2O, and BaO, the oxygen atoms are given the oxidation number -1, whereas at OF2 is given a +2 oxidation number
The more universal concept of redox reactions to explain reactions involving covalent compounds is the concept of redox reactions based on changes in oxidation numbers.
The redox reactions that are difficult to explain with the concept of oxygen and the concept of electrons can be easily explained using the concept of oxidant numbers.
Oxidation number
The oxidation or oxidation state of an element is a positive or negative integer given to an element in forming a compound. The oxidation number of an element is determined by considering the following.
A) The ionic compound
The oxidation number of elements in monoatomic ions is the real charge of the compound ions.
Example:
The NaCl compound, formed from Na + and Cl- ions, then the oxidation number of Na atoms in NaCl is +1, and the oxidation number Cl is -1.
B) Covalent compounds
The thing to note in determining the oxidation number in the covalent compound is the price of the electronegativity scale of each of the constituent atoms.
The atomic elements having a higher electronegative scale value indicate that the attraction of the atoms to the bonding electron pair is stronger. Because it is stronger to attract an electron pair, it becomes negative, and therefore the oxidation number is given a negative number. Atoms that have lower electronegativity prices are given positive oxidation numbers.
Example:
HCl compounds are formed from hydrogen atoms (electronegativity H = 2.0) and chlorine atoms (electronegativity Cl = 3.0) by using a common electron pair. These joint electrons are more attracted to Cl atoms, then the chlorine atom is oxidized -1, whereas the hydrogen atom is given the +1 oxidation number.
Determination of oxidation number
To determine the oxidation number of suatau atoms in a compound can be used some of the following provisions.
1. The oxidation number of free element (not compound) is 0 (zero).
2. The number of algebraic oxidation numbers of all atoms in a compound is 0 (zero).
3. The number of algebraic oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.
4. Certain elements in forming compounds have certain oxidation numbers, for example:
Group IA atoms (Li, Na, K, Rb, Cs, and Fr) in the compound have +1 oxidation states.
Group IIA atoms (Be, Mg, Ca, Sr, and Ba) in the compound have a +2 oxidation number.
Class IIIA (B, Al, and Ga) atoms in the compound have +3 oxidation numbers.
The hydrogen atom (H) in the compound generally has a +1 oxidation number, except in the metal hydride. Metal hydrides are compounds formed from metal and hydrogen elements. In metal hydrides, such as LiH, NaH, CaH2, MgH2, and AlH3, the hydrogen atom is given a -1 oxidation number.
The oxygen atom (O) in the compound generally has an oxidation number of -2, except in the peroxide compound and OF2.
In peroxides, such as H2O2, Na2O, and BaO, the oxygen atoms are given the oxidation number -1, whereas at OF2 is given a +2 oxidation number
Redox reactions
A redox reaction is a reaction involving a reduction reaction and an oxidation reaction. The meaning of the oxidation reaction and the reduction reaction develops in accordance with the development of chemistry. Reduction reactions and oxidation reactions occur in everyday life, such as combustion reactions, vinegar making from alcohol, glucose breaking events in the body, iron filings, and so on.
Understanding Redox Reactions
Initially the concept of reduction and oxidation (redox) is limited to reactions involving the release and binding of oxygen. The oxidation reaction is the reaction of oxygen binding by a substance.
A redox reaction is a reaction involving a reduction reaction and an oxidation reaction. The meaning of the oxidation reaction and the reduction reaction develops in accordance with the development of chemistry. Reduction reactions and oxidation reactions occur in everyday life, such as combustion reactions, vinegar making from alcohol, glucose breaking events in the body, iron filings, and so on.
Understanding Redox Reactions
Initially the concept of reduction and oxidation (redox) is limited to reactions involving the release and binding of oxygen. The oxidation reaction is the reaction of oxygen binding by a substance.
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How about equalizing redox reactions, specify how to do it
BalasHapusThe steps or steps of equalization are as follows:
HapusDivide the reaction into 2 parts half-reaction, collect the species that have atomic similarities (except O and H need not be noticed). It is permissible to add the same substance to two half-part reactions if necessary;
Equalizes the number of atoms other than atoms O and H;
Equalizes the number of O atoms by adding OH- to the deficient side O in any ambience, and equalizes the number of H atoms by adding H + to the H-deficient side;
Equalize the amount of charge by adding e-;
If necessary multiply every half of the reaction with an integer so that the electron removed is the same as the one received. Remember the redox reaction of the electron-transferring reaction.
Sums up the two half-reactions, and lists the remaining margin if the same species are found on the left and right sides;
Adding H + or OH- (according to the requested atmosphere) on the left and right if necessary;
If on one segment there are H + ions and also OH - it is necessary to convert them to H2O molecules;
Ensure the number of atoms and cargo is equal.
What characteristic features of oxidation and reduction reactions, explain please?
BalasHapusCharacteristics - REDUCE REACTIONS AND OXIDATIONS
HapusReaction 2 H2 (g) + O2 (g) à 2 H2O (g)
H2 is oxidized because the oxidation of H atoms in H2 changes from 0 to +1 in H2O
O2 is reduced because the oxidation of the O at O2 is changed from 0 to 2 in H2O
Oxidation reaction (release electrons): 2 H2 (g) à 4 H + (aq) + 4 electrons (e-)
Reduction (electron capture) reaction: O2 (g) + 4 electrons (e-) à 2 O2- (aq)
The oxidation reaction + the reduction reaction = the redoxs reaction
4 H + (aq) + 2 O2- (aq) à 2 H2O (l)
Reaction H2 (g) + Cl2 (g) à 2 HCl (g)
H2 is oxidized because the oxidation of H atoms changes from 0 in H2 to +1 in HCl
Cl2 is reduced because the oxidation of Cl atoms changes from 0 in Cl2 to - 1 in HCl
Oxidation reaction (electron release event): H2 (g) à H + (aq) + 2 e-
Reduction reaction (electron capture event): Cl2 (g) + 2 e- à 2 Cl- (aq)
The oxidation reaction + the reduction reaction = the redoxs reaction
2 H + (aq) + 2 Cl- (aq) à 2 HCl (g)
What is the Oxidation Reduction process ...?
BalasHapusA redox reaction is a reaction involving a reduction reaction and an oxidation reaction. The meaning of the oxidation reaction and the reduction reaction develops in accordance with the development of chemistry. Reduction reactions and oxidation reactions occur in everyday life, such as combustion reactions, vinegar making from alcohol, glucose breaking events in the body, iron filings, and so on.
HapusHow to calculate the oxidation by half the reaction?
BalasHapusREDOXETING EQUIPMENT WITH HOW TO USE REACTIONS
Hapus1. REACTIONS IN ATMOSPHERE ACTIVITIES:
A.tulis equation with the reaction: oxidation and reduction separately
B equate the number of atoms except H and O
C. The deficiency of O plus H2O, which is less H plus H +
D. Let the number of electrons of both reactions multiply by the number of each reaction, then summed and refined.
Example: Apply the following redox reaction to the half-reaction method:
Cu (s) + NO3- (aq) + H + → Cu2 + (aq) + NO (g) + H2O (l)
A. Write the oxidation and reduction
Oxidation: Cu → Cu2 +
Reduction: NO3- → NO2
B.equated the number of atoms other than H and O (already equal)
C. Parties who lack O plus H2O, which is less H plus H +
Cu → Cu2 +
NO3- + 4H + → NO + 2H2O
D. Let the number of electrons of both reactions multiply by the number of each reaction, then summed and refined
Cu → Cu2 + + 2e x3
NO3- + 4H + + 3e → NO + 2H2O x2
3 Cu + 2NO3- + 8H + - 3 Cu2 + + 2NO +4 H2O
Exercise: Apply the following redox reaction with the half-reaction method (acid)
1.CuS + NO3- → Cu2 + + S + NO
2.MnO + PbO2 - → MnO4- + Pb2 +
3.Cr2O72- + C2O42- + H + - Cr3 + + CO2 + H2O
4. MnO4- + H2S + H + --- Mn2 + + S + H2O
2.REAKSI IN BASIC ATMOSPHERE
A.tulis equation with the reaction: oxidation and reduction separately
B equate the number of atoms except H and O
C. The excess O plus H2O, the less Hydrogen plus OH-
D. Let the number of electrons of both reactions multiply by the number of each reaction, then summed and refined.
Example: Apply the following redox reaction to the half-reaction method
Bi2O3 + NaOH + NaClO - NaBiO3 + NaCl + H2O
A.tulis equation with the reaction: oxidation and reduction separately
Oxidation: Bi2O3 - BiO3-
Reductions: ClO- - Cl-
B equate the number of atoms except H and O
Oxidation: Bi2O3 - 2BiO3-
Reduction: ClO- - Cl-
C. The excess O plus H2O, the less Hydrogen plus OH-
Oxidation: Bi2O3 + 6OH- - 2BiO3- + 3H2O
Reduction: ClO- + H2O - Cl- + 2OH-
D. Let the number of electrons of both reactions multiply by the number of each reaction, then summed and refined.
Bi2O3 + 6OH- - 2BiO3- + 3H2O + 4e
ClO- + H2O + 2e - Cl- + 2OH- x 2e
Redox: Bi2O3 + 2OH- + 2 ClO- - 2BiO3- + 2 Cl- + H2O
Reaction Formula: Bi2O3 + 2NaOH + 2NaClO - 2NaBiO3 + 2NaCl + H2O
Exercise: Apply the following redox reaction with the half-reaction method (base)
1. MnO4- + C2O4- --- MnO2 + CO2
2. CrO4- + Fe (OH) 2 --- Cr2O3 + Fe (OH) 3
3. Cl2 + IO3- --- Cl- + IO4-
4. Zn + NO3- --- ZnO2- + NH3
REDOXAGE DEVELOPMENT WITH THE BILOKS METHOD
1. Determine the element undergoing change of oxidation
2.Setarakan elements undergoing changes in oxidation with appropriate coefficients
3. Determine the rise of oxidation and the decrease of oxidation
4, Simulate the number of changes of the oxidation by multiplying with the appropriate coefficients.
5. Align the coefficients of other unsure reactions.
Example: Apply the following redox reaction with the oxidation method.
KMnO4 + H2SO4 + KNO2 → K2SO4 + MnSO4 + H2O + KNO3
Answer:
1. Determine the element undergoing change of oxidation
+7 + 3 + 2 + 5
KMnO4 + H2SO4 + KNO2 - K2SO4 + MnSO4 + H2O + KNO3
2.Setarakan elements undergoing changes in oxidation with appropriate coefficients
(Mn and N, already equivalent)
3. Determine the rise of oxidation and the decrease of oxidation
+7 + 3 + 2 + 5
KMnO4 + H2SO4 + KNO2 - K2SO4 + MnSO4 + H2O + KNO3
Reduction (release 5e)
Oxidation (receive 2e)
4, Simulate the number of changes of the oxidation by multiplying with the appropriate coefficients
+7 + 3 + 2 + 5
KMnO4 + H2SO4 + KNO2 → K2SO4 + MnSO4 + H2O + KNO3
(5e) x 2 (2e) x 5
Once multiplied then the coefficient becomes:
2KMnO4 + H2SO4 + 5KNO2 → K2SO4 + 2MnSO4 + H2O + 5KNO3
5. Align the coefficients of other unsure reactions.
2KMnO4 + 3H2SO4 + 5KNO2 → K2SO4 + 2MnSO4 + 3H2O + 5KNO3