Redox reaction(??????), Oxidation number and Electrolysis(??)

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Redox reaction(??????), Oxidation number and
Electrolysis(??)
International Junior Science Olympiad (IJSO)
Dr. Yu-San Cheung yscheung_at_cuhk.edu.hk Department
of Chemistry The Chinese University of Hong Kong
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Oxidation(??) and Reduction(??)
Oxidation Reaction of an element or a compound
with O2 to give an oxide e.g., 4 Na(s) O2(g)
? 2 Na2O(s) (sodium oxide) 2 H2(g)
O2(g) ? 2 H2O(l) (hydrogen
oxide?) Reduction Reverse of oxidation e.g.,
2 CuO(s) ? 2 Cu(s) O2 (g)
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Oxidation(??) and Reduction(??)
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 (??) and Electron Transfer(????)
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(???)
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 (cont)

• 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 O.N. for O 2 ? O.N. for S
6 SO2 overall 0 O.N. for O 2 ? O.N. for
S 4 SF2 overall 0 O.N. for F 1 ? O.N.
for S 2 H2S overall 0 O.N. for H 1 ?
O.N. for S 2 PCl5 overall 0 O.N. for Cl
1 ? O.N. for P 5 PO4 3 overall 3 O.N.
for O 2 ? O.N. for P 5 PO3 3 overall 3
O.N. for O 2 ? O.N. for P 3 PH3 overall
0 O.N. for H 1 ? O.N. for S 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 O.N. for O 2 ? O.N. for Cl
7 ClO3 overall 1 O.N. for O 2 ? O.N.
for Cl 5 ClO2 overall 1 O.N. for O 2 ?
O.N. for Cl 3 ClO overall 1 O.N. for O
2 ? O.N. for Cl 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(???)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 O.N. for O atom 1
• Superoxide O2 O.N. for O atom 1/2

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Redox reaction(??????) (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(????)
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(???) and Reducing Agent(???)

• 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 (???)
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(???)
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)
oxidation / reduction SO2(g) ? SO42(aq)
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 1 1
1 O 2 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(??)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 (???) 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(??) in Various Species
Exercise verify the O.N.
Oxidation number Sulphur Nitrogen Carbon Iron Copper Manganese Chromium
7 KMnO4
6 H2SO4 K2MnO4 K2Cr2O7
5 HNO3
4 SO2 NO2 CaCO3 MnO2
3 HNO2 FeCl3 Mn2O3 CrCl3
2 SCl2 NO CO FeSO4 CuSO4 MnSO4 CrCl2
1 N2O CuCl
0 S N2 C Fe Cu Mn Cr
-1 NH2OH C2H2
-2 H2S N2H4 C2H4
-3 NH3 C2H6
-4 CH4
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Electrolysis(??)

• 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(??) of molten PbBr2
Anode (, attracting anions) Br - ?Br
e- Cathode (-, attracting cations) Pb2 2e-
? Pb An Ox anode - oxidation Red Cat
reduction - cathode
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Electrolysis(??) of CuSO4 solution
Ions attracted to anode OH-(aq)
SO42-(aq) Oxidation 4OH-(aq) ? O2(g) 2H2O(l)
4e- SO42-(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

• Electrochemical series
• Na ? Na e- (More easily)
• Ag ? Ag e- (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_page29
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Characteristics

• The half-reactions are reductions (i.e., gaining
  electrons on the left-
• hand-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) e- Eo 2.71 V
• Ag(s) ? Ag(aq) e- 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-hand-
• side, 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)
  4e- (1)
• cathode Cu2(aq) 2e- ? Cu(s) (2)
• They are combined to form a complete reaction, in
  which no
• electron shows up.

(1) (2)x2 4OH-(aq) ? O2(g) 2H2O(l) 4e- 2Cu2(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) 4OH-(aq) ? O2(g) 2H2O(l) 4e- -0.40 V
• (2)x2 2Cu2(aq) 4e- ? 4Cu(s) 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,
• 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.

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Characteristics
e.g., 2H(aq) 2e- ? H2(g) E1o 0 V
Zn2(aq) 2e- ? Zn(s) E2o
? Consider Zn(s) ? Zn2(aq) 2e-
-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) 4e-
  -0.40 V
• 2SO42-(aq) ? S2O82-(aq) 2e-
  -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) 4e- -0.40 V 2Cl-(aq) ? Cl2(g)
2e- -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 (??)

• Commonly used graphite and platinum electrodes
  are inert and have no effect on the preference of
  ion discharge. But some may.
• Mercury electrode(???)
• 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 (??)

• 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(??)
cf. Cu(s) ? Cu2(aq) 2e-
-0.34 V 4OH-(aq) ? O2(g) 2H2O(l)
4e- -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 Solution Electrodes Electrodes Main products at Main products at
Solution Solution Anode Cathode Anode Cathode
NaNO3 or NaSO4 NaNO3 or NaSO4 Graphite Graphite O2 H2
H2SO4 H2SO4 Graphite Graphite O2 H2
NaOH NaOH Graphite Graphite O2 H2
NaCl (Dil) Graphite Graphite O2 H2
NaCl (Conc) Graphite Graphite Cl2 H2
NaCl (Conc) Graphite Mercury Cl2 Na
CuSO4 CuSO4 Graphite Graphite O2 Copper Copper
CuSO4 CuSO4 Copper Graphite Cu2 Copper Copper
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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 half-cells.
• providing ions to keep the half-cells
  electrically neutral
• simplest version a strip of filter-paper soaked
  in KNO3 or NH4NO3

Electrical current
Electron flow
NO3
K
Cu
Ag
Salt bridge
Cu2
NO3
Ag
Cu2
1 M Cu(NO3)2(aq)
1 M AgNO3(aq)
Modified from http//www.yorku.ca/skrylov/Teachin
g/Chemistry1001/electrochemistry.pdf
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Fuel Cell(????)

• 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 0.00 V O2(g) 4H(aq)
4e ? 2H2O(l) 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(????)

• 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(?)Battery(1.8V)
Half reactions of discharge
Reduction Oxidation Ag2O(s) H2O(l) 2e- ? 2Ag(s) 2OH- (aq) 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
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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(????))
nFEo RTln(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 0.34 V 2 Ag(aq) 2 e? 2 Ag(s)
0.80 V Eo 0.34 V 0.80 V 0.46 V
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Eo and K (Equilibrium Constant(????))
n no. of electrons in the half-reactions 2 F
96485 C mol1 (Faraday constant) Put into the
equation, nFEo RTln(K), (2) (96485 C mol1)
(0.46 V) (8.314 J mol1 K1) (298 K) ln(K) K
3.6 x 1015 dm3 mol1 Note the unit of K is
determined by the expression K
Cu2(aq)/Ag(aq)2 (mol dm3 for
concentration, atm for pressure)