Booklet on the topic of redox reactions. Redox reactions in nature. A range of standard electrode potentials for metals
Department of Analytical and Organic Chemistry
General and inorganic chemistry
Redox reactions
Lecture for 1st year students of the Faculty of Pharmacy
Classification of reactions
2All chemical reactions are possible
divided into 2 groups, one
reactions, oxidation state of atoms
remains unchanged (exchange
reactions), and in other reactions it
changes - these are redox reactions.
Their occurrence is associated with the transition
electrons from some atoms (ions) to
to others.
2The process of electron loss, oxidation, is accompanied by an increase
decreasing negative. Process
acceptance of electrons - reduction,
accompanied by a decrease
positive oxidation state or
increasing negative.
3
3Atoms, molecules or ions,
adding electrons are called
oxidizing agents. Atoms, molecules or ions,
donating electrons are called
restorers.
Oxidation is always accompanied
restoration. Redox reactions are
is the unity of two opposites
processes - oxidation and reduction.
4
Oxidizing agents are:
simple substances whose atoms havehigh electronegativity. This
elements of VII, VI, V groups of main subgroups, of which
the most active are fluorine, oxygen, chlorine.
complex substances whose cations are found in
highest degree of oxidation.
For example: SnCl4, FeCl3, CuSO4.
complex substances whose anions contain an atom
metal or non-metal are in the highest
oxidation states
For example: K2Сr2O7, KMnO4, KNO3, H2SO4.
5
5
Reducers are:
Elements of groups I, II, III of the main subgroups. For example:Na, Zn, H2, Al.
Complex substances whose cations are found in
lowest oxidation state. For example: SnCl2, FeCl2.
Complex substances in which anions reach
extreme negative oxidation state.
For example:
KI, H2S, NH3.
Substances whose ions are in intermediate
oxidation states can be both an oxidizing agent and
reducing agent For example: Na2SO3.
The measure of reducing properties is the value
ionization energy (this is the energy required for
6successive separation of electrons from an atom.)6
Three types of redox reactions.
Three types of redox reactions.- intermolecular,
- intramolecular,
- disproportionation
- In intermolecular ORRs
elements
the oxidizing agent and the reducing agent are in
different substances. For example:
SnCl2 + 2FeCl3 → SnCl4 + 2FeCl2
2 Fe 3+ + e = Fe 2+
- recovery
1 Sn 2+ - 2е = Sn 4+
- oxidation
7Intramolecular
reactions
occur with a change in degree
oxidation of different atoms in the same
same molecule. For example:
2 KClO3 → 2KCl + 3O2
2 Cl5+ + 6e = Cl 3 2O2- - 4e- = O2
8
- recovery
- oxidation
8
Disproportionation reactions
occur with simultaneousdecrease and increase
oxidation states of atoms of one
and the same element.
3HNO2 → HNO3 + 2NO + H2O
2 N 3+ + e = N 2+ - recovery
1 N 3+ - 2е = N 5+ - oxidation
9
Influence of the environment on the nature of the course of OVR
- OVR can occur in differentenvironments: acidic (excess H3O+ - ions),
neutral (H2O) and alkaline (excess
OH- - ions).
Depending on the environment, it may
change the nature of the reaction
between the same substances.
The environment affects the change in degree
oxidation of atoms.
10
Let's look at a few examples.
KMnO4 (potassium permanganate) isstrong oxidizing agent, in strongly acidic
environment is reduced to Mn2+ ions,
in a neutral environment - up to MnO2 (oxide
manganese IV) and in a highly alkaline environment
- to MnO42- (manganate ion).
1.
11Schematically:
Oxidized
form
Restored
form
H3O+
KMnO4
H2O
HE-
Mn 2+ (colorless solution)
MnO2 (brown precipitate)
МnО42- (green solution)
12
Redox duality of hydrogen peroxide
Redoxduality of hydrogen peroxide
Hydrogen peroxide as an oxidizing agent.
BUT
H–O
H+
+
HE-
2H2O
2OH-
Н2О2 + 2Н3О+ + 2е = 4Н2О
H2O2 + 2e = 2OH-
Hydrogen peroxide as a reducing agent.
BUT
H–O
13
H+
O2 + 2H3O+ ; Н2О2 - 2е + 2Н2О = O2 + 2Н3О+
+
OH- O2 + 2H2O;
H2O2 + 2OH- - 2e = O2 + 2H2O
13
Oxidative properties of K2CrO4 and K2Cr2O7
3. Potassium chromate K2CrO4 and potassium dichromateK2Сr2О7 are strong oxidizing agents. In sour and
alkaline solutions of Cr(III) compound and
Cr(VI) exists in different forms.
Oxidized
Restored
form
form
Cr2O72- + H3O+
2 Cr 3+
CrO42- + OHCr(OH)3, CrO2-, 3
14
14
К2Сr2О7
1515
Electron-ion balance method (half-reaction method).
Electron-ion methodbalance (half-reaction method).
Reactions occurring in an acidic environment.
Rule: if a reaction occurs in an acidic
environment, then you can operate with H3O+ ions
(H+) and water molecules. H3O+ ions (H+)
written on that side of the equation
half-reactions where there is excess oxygen,
water molecules write
respectively in the part where oxygen
no or there is a lack of it. Moreover
the amount of H3O+ (H+) is taken twice
greater than the number of excess atoms
16
oxygen. Example 1.
КМnO4 + Na2SO3 + H2SO4 = MnSO4 + K2SO4 + …
OK
sun
Wednesday
Solution
2
MnO4- + 8H+ + 5e = Mn2+ + 4H2O
5
SO32- + H2O – 2e = SO42- + 2H+
2MnO4- +16H++5SO32-+5H2O=2Mn2++8H2O+5SO42- +10H+
2КMnО4 + 5Nа2SO3 + 3Н2SO4 = 2MnSO4 + К2SO4 + 5Na2SO4
+ 3H2O
KMnO4 – oxidizing agent, oxidizing agent; Na2SO3 – reducing agent, oxide
17
Example 2.
Na2Cr2O7 + KBr + H2SO4 = Cr2(SO4)3 + Br2 + …OK.
sun
Wednesday
Solution.
1| Cr2O72- + 14H+ + 6e = 2Cr3+ + 7H2O
3| 2Br- - 2e = Br2
Cr2O72- + 14H+ + 6Br- = 2Cr3+ + 7H2O + 3Br2
Na2Cr2O7 + 6KBr + 7 H2SO4 = Cr2(SO4)3 + 3Br2 +
3 K2SO4 + Na2SO4 + 7H2O
Na2Cr2O7 - oxidizing agent, reduced;
18KBr is a reducing agent, oxidizes.
18
Reactions occurring in an alkaline environment.
Rule: if a reaction occurs inalkaline environment, then you can operate
OH- ions and water molecules. OH ions are written in that part of the equation
half-reactions where there is a deficiency
oxygen, water molecules are written
respectively in the part where
more oxygen. Moreover, for every
the missing oxygen atom is written down
two OH- ions.
19
19
Example 1.
Cr2O3 + KNO3 + KOH = K2CrO4 + KNO2 + …sun
OK.
Wednesday
Solution.
3 | NO3- + H2O + 2e = NO2- + 2OH1 | Cr2O3 + 10 OH- -6e = 2CrO4 2- + 5H2O
3NO3-+3H2O+Cr2O3+10OH-=3NO2-+6OH-+ 2CrO42- + 5 H2O
Cr2O3 + 3KNO3 + 4 KOH = 2 K2CrO4 + 3 KNO2 + 2 H2O
Cr2O3 - reducing agent, oxidizes;
KNO3 is an oxidizing agent and is reduced.
20
Example 2.
КMnО4 + Na2SO3 + KOH = K2MnO4 + Na2SO4 +OK.
sun
Wednesday
Solution.
…
2 | MnO4- + 1e = MnO4 21 | SO32- + 2OH- - 2e = SO4 2- + H2O
2MnO4- + SO3
2-
+ 2 OH- = 2 MnO4 2- + SO4 2- + H2O
2 KMnO4 + Na2SO3 + 2 KOH = K2MnO4 + Na2SO4 + H2O
21
Reactions occurring in a neutral environment.
22Rule: if a reaction occurs in a neutral environment,
should operate only with water molecules. Moreover
excess oxygen in the oxidizing agent is bound by molecules
water, due to H3O+ (H+) ions, for each excess atom
oxygen, one molecule of water is consumed, which
placed on the left side of the half-reaction equation, in solution
OH- ions accumulate and are placed on the right side
half-reaction equations. Lack of oxygen
the reducing agent is replenished from water molecules at the expense of OH ions, for each missing oxygen atom it is consumed
one molecule of water, which is placed on the left side
half-reaction equations, ions accumulate in solution
H3O+ (H+) and they are placed on the right side of the equation
half-reactions.
22
Example 1.
KMnO4 + Na2SO3 + H2O = MnO2 + Na2SO4 + …OK.
sun
Solution.
2 | MnO4- + 2H2O +3e = MnO2 + 4 OH3 | SO32- + H2O -2e = SO42- + 2 H+
2 MnO4-+4H2O+3SO32-+3H2O=2MnO2 +8OH- + 6H++ 3SO42
2KMnO4 + 3Na2SO3 + H2O = 2 MnO2 + 3 Na2SO4 + 2 KOH
KMnO4–oxidizing agent, oxidizing agent;
23
Na2SO3–reducing agent, oxide
23
Example 2.
MnSO4 + KMnO4 + H2O = MnO2 + K2SO4 + …sun
OK.
Wednesday
Solution.
2 | MnO4- + 2 H2O + 3e = MnO2 + 4 OH3 | Mn2+ + 2 H2O - 2e = MnO2 + 4 H+
2MnO4- +4H2O+3Mn2++6H2O=2MnO2+8OH-+3MnO2+12H+
3MnSO4+2KMnO4+2H2O=5MnO2+K2SO4+2H2SO4
MnSO4 - reducing agent, oxidizes;
KMnO4 is an oxidizing agent and is reduced.
24Theory of origin
equilibrium electrode and
redox
potentials
Determining direction
redox
process
Mechanism of occurrence of electrode potential
Me Men+ + n eWhen immersing metal in water...
Ме + m Н2О Мen+(Н2О)m+n eМen+(Н2О)m+ne Ме + m Н2О
Me +m H2O Men+ (H2O)m+
ne
The potential established under conditions of equilibrium of the electrode reaction is called the equilibrium electrode potential.
If a metal is immersed in a solution of its salt, then the processes occurring at the metal-solution boundary will be similar.
ZnCu
To compare electrode
potentials of various
metals are chosen
standard conditions:
temperature - 250 C, pressure
- 101.3 kPa, activity
ion of the same name - 1 mol/l.
Potential difference,
arising between
metal and solution in
such conditions are called
standard electrode
potential.
Standard electrode potential
The standard electrode potential (E0) is the EMF of a galvanic cell composed of a given electrode and a reference electrode. In quality
Standard electrode potential (E0) is emfgalvanic cell composed of a given electrode and
reference electrode. As a reference electrode
use a normal hydrogen electrode (NHE):
H2 2H+ + 2e
Pt(H2) | 2H+
H2
Platinum electrode,
platinum plated
powder, in water
acid solution with
c(H+) = 1 mol/l and
washed
hydrogen gas
(p = 1 atm)
at 298 K
A range of standard electrode potentials for metals
LiBa
Na
Zn
Fe
Pb
-3,04
-2,90
-2,71
-0,76
-0,44
-0,13
Li+
Ba2+
Na+
Zn2+
Fe2+
Pb2+
H2
0
2H+
Cu
Ag
Au
+0,34
+0,80
+1,5
Cu2+
Ag+
Au3+ The magnitude of the potential in real conditions
calculated using the Nernst equation:
E Me n / Me E
0
Me n/Me
RT
ln a Me n
nF
Transition factor from ln to lg
RT
at 20 C:
2,303 0,058
F
RT
0
at 25 C:
2,303 0,059
F
0
E Me n / Me E
0
Me n/Me
0,059
lg a Men
n E
0
Men/Me
- standard electrode potential,
measured under standard conditions:
T 298 K
aMen 1 mol/l
F 96500 C/mol
J
R 8.314
mole K
If the potential of the hydrogen electrode is known, the pH of the solution can be calculated:
E2 H/H E2
0
2H/H2
0.059 lg aH
=0
lg a H pH
pH
E2 H/H 0
2
0,059
Silver chloride electrode (SSE)
Ag, AgCl | KClElectrode of the second kind
AgCl
KCl
Ag
When immersed in solution
salts of the same name
anion its potential
will be determined
activity of the anion in
solution. Ag Ag+ + e
(1)
Ks
AgCl Ag+ + Cl-
(2)
KCl K+ + Cl-
(3)
The higher the KCl concentration, the higher the Cl- concentration, the more
the solubility of AgCl is lower and the concentration of Ag+ is lower. in these
conditions is very small and practically undetectable. Potential,
arising at the Ag|Ag+ boundary is determined by the Nernst equation:
E h.s. E
0
Ag
Ag
RT
ln a Ag
nF K s a Ag aCl ; a Ag
Ex.s. E
Ex.s. E
0
Ag
0
Ag
Ag
Ag
Ks
aCl
RT K s
ln
nF aCl
RT
RT
ln K s
ln aCl
nF
F
0,222
E h.s. 0.222 0.059 lg a Cl
E h.s.
E h.s.Silver chloride potential value
electrode at different concentrations of aqueous
KCl solution at T = 298 K
Galvanic cells
IsometallicBimetallic
Galvanic cell (bimetallic)
Anode: Zn - 2e = Zn2+Cathode: Cu2++2e = Cu
Zn + Cu2+ = Zn2+ + Cu
Interface
-Zn|ZnSO4||CuSO4 |Cu +
Eliminated diffusion
potential
solution ZnSO4
CuSO4 solution
A measure of the performance of a GE element is the EMF or potential difference of the electrodes:
Unified State Exam Ekatoda Eanoda;Unified State Examination E
0
0
cat.
E
0
if E0Zn 2 / Zn 0.76 B; ECu
0,34,
2
/Cu
then, E
0
GE
0.34 (0.76) 1.1 B
0,059
E Zn 2 / Zn E
lg a Zn 2
n
0,059
0
ECu2 / Cu ECu2 / Cu
lg a Cu2
n
0
Zn2/Zn
E GE
0.059 a Cu2
1,1
lg
n
a Zn 2
0
an.
Concentration galvanic cell (isometallic)
Anode: Zn Zn2+(0.1n) +2eCathode: Zn2+(1n) +2e Zn
Zn2+(1n) Zn2+(0.1n)
- Zn|Zn2+(0.1n)||Zn2+(1n)|Zn +
p-p ZnSO4 0.1 n (a1)
p-p ZnSO4 1 n (a2)
a1< a 2E Zn 2 / Zn E
0
Zn2/Zn
E Zn 2 / Zn E
0
Zn2/Zn
E GE
0,059
log a Zn 2 (a 2)
n
0,059
log a Zn 2 (a1)
n
0.059 a 2
lg
n
a1
Redox potentials
PtFe 2+(solution) Fe 3+(solution)+e (Pt pl-ka)
Red Ox + ne
Red - restored form
Ox – oxidized form
Nernst equation:
FeCl2, FeCl3
E ok. f./v.f. E
0
OK. f./v.f.
RT Acid. f-ma
ln
nF
Svost. f-ma
Standard RF potential
Walter Friedrich Hermann Nernst (1864-1941)
RH potential depends on:
temperaturenature of the oxidizing agent and reducing agent
concentrations of oxidized and
restored forms
pH of the environment
Standard RF potential
EMF of HE, composed of redoxsystems,
containing
oxidized and reduced forms in
concentrations of 1 mol/l and NVE – yes
standard OM potential of a given OB
systems If we compose the GE from MnO4-/Mn2+ and (Pt),H2|2H+,
then the standard OF potential = +1.51 V.
MnO4- + 8H+ +5e Mn2+ + 4H2O
a(MnO4-)= a(Mn2+)=1 mol/l
a(H+)= 1 mol/l
In real conditions, the calculation of the RH potential of the MnO4-/Mn2+ system is carried out using the Nernst equation:
E MnO / Mn 24
4
8
RT[MnO][H]
1,51
ln
2
5F
[Mn] The higher the standard RH
potential of the system, the more
its oxidative effects are expressed
properties under standard conditions.
For example,
MnO4-/Mn2+
Fe3+/Fe2+
Sn4+/Sn2+
E0= 1.51 V
E0= 0.77 V
E0= 0.15 V
Criteria for spontaneous occurrence of OM reactions
G 0G reactions Gprod. Gref. V.
G Apoloznaya Ael.
Ael. q E
qnF
Portable email
charge
Email work on
electron transfer
Potential difference
between electrodes
Number of electrons transferred to
elementary act of OVR
E Ok la Ev la
G nF E
if G 0 then E 0
Example:
3Co/Co
2
E
0
(ok., v.)
1.84 V
Fe 3 / Fe 2 E (0 approx., v.) 0.77 V
Co
3
oxidizer
Fe
2
2
reducing agent
HER
E 0, therefore, the reaction proceeds
0
OK.
E
Co Fe
3
0
voss.
1,84 0,77 1,07
spontaneously from left to right
Depth of OM reactions
A B C DK x. R.
[D]
; G 0 RT ln K x. R.
[A][B]
0
G
nF E
RT ln K x. R. nF E nF (Eok0 l I Ev0 l I)
nF (Eok0 l I Ev0 l I)
ln K x. R.
RT
ln K x. R. the greater, the greater the difference Eok0 l i Eв0 l i,
and K x. R. assesses the depth of chemical flow. reactions
Redox GE
Redox GE2KI + 2FeCl3 I2 + 2FeCl2+2КCl
2KI + 2FeCl3 I2 + 2FeCl2+2КCle
Pt
Pt
e
KI
2I- -2e I2
I2 | 2I-
e
FeCl3
Fe3++e Fe2+
Fe3+ | Fe2+
When the circuit is closed in
the left half-element goes
oxidation process - giving away electrons
platinum, turn into
I2, resulting plate
charged conditionally
negative.
In the right half-element
Fe3+ takes electrons from
records turning into
Fe3+, the plate is charging
conditionally positive.
The system strives
equalize the charges on
records due
moving electrons
via an external circuit.
Ion selective electrodes
Glass electrode
R(Na+, Li+) + H+ R(H+) + Na+, Li+Glass
electrode body
membrane
solution
membrane
solution
Ag AgCl, 0.1 M HCl glass H+, solution
1
2
3
glass = 1+ 2+ 3
Internal solution
0.1 M HCl
1- internal silver chloride potential
electrode (const)
2- inner surface potential
glass membrane (const)
HSE
3 - potential of the outer surface
glass membrane (variable)
1+ 2 = K
glass = K + 0.059 lg a(H+) or
Electrode glass
(membrane)
glass = K - 0.059 pH
Determination of pH in a laboratory workshop
To the measuringdevice
EMF of the presented Ecircuit circuit:
E chain = E h.s. - Eating.
Echain = E h.s. – K + 0.059рН
pH
E chains E h.s. TO
0,059
E chain const
Oxidation is the process of losing electrons by an atom, molecule or ion. The atom turns into a positively charged ion: Zn 0 – 2e Zn 2+ the negatively charged ion becomes a neutral atom: 2Cl - -2e Cl 2 0 S 2- -2e S 0 The size of the positively charged ion (atom) increases according to the number of electrons given up: Fe 2 + -1e Fe 3+ Mn +2 -2e Mn +4
Reduction is the process of gaining electrons by an atom, molecule or ion. The atom turns into a negatively charged ion S 0 + 2e S 2 Br 0 + e Br The size of a positively charged ion (atom) decreases according to the number of attached electrons: Mn e Mn +2 S e S +4 or it can transform into a neutral atom: H + + e H 0 Cu e Cu 0
Reducing agents are atoms, molecules or ions that donate electrons. They are oxidized during the redox process. Typical reducing agents: metal atoms with large atomic radii (I-A, II-A groups), as well as Fe, Al, Zn, simple non-metal substances: hydrogen, carbon, boron; negatively charged ions: Cl, Br, I, S 2, N 3. Fluoride ions F are not reducing agents. Metal ions in the lowest concentration: Fe 2+, Cu +, Mn 2+, Cr 3+; complex ions and molecules containing atoms with intermediate residues: SO 3 2, NO 2; CO, MnO 2, etc.
Oxidizing agents are atoms, molecules or ions that gain electrons. They are reduced in the process of ORR. Typical oxidizing agents: atoms of non-metal groups VII-A, VI-A, V-A in the composition of simple substances, metal ions in the highest concentration: Cu 2+, Fe 3+, Ag + ... complex ions and molecules containing atoms with the highest and highest d.o.: SO 4 2, NO 3, MnO 4, СlО 3, Cr 2 O 7 2-, SO 3, MnO 2, etc.
Sulfur oxidation states: -2.0,+4,+6 H 2 S -2 - reducing agent 2H 2 S+3O 2 =2H 2 O+2SO 2 S 0,S +4 O 2 - oxidizing agent and reducing agent S+O 2 =SO 2 2SO 2 +O 2 =2SO 3 (reducing agent) S+2Na=Na 2 S SO 2 +2H 2 S=3S+2H 2 O (oxidizing agent) H 2 S +6 O 4 - oxidizing agent Cu+2H 2 SO 4 =CuSO 4 +SO 2 +2H 2 O
Determination of oxidation states of atoms of chemical elements С.о. chemical atoms in the composition of a simple substance = 0 Algebraic sum of s.o. of all elements in the ion is equal to the charge of the ion Algebraic sum s.o. of all elements in the composition of a complex substance is 0. K +1 Mn +7 O x+4(-2)=0
Classification of redox reactions Intermolecular oxidation reactions 2Al 0 + 3Cl 2 0 2Al +3 Cl 3 -1 Intramolecular oxidation reactions 2KCl +5 O KCl O 2 0 Reactions of disproportionation, dismutation (auto-oxidation-self-reduction): 3Cl KOH (hor.) KCl + 5 O 3 +5KCl -1 +3H 2 O 2N +4 O 2 + H 2 O HN +3 O 2 + HN +5 O 3
This is useful to know. The oxidation states of the elements in the composition of the salt anion are the same as in the acid, for example: (NH 4) 2 Cr 2 +6 O 7 and H 2 Cr 2 +6 O 7 The oxidation state of oxygen in peroxides is -1 Oxidation state sulfur in some sulfides is -1, for example: FeS 2 Fluorine is the only non-metal that does not have a positive oxidation state in compounds. In compounds NH 3, CH 4 and others, the sign of the electropositive element hydrogen is in second place
Oxidative properties of concentrated sulfuric acid Products of sulfur reduction: H 2 SO 4 + very active. metal (Mg, Li, Na...) H 2 S H 2 SO 4 + act. metal (Mn, Fe, Zn...) S H 2 SO 4 + inactive. metal (Cu, Ag, Sb...) SO 2 H 2 SO 4 + HBr SO 2 H 2 SO 4 + non-metals (C, P, S...) SO 2 Note: it is often possible to form a mixture of these products in different proportions
Hydrogen peroxide in redox reactions Solution medium Oxidation (H 2 O 2 -reducing agent) Reduction (H 2 O 2 -oxidizing agent) acidic H 2 O 2 -2eO 2 + 2H + (O – 2eO 2 0) H 2 O 2 + 2H + +2e2H 2 O (O e2O - 2) alkaline H 2 O 2 +2OH -O 2 +2H 2 O (O – 2eO 2 0) H 2 O 2 +2e2OH - (O e2O - 2) neutral H 2 O 2 - 2eO 2 + 2H + (O – 2eO 2 0) H 2 O 2 + 2e2OH - (O e2O - 2)
Nitric acid in redox reactions Nitrogen reduction products: Concentrated HNO 3: N +5 +1e N +4 (NO 2) (Ni, Cu, Ag, Hg; C, S, P, As, Se); passivates Fe, Al, Cr Dilute HNO 3: N +5 +3e N +2 (NO) (Metals in EHRNM Al...Cu; non-metals S, P, As, Se) Dilute HNO 3: N +5 +4e N +1 (N 2 O) Ca, Mg, Zn Dilute HNO 3: N +5 +5e N 0 (N 2) Very dilute: N e N -3 (NH 4 NO 3) (active metals in EHRNM to Al)
The Meaning of OVR OVR is extremely common. They are associated with metabolic processes in living organisms, respiration, decay, fermentation, photosynthesis. OVRs ensure the circulation of substances in nature. They can be observed during fuel combustion, corrosion and metal smelting. With their help, alkalis, acids and other valuable chemicals are obtained. OVRs underlie the conversion of the energy of interacting chemicals into electrical energy in galvanic batteries.
Redox reactions are the most common and play a large role in nature. They are the basis of life on Earth, since they are associated with respiration and metabolism in living organisms, decay and fermentation, photosynthesis in the green parts of plants and the nervous activity of humans and animals.
Respiration In the process of respiration, carbohydrates, fats and proteins, in reactions of biological oxidation and gradual restructuring of the organic skeleton, give up hydrogen atoms to form reduced forms. The latter, when oxidized in the respiratory chain, release energy, which is accumulated in active form in the coupled reactions of ATP synthesis.
Chemical corrosion of metals After the destruction of the metal bond, the metal atoms and the atoms that make up the oxidizing agents form a chemical bond. This type of corrosion is inherent in media that are not capable of conducting electric current - these are gases and liquid non-electrolytes.
Description of the presentation by individual slides:
1 slide
Slide description:
Completed by: Chemistry teacher Baimukhametova Batila Turginbaevna Oxidation-reduction reactions
2 slide
Slide description:
The motto of the lesson is “Someone loses, and someone finds...” By working, you will do everything for your loved ones and for yourself, and if there is no success during your work, failure is not a problem, try again. D. I. Mendeleev.
3 slide
Slide description:
4 slide
Slide description:
Lesson topic: “Redox reactions” Purpose: To get acquainted with redox reactions and find out what is the difference between metabolic reactions and redox reactions. Learn to identify oxidizing and reducing agents in reactions. Learn to draw diagrams of the processes of giving and receiving electrons. Learn about the most important redox reactions found in nature.
5 slide
Slide description:
Perhaps these electrons are Worlds where there are five continents, Arts, knowledge, wars, thrones And the memory of forty centuries! Also, perhaps, every atom is a Universe with a hundred planets; There is everything that is here, in a compressed volume, But also what is not here. V. Brusosova.
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What is oxidation state? The oxidation state is the nominal charge of an atom of a chemical element in a compound, calculated based on the assumption that all compounds consist only of ions. The oxidation state can be positive, negative or zero, depending on the nature of the compounds involved. Some elements have constant oxidation states, others have variable ones. Elements with a constant positive oxidation state include - alkali metals: Li+1, Na+1, K+1, Rb+1, Cs+1, Fr+1, the following elements of group II of the periodic table: Be+2, Mg+2, Ca+2, Sr+2, Ba+2, Ra+2, Zn+2, as well as element III A of group - A1+3 and some others. Metals in compounds always have a positive oxidation state. Of non-metals, F has a constant negative oxidation state (-1). In simple substances formed by atoms of metals or non-metals, the oxidation states of elements are zero, for example: Na°, Al°, Fe°, H2, O2, F2, Cl2, Br2. Hydrogen is characterized by oxidation states: +1 (H20), -1 (NaH). Oxygen is characterized by oxidation states: -2 (H20), -1 (H2O2), +2 (OF2).
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The most important reducing agents and oxidizing agents Reducing agents: Oxidizing agents: Elementary metals Hydrogen Carbon Carbon monoxide (II) (CO) Hydrogen sulfide (H2S) Sulfur oxide (IV) (SO2) Sulfurous acid H2SO3 and its salts Hydrohalic acids and their salts Metal cations in intermediate states oxidation: SnCl2, FeCl2, MnSO4, Cr2(SO4)3 Nitrous acid HNO2 Ammonia NH3 Nitrogen oxide (II) (NO) Halogens Potassium permanganate (KMnO4) Potassium manganate (K2MnO4) Manganese oxide (IV) (MnO2) Potassium dichromate (K2Cr2O7) Nitric acid (HNO3) Sulfuric acid (conc.H2SO4) Copper(II) oxide (CuO) Lead(IV) oxide (PbO2) Hydrogen peroxide (H2O2) Iron(III) chloride (FeCl3) Organic nitro compounds
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The degree of oxidation of manganese in the potassium permanganate compound KMnO4. 1. Oxidation state of potassium +1, oxygen -2. 2. Let's count the number of negative charges: 4 (-2) = - 8 3. The number of positive charges on manganese is 1. 4. We make up the following equation: (+1) + x+ (-2)*4 =0 1+ x - 8=0 X = 8 - 1 = 7 X= +7 +7 is the oxidation state of manganese in potassium permanganate.
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Rules for determining oxidation states 1. The oxidation state of an element in a simple substance is 0. For example: Ca, H2, Cl2, Na. 2. The oxidation state of fluorine in all compounds except F2 is – 1. Example: S+6F6-1 3. The oxidation state of oxygen in all compounds except O2, O3, F2-1O+2 and peroxide compounds Na2+1 O- 12; H2+1O-12 is equal to –2 Examples: Na2O-2, BaO-2, CO2-2. 4. The oxidation state of hydrogen is +1 if the compounds contain at least one non-metal, -1 in compounds with metals (hydrides) 5. The oxidation state of O in H2 Examples: C-4H4+1 Ba+2H2-1 H2 The oxidation state of metals always positive (except for simple substances). The oxidation state of metals of the main subgroups is always equal to the group number. The degree of oxidation of side subgroups can take different values. Examples: Na+ Cl-, Al2+3O3-2, Cr2+3 O3-2, Cr+2O-2. 6. The maximum positive oxidation state is equal to the group number (exceptions: Cu+2, Au+3). The minimum oxidation state is equal to the group number minus eight. Examples: H+1N+5O-23, N-3H+13. 7. The sum of the oxidation states of atoms in a molecule (ion) is equal to 0 (the charge of the ion).
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Laboratory work Safety rules. Experiment 1. Carry out a chemical reaction between solutions of copper (II) sulfate and sodium hydroxide. Experiment 2. 1. Place an iron nail in a solution of copper (II) sulfate. 2. Make up equations for chemical reactions. 3. Determine the type of each chemical reaction. 4. Determine the oxidation state of the atom of each chemical element before and after the reaction. 5. Think about how these reactions differ?
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Answers: Cu+2S+6O4-2 +2Na +1O-2H+1Cu +2(O -2H+1)2+Na2 +1S +6O4-2 – exchange reaction Cu+2S+6O4-2 + Fe0 Fe+2 S+6O4 -2+Сu0 – substitution reaction Reaction No. 2 differs from reaction No. 1 in that in this case the oxidation state of the atoms of chemical elements changes before and after the reaction. Note this important difference between the two reactions. The second reaction is OVR. Let us underline in the reaction equation the symbols of the chemical elements that changed the oxidation state. Let's write them down and indicate what the atoms did with their electrons (Give or receive?), i.e. electron transitions. Cu+2 + 2 e- Cu0 – oxidizing agent, reduced Fe0 - 2 e- Fe+2 – reducing agent, oxidized
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Classification of redox reactions 1. Intermolecular redox reactions The oxidizing agent and the reducing agent are found in different substances; the exchange of electrons in these reactions occurs between different atoms or molecules: 2Ca0 + O20 → 2 Ca+2O-2 Ca - reducing agent; O2 - oxidizing agent Cu+2O + C+2O → Cu0 + C+4O2 CO - reducing agent; CuO – oxidizing agent Zn0 + 2HCl → Zn+2Cl2 + H20 Zn – reducing agent; HСl - oxidizing agent Mn+4O2 + 2KI-1 + 2H2SO4 → I20 + K2SO4 + Mn+2SO4 + 2H2O KI - reducing agent; MnO2 is an oxidizing agent.
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2. Intramolecular redox reactions In intramolecular reactions, the oxidizing agent and the reducing agent are in the same molecule. Intramolecular reactions usually occur during the thermal decomposition of substances containing an oxidizing agent and a reducing agent. 4Na2Cr2O7 → 4Na2CrO4 + 2Cr2O3 + 3O2 Cr+6- oxidizing agent; O-2 - reducing agent
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3. Disproportionation reactions Redox reactions in which one element simultaneously increases and decreases the oxidation state. 3S + 6NaOH → Na2SO3 + 2Na2S + 3H2O Sulfur in oxidation state 0 is both an oxidizing agent and a reducing agent. 4. Comporportionation reactions Redox reactions in which atoms of one element in different oxidation states acquire one oxidation state as a result of the reaction. 5NaBr + NaBrO3 + 3H2SO4 → 3Na2SO4 + 3Br2 + 3H2O Br+5 – oxidizing agent; Br-1 – reducing agent
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Algorithm for composing equations of redox reactions using the electronic balance method 1. Write down the reaction scheme KMnO4+KI+H2SO4→MnSO4+ I2+K2SO4+H2O 2. Enter the oxidation states of the atoms of the elements for which it changes KMn+7O4+ KI-+ H2SO4→ Mn+2SO4+ I20+ K2SO4+ H2O 3. Elements that change oxidation states are identified and the number of electrons accepted by the oxidizing agent and donated by the reducing agent is determined. Mn+7 + 5ē → Mn+2 2I-1 - 2ē → I20 4. Equalize the number of accepted and donated electrons, thereby establishing coefficients for compounds that contain elements that change the oxidation state. Mn+7 + 5ē → Mn+22 2I-1 - 2ē → I205 2Mn+7 + 10I-1 → 2Mn+2 + 5I20 5. Select coefficients for all other participants in the reaction. 2KMnO4+10KI+8H2SO4→2MnSO4+5I2+6K2SO4+ 8H2O
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Electronic balance is a method of finding coefficients in the equations of redox reactions, which considers the exchange of electrons between atoms of elements that change their oxidation state. The number of electrons donated by the reducing agent is equal to the number of electrons accepted by the oxidizing agent.
