Title: Electrochemistry- study of interchange of chemical and electrical energy
1Electrochemistry-study of interchange of
chemical and electrical energy
- Generating electric current from spontaneous
chemical reactions and use of current to produce
chemical change - Labs
- 31 The Thermodynamics of the Dissolution of
Borax - 33 Electrolytic Cells Avogadros Number
- Chemical Equations
- Chapter 12
2Oxidation-Reduction Reactions (Redox Reactions)
- Oxidation (LEO)
- Loss of electrons-atoms or ions undergo increase
in oxidation state (is oxidized) - Electrons given to another atom which is being
reduced (reducing agent or reductant) - Reduction (goes GER)
- Gain of electrons-atoms/ions undergo decrease in
oxidation state (is reduced) - Takes electrons away from another atom which is
being oxidized (oxidizing agent or oxidant)
3- Redox reaction can be written as two
half-reactions (one reduction, one oxidation) - To generate current
- Separate oxidizing agent from reducing agent
- Electron transfer occurs through wire
- Electron transfer directed through device to
provide useful work - Reactants separated by salt bridge/porous
partition - Each half-reaction has a potential, or voltage,
associated with it - Given as reduction half-reactions
- Read in reverse and change sign on voltage to get
oxidation potentials
4Electrochemical cellsdevice associated with
redox reaction (galvanic cells, electrolytic
cells)
- Electrochemical process involves electron
transfer at interface between electrode and
solution - Species undergoing reduction receive electrons
from cathode - Species in solution act as oxidizing agent
- Species undergoing oxidation donate electrons to
anode - Species in solution act as reducing agent
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5Galvanic Cell (voltaic cell)
- Chemical energy ? electrical energy
- Harnesses energy of spontaneous redox reactions
- Physically separate chemicals in 2 half-reactions
- Electrons generated by oxidation half-reaction
flow through electrical conductor before being
used in reduction half-reaction - Flow diverted through meters, motors, light bulbs
to perform useful work before reaching
destination - Current (defined by physicists as flow of
positive charge) always in opposite direction
from flow of electrons (always from anode to
cathode)
6- Electrodes are metal strips
- Sign of electrodes determined by
- Since electrons flow out of anode and into
external circuit, anode is negative - Since electrons flow from external circuit into
cathode, cathode is positive - Opposite is true for electrolytic cells
- Where reactions occur (The Red Cat ate An Ox)
- Oxidation occurs at anode (AN OX)-on left
- Reduction occurs at cathode (RED CAT)-on right
7Porous barrier separate two compartments
- Allows for migration of positive/negative ions
between half-cells, completing electric circuit - Problem with porous barriers
- Inside barriers, ionic solutions mix
- Has effect on operation of cell
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8Galvanic cells with salt bridges
- Inverted tube contains electrolyte
- Gel (agar) added to provides
firmness but permits ion flow - Prevents two reacting solutions from mixing
- Ions dont react with other ions in cell or with
electrode material - Maintains electrical neutrality in system
- Provides - ions to equal ions created at anode
(during oxidation) and ions to
replace - ions being used up at
cathode (during reduction) - Anions always migrate toward anode
- Cations toward cathode
9Electron Flow
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12Cell Potential (Ecell) or Electromotive force
(emf)
- Force with which electrons flow from - electrode
(anode on left) to electrode (cathode on right)
through external wire - Due to PE difference of electrons before/after
transfer - In electrochemical cell, electric potential
created between two dissimilar metals - Greater tendency or potential of two
half-reactions to occur spontaneously, greater
emf of cell - Measured in volts (V-why called cell voltage)
- 1 V 1 J/coulomb (of charge transferred)
- Measured with voltmeter which draws current
through known resistance (heat is produced)
13- Voltage of voltaic cