Transcript Redox reaction( ), Oxidation number and Electrolysis( 反應

Redox reaction(氧化還原
反應), Oxidation number
and Electrolysis(電解)
International Junior
Science Olympiad (IJSO)
Dr. Yu-San Cheung
[email protected]
Department of Chemistry
The Chinese University of Hong Kong
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Oxidation(氧化) and Reduction(還原
Oxidation:
Reaction of an element or a compound with O2 to give an oxide
e.g., 4 Na(s) + O2(g)  2 Na2O(s)
2 H2(g) + O2(g)  2 H2O(l)
(sodium oxide)
(“hydrogen oxide”?)
Reduction:
Reverse of oxidation
e.g., 2 CuO(s)  2 Cu(s) + O2 (g)
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Oxidation(氧化) and Reduction(還原)
Reduction:
Reverse of oxidation
e.g., 2 CuO(s)  2 Cu(s) + O2 (g)
CuO(s) can also be converted to Cu(s) with hydrogen:
e.g., CuO(s) + H2(g)  Cu(s) + H2O(l)
Therefore, we may also say that:
Reduction is a reaction of an element or a compound with H2.
e.g., H2(g) + Cl2(g)  2 HCl(g)
[Cl2(g) is reduced]
Here, we consider that H2 and O2 have opposite properties.
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Oxidation (氧化) and Electron Transfer(電子轉移)
In the following reaction, copper is oxidized and loses electrons to
have a positive charge:
e.g., 2 Cu(s) + O2(g)  2 CuO(s)
Therefore, we may also say that:
In an oxidation, an element or a compound loses electron(s) to have
a positive charge.
Similarly: in an reduction, an element or a compound gains
electron(s) to have a positive charge.
e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s)
Which ion is oxidized? __________
Which ion is reduced? __________
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Oxidation Number(氧化數)
Consider: H+(aq) + OH–(aq)  H2O(l)
Is OH–(aq) oxidized or reduced?
(Gaining “hydrogen” but with charge increased)
Oxidation Number/Oxidation State:
Numbers assigned to elements, ions, and compounds to help us to
tell whether they are oxidized or reduced in a reaction.
It is also assigned to individual atoms in ions and molecules.
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Rules of Assigning Oxidation
Number (O.N.)
From high to low priority:
•
O.N. for atoms in elements: 0
Overall O.N. for neutral molecules and compounds: 0
e.g., H2(g), Na(s), NaCl(s), CO2(g)
•
Overall O.N. for ions: equal to the charges
e.g., Na+ & NH4+(O.N.= +1), SO4 2–(O.N.= –2)
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Rules of Assigning Oxidation
Number (con’t)
For individual atoms in neutral molecules and compounds:
•
F: –1
•
Group 1A metals (Li, Na, K, …): +1
•
Group 2A metals (Be, Mg, Ca, …): +2
•
H: +1 (except in metal hydride, see below)
•
O: –2
•
Cl: –1
•
Br, I: –1;
N, P: –3;
S: –2
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Examples:
SO4 2–: overall: –2
SO2: overall: 0
SF2: overall: 0
H2S: overall: 0
O.N.
O.N.
O.N.
O.N.
for
for
for
for
O: –2
O: –2
F: –1
H: +1




O.N.
O.N.
O.N.
O.N.
for
for
for
for
S
S
S
S
=
=
=
=
+6
+4
+2
–2
PCl5: overall: 0
PO4 3–: overall: –3
PO3 3–: overall: –3
PH3: overall: 0
O.N.
O.N.
O.N.
O.N.
for
for
for
for
Cl: –1
O: –2
O: –2
H: +1




O.N.
O.N.
O.N.
O.N.
for
for
for
for
P
P
P
S
=
=
=
=
+5
+5
+3
–3
H2O: overall: 0
O.N. for O: –2  O.N. for H = +1
CaH2: containing Ca2+ and H–
 O.N. for H = –1
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Examples:
ClO4–: overall: –1
ClO3–: overall: –1
ClO2–: overall: –1
ClO–: overall: –1
O.N.
O.N.
O.N.
O.N.
for
for
for
for
O:
O:
O:
O:
–2
–2
–2
–2




O.N.
O.N.
O.N.
O.N.
for
for
for
for
Cl
Cl
Cl
Cl
=
=
=
=
+7
+5
+3
+1
MnO4–: overall: –1
O.N. for O: –2  O.N. for Mn = +7
Note: Mn has bonding with oxygen atoms. It does not exist
as Mn7+ ion.
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Oxidation Number(氧化數)for Oxygen
O.N. for O is usually –2, except (including but not limited to):
•
In fluorine-oxygen compounds
e.g.,: OF2
Neutral molecule: overall O.N. = 0
Then assign O.N. = –1 for F
So, O.N. = +2 for O
How about FIO3?
O.N. = –1 for F, O.N. = –2 for O, finally we have
+7 for I
•
In some ions:
Peroxide:
O2 2–
Superoxide:
O2–
O.N. for O atom: –1
O.N. for O atom: –1/2
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Redox reaction(氧化還原反應)
(Oxidation-Reduction)
Oxidation of an atom: O.N. increases
Reduction of an atom: O.N. decreases
In a reaction, O.N. increase of an atom must be accompanied with
O.N. decrease of another atom
e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag (s)
O.N. change:
Fe: +2  +3
Ag: +1  0
We say that: Fe2+(aq) is oxidized to Fe3+(aq)
Ag+(aq) is reduced to Ag(s)
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O.N. and Electron Transfer(電子轉移)
e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag (s)
Ag+: gaining electron, being reduced
Fe3+: losing electron, being oxidized
In general:
- gaining electron, being reduced
- losing electron, being oxidized
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Oxidizing Agent(氧化劑) and
Reducing Agent(還原劑)
Oxidizing agent (oxidant):
- oxidizing another species
- being reduced (O.N. decreased)
Reducing agent (reductant) :
- reducing another species
- being oxidized (O.N. increased)
e.g., Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag (s)
Oxidizing Agent: Ag+(aq)
Reducing Agent: Fe2+(aq)
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Common Oxidizing Agents (氧化劑)
Ions of metal at low position in the reactivity series
(e.g., Ag+  Ag)
MnO4– & Cr2O7– in acidic medium:
MnO4–  Mn2+;
Cr2O7–  Cr3+
Conc. HNO3(aq): NO3–  NO2(g)
Conc. H2SO4(aq): SO42–  SO2(g)
Cl2(g)  2 Cl–;
Br2(l)  2 Br–
O2(g)  O2–; H2O2(aq)  H2O(l)
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Common Reducing Agents(還原劑)
Metal at high position in the reactivity series (e.g., Na  Na+)
C(s)  CO(g) or CO2(g)
CO(g)  CO2(g)
SO32–  SO42–
Fe2+  Fe3+
2I–  I2
H2(g)  2 H+
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Remarks:
Some chemical can act as both oxidizing agent and reducing agent.
e.g.,
SO2(g)  S(s)
SO2(g)  SO42–(aq)
oxidation / reduction
oxidation / reduction
Species with high O.N. atom has higher chance to be an oxidizing
agent.
Similarly, species with low O.N. atom has higher chance to be a
reducing agent.
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Acid-Base Reaction & Redox
Reaction
H+(aq) + OH–(aq)  H2O(l)
O.N.:
H
O
+1
+1
–2
+1
–2
No atom has its O.N. change.
In general, acid-base reaction is NOT a redox reaction.
Exercise
Verify the conclusion for the following:
HCl(aq) + NaHCO3(aq)  NaCl(aq) + CO2(g) + H2O(l)
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Concentration(濃度)Effect on
Oxidizing/Reducing Power
Example
Dilute nitric acid (HNO3) reacts with magnesium but not copper.
Concentrated nitric acid reacts with copper.
• Dil. HNO3(aq) with Mg:
2 H+(aq) + Mg(s)  H2(g) + Mg2+(aq)
• Dil. HNO3(aq) with Cu: no reaction
• Conc. HNO3(aq): (unbalanced equation)
NO3–(aq) + Cu(s)  NO2(g) + Cu2+(aq)
• Dil. HNO3(aq) acts as an acid, conc. HNO3(aq) can act as an
oxidizing agent. Similarly for H2SO4(aq).
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Oxidation Number (氧化數) in
Chemical Naming
Roman number is sometimes used for O.N./O.S.
e.g., “The Oxidation State of Mn in MnO4– is +VII.”
The Romanic number system is also used in the “Stock system”
to distinguish different compounds.
e.g.,
Cu2O, containing Cu+ ion: copper(I) oxide
CuO, containing Cu2+ ion: copper(II) oxide
SO42–, sulphate(VI) ion (more common: sulphate)
SO32–, sulphate(IV) ion (more common: sulphite)
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O.N. of Atoms(原子) in Various Species
Exercise: verify the O.N.
Oxidation
number
Sulphur
Nitrogen
Carbon
Iron
Copper
+7
+6
KMnO4
H2SO4
+5
+4
SO2
SCl2
-3
-4
CaCO3
K2Cr2O7
NO
MnO2
FeCl3
CO
FeSO4
N2 O
S
-1
-2
NO2
HNO2
+1
0
K2MnO4
HNO3
+3
+2
Manganese Chromium
H2 S
CuSO4
Mn2O3
CrCl3
MnSO4
CrCl2
Mn
Cr
CuCl
N2
C
NH2OH
C2 H 2
N2 H4
C2 H 4
NH3
C2 H 6
CH4
Fe
Cu
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Electrolysis(電解)
Charging and discharge of rechargeable battery
• Charging: applying voltage  new substances
(electrical energy  chemical energy)
• Discharge: giving out electrical energy
(chemical energy  electrical energy)
Electrolysis: chemical reaction by applying voltage
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Electrolysis(電解) of molten
PbBr2
Anode (“+”, attracting anions): Br - Br + eCathode (“-”, attracting cations): Pb2+ + 2e-  Pb
“An Ox”:
anode - oxidation
“Red Cat”:
reduction - cathode
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Electrolysis(電解) of CuSO4 solution
Ions attracted to anode: OH-(aq) & SO42-(aq)
Oxidation:
4OH-(aq)  O2(g) + 2H2O(l) + 4eSO42-(aq): no reaction
Ions attracted to cathode: Cu2+(aq) & H+(aq)
Reduction:
Cu2+(aq) + 2e-  Cu(s)
H+(aq): no reaction
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Preference of ion discharge
(1) Electrochemical series
Na  Na+ + eAg  Ag+ + e-
(More easily)
(More difficultly)
Therefore, we can infer that sodium ions gain electrons to form
atoms more difficultly than silver ions.
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Different species have different abilities to gain electrons to be
reduced.
These abilities are summarized in the electrochemical series.
Example:
http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29
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Characteristics
• The half-reactions are reductions (i.e., gaining electrons on the lefthand-side).
• The half-reactions are equilibriums.
• The abilities of gaining electrons are quantified by “standard
electrode potentials”. The smaller the potentials (near the top) are,
the more difficultly the reductions occur.
Compare
Na(s)  Na+(aq) + eAg(s)  Ag+(aq) + e-
Eo = +2.71 V
Eo = -0.80 V
(Note that when the reactions are reversed, the signs of Eo are
changed.)
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Characteristics
• Higher concentration or pressure: more favorable to go to the
opposite side. “Standard” values: 1 M for concentration and 1 atm
pressure for pressure.
e.g., Ca2+(aq) + 2e-  Ca(s)
Eo = -2.87 V
If increasing Ca2+ concentration, more favorable to the right-handside, larger E (less negative or even positive).
e.g., NO3-(aq) + 2H+(aq) + e-  NO2(g) + H2O(l) Eo = +0.78 V
If increasing NO2(g) pressure, more favorable to the left-hand-side,
smaller E.
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Characteristics
• A complete reaction consists of a reduction and an oxidation.
e.g., anode: 4OH-(aq)  O2(g) + 2H2O(l) + 4ecathode: Cu2+(aq) + 2e-  Cu(s)
(1)
(2)
They are combined to form a complete reaction, in which no
electron shows up.
(1):
(2)x2:
4OH-(aq)  O2(g) + 2H2O(l) + 4e2Cu2+(aq) + 4e-  2Cu(s)
Adding: 4OH-(aq) + 2Cu2+(aq) 
O2(g) + 2H2O(l) + 2Cu(s)
(3)
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Characteristics
•
Voltage of an electrochemical cell:
(1):
(2)x2:
4OH-(aq)  O2(g) + 2H2O(l) + 4e2Cu2+(aq) + 4e-  4Cu(s)
-0.40 V
+0.34 V
IMPORTANT: Eo does not change when the equation is “doubled”.
Adding: Eo = -0.40 + 0.34 = -0.06 V
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http://www.yorku.ca/skrylov/
Teaching/Chemistry1001/
electrochemistry.pdf
Characteristics
•
An electrochemical cell consists of two
half-reactions. A single half-reaction
does not exist alone and the absolute
values of Eo for half-reactions cannot
be measured. Therefore, the Eo of
one of the half-reactions,
Metal wire
H2(g) at 1 atm
Glass tube
2H+(aq) + 2e-  H2(g),
is set to zero. The electrode for this
half-reaction is shown on the right
and is called “standard hydrogen
electrode (SHE)”. The Eo of all the
others can be determined as values
relative to this “standard”.
Pt electrode
H+(aq, 1 M)
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Characteristics
e.g.,
2H+(aq) + 2e-  H2(g)
Zn2+(aq) + 2e-  Zn(s)
Consider Zn(s)  Zn2+(aq) + 2e-
E1 o = 0 V
E2 o = ?
-E2o
Adding: 2H+(aq) + Zn(s)  H2(g) + Zn2+(aq)
Ecello = E1o + (-E2o) = -E2o
Ecello of the last electrochemical cell is measured as +0.76 V.
Therefore, E2o = -0.76 V.
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Characteristics
• If the voltage is positive, the reaction occurs spontaneously. If
the voltage is negative, the reaction does not occur
spontaneously and an external voltage must be applied. The
external voltage must be larger than the magnitude of the cell
voltage.
e.g., H2(g) + Zn2+(aq)  2H+(aq) + Zn(s)
Eo = –0.76 V
The external voltage applied must be at least 0.76 V.
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Characteristics
• In electrolysis, the preference of ion discharge depends on Eo of
the relevant half-reaction potential. For example,
4OH-(aq)  O2(g) + 2H2O(l) + 4e2SO42-(aq)  S2O82-(aq) + 2e-
-0.40 V
-2.01 V
The first half-reaction is preferred because its Eo is larger.
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Concentrations of species
Increasing concentration increases the discharge tendency of an
ion. For example,
4OH-(aq)  O2(g) + 2H2O(l) + 4e2Cl-(aq)  Cl2(g) + 2e-
-0.40 V
-1.36 V
For dilute NaCl solution, OH- is discharged because the Eo value of
the first half-reaction is preferred. But for concentrated NaCl
solution, Cl- concentration is high enough for Cl- to be discharged.
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Electrodes (電極)
Commonly used graphite and platinum electrodes are inert and
have no effect on the preference of ion discharge. But some may.
• Mercury electrode(汞電極)
If graphite or platinum electrodes are used in the electrolysis of
concentrated NaCl solution, only H+ is discharged at the cathode.
But if mercury electrode is used for the cathode, Na+ is
discharged because sodium metal forms an alloy with mercury.
(This method is used in industry for the production of sodium.)
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Electrodes (電極)
• Metal electrode
When an anion discharges at anode, it gives out electrons. If a
metal electrode is used as the anode, the metal atoms may
also give out electrons to form metal ions, i.e., the metal
electrode may compete with the anion in giving out electrons.
For example, if copper electrode is used as the anode in the
electrolysis of copper sulfate solution, the copper electrode
becomes thinner and thinner.
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Electrodes(電極)
cf.
Cu(s)  Cu2+(aq) + 2e4OH-(aq)  O2(g) + 2H2O(l) + 4e-
-0.34 V
-0.40 V
Copper metal of the anode completes with OH-. The potential
of the first half-reaction is larger. Copper metal, rather than OH-,
gives out electrons.
In principle, if platinum electrode is used, platinum may also
give out electrons to form platinum ion. But in practice, it seldom
happens due to the very negative value of Eo for this process:
Pt(s)  Pt2+(aq) + 2e-
-1.20 V
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Summary of common cases:
Solution
Electrodes
Anode
Cathode
Main products at
Anode
Cathode
NaNO3 or
NaSO4
H2SO4
O2
Graphite
NaOH
H2
(Dil)
NaCl (Conc)
(Conc)
CuSO4
Cl2
Graphite
Graphite
Copper
Mercury
Graphite
Cl2
Na
O2
Copper
Copper
Cu2+
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Salt Bridge
The container in which a half-reaction occurs is called a “half-cell”.