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Oxidation-reduction reactions are reactions in which oxidation and reduction processes occur simultaneously and, as a rule, the oxidation states of elements change. Let's consider the process using the example of the interaction of zinc with dilute sulfuric acid:
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Let's remember: 1. Oxidation-reduction reactions are reactions in which electrons transfer from one atom, molecule or ion to another. 2. Oxidation is the process of losing electrons, and the degree of oxidation increases. 3. Reduction is the process of adding electrons, the oxidation state decreases. 4. Atoms, molecules or ions that donate electrons are oxidized; are reducing agents. 5.Atoms, ions or molecules that accept electrons are reduced; are oxidizing agents. 6. Oxidation is always accompanied by reduction; reduction is associated with oxidation. 7. Oxidation-reduction reactions are the unity of two opposite processes: oxidation and reduction.
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Recovery reactions. Classification of OVR. Lesson objectives: 1. educational - to systematize students’ knowledge about the classification of chemical reactions in the light of electronic theory; - teach to explain the basic concepts of OVR; - give a classification of ODD 2. developing - develop the ability to observe, draw conclusions; - continue the development of logical thinking, ability to analyze and compare; 3. educational - to form the scientific worldview of students, improve work skills; -develop the ability to listen to each other, analyze the situation, improve the culture of interpersonal communication
Basic concepts: redox reactions, oxidizer, reducer, oxidation processes, reduction reactions, intermolecular intramolecular disproportionations Equipment: PSHE D. I. Mendeleeva
When certain types of chemical bonds are formed, the process of adding electrons to an atom or giving them away occurs, so the formation of common electron pairs or charged particles - cations and anions - is possible. The reduction process is the process of accepting electrons by an atom (particle) +n. As a result, a decrease in the oxidation state is observed. during restoration - s.o. decreases For example +2 Task. Write the process of copper reduction () The oxidation process is the process of giving up electrons by an atom (particle) n As a result, an increase in the degree of oxidation is observed. during oxidation - s.o. increases For example Task. Write the oxidation process of aluminum ()
Oxidizing agent and reducing agent. The ability to determine the functions of a substance/particle (oxidizing or reducing) by s.o. element Reducing agent - particle, atom, molecule that donates electrons (electron donor). The reducing agent always increases the d.o. An oxidizing agent is a particle, atom, molecule that accepts electrons (electron recipient). The oxidizing agent always lowers the s.o. 1. So, if in a compound the element is in the minimum r.o., like sulfur in (-2 is the minimum r.o. of sulfur / group number -8 /), then the compound acts as a reducing agent. For example: ... 2. If in the compound the element is at maximum s. o., like sulfur in - the compound acts as an oxidizing agent. For example: H ...
The most important Oxidizing and Reducing Agents Oxidizing agents: K H And also some simple substances Reducing agents H H And also some simple substances Metals, CO, C Task: Find oxidizing and reducing agents HN S CuO among the proposed compounds
All chemical reactions that occur with a change in d.o. elements are called redox.
Intermolecular ORR - exchange of electrons occurs between different atoms (molecules, ions) - the oxidizing agent and the reducing agent are in different molecules: + = Reactions of intramolecular oxidation and reduction - the oxidizing agent and the reducing agent are in the same substance (molecule, particle) = + 2 Reactions disproportionation (dismutation) - reactions in which the same element acts both as an oxidizing agent and as a reducing agent, and as a result of the reaction, compounds are formed that contain the same chemical element in different d.o. K _________________________________________________________________ Assignment What type of OVR is the reaction: N + + HN
PIN 2 𝑆+𝑆 = 3S + 2 O Is the reaction ORR? Determine the oxidation state of elements Find an oxidizing agent, a reducing agent Determine the type of ORR HOMEWORK 1. p. 11, learn 2. write out ORR of all types from the text (two examples each)