In the diagram shown, the two half-cells are in the same beaker.
But in some cases they
must be separated
physically, because the
species of the two
half-cells react directly
without electrons
going through the
external circuit.
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Salt Bridge
For example, an electrochemical cell can be constructed for
the following reaction:
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
We can break it into two half-reactions:
Cu(s)  Cu2+(aq) + 2e–
–0.34 V
2Ag+(aq) + 2e–  2Ag(s)
+0.80 V
If we put everything into the same beaker, it does not work
as expected.
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Salt Bridge
Reason: when Ag+ is in contact with Cu, they react on the
surface of the copper plate. Electrons are given out by Cu to
Ag+ directly and they do not go through the external circuit.
Voltmeter
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
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Salt Bridge
Therefore, the Ag+ solution and Cu2+ solution must be separated in
two beakers.
Voltmeter
Salt-bridge:
• “connecting” the two halfcells.
• providing ions to keep the
half-cells electrically neutral
• simplest version: a strip of
filter-paper soaked in KNO3 or
NH4NO3
Electron
flow
Electrical
current
NO3–
K+
Cu
Ag
Salt
bridge
Cu2+
Cu2+
1 M Cu(NO3)2(aq)
NO3–
Ag+
1 M AgNO3(aq)
Modified from:
http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf
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Fuel Cell(燃料電池)
• Burning fuel  …  electrical energy:
some energy is lost in heating
• Fuel cell: chemical energy  electrical energy
Example: H2 fuel cell
H2 fuel cell is reversible and it can act as a rechargeable battery
Charging (storing up electrical energy):
2H2O(l)  2H2(g) + O2(g)
Discharging (releasing electrical energy):
2H2(g) + O2(g)  2H2O(l)
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H2 fuel cell
Half-reactions: (discharging)
H2(g)  2H+(aq) + 2e–
O2(g) + 4H+(aq) + 4e–  2H2O(l)
0.00 V
+1.23 V
The cell gives an electrical potential of 1.23 V.
Fuel Cell – Car & Experiment Kit Lab
Manual, Thames & Kosmos (2000)
H2
O2
H2O
Anode
PEM
Cathode
PEM: proton exchange membrane
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Fuel Cell(燃料電池)
Other than H2, some other compounds can also be used
for fuel cell. For example:
• Methanol (CH3OH)
• Ethanol (CH3CH2OH)
These kinds of fuel cell may not be reversible, but easier
to handle and more energy-rich.
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Silver-Zinc(鋅)Battery(1.8V)
Half reactions of discharge:
Reduction: Ag2O(s) + H2O(l) + 2e-  2Ag(s) + 2OH- (aq)
Oxidation: Zn(s) + 2OH-(aq)  ZnO(s) + H2O(l) + 2e-
Overall:
Zn(s) + Ag2O(s)  ZnO(s) + 2Ag(s)
http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf
46
The Nickel-Cadmium
Rechargeable Battery (1.4 V)
Half reactions of discharge:
Reduction:
Oxidation:
2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH- (aq)
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e-
Overall:
Cd(s) + 2NiO(OH)(s) + 2H2O(l)  2Ni(OH)2(s) + Cd(OH)2(s)
http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf
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Eo and K (Equilibrium Constant(平衡常數))
nFEo = RT·ln(K)
e.g,
Calculate K for the equilibrium:
Cu(s) + 2 Ag+(aq)
Cu2+(aq) + 2 Ag(s)
Solution:
Cu(s)  Cu2+(aq) + 2 e–
2 Ag+(aq) + 2 e– 2 Ag(s)
–0.34 V
+0.80 V
Eo = –0.34 V + 0.80 V = +0.46 V
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Eo and K (Equilibrium Constant(平衡常數))
n = no. of electrons in the half-reactions = 2
F = 96485 C mol–1 (Faraday constant)
Put into the equation, nFEo = RT·ln(K),
(2) (96485 C mol–1) (0.46 V) = (8.314 J mol–1 K–1) (298 K) ln(K)
K = 3.6 x 1015 dm3 mol–1
Note: the unit of K is determined by the expression:
K = [Cu2+(aq)]/[Ag+(aq)]2
(mol dm–3 for concentration, atm for pressure)
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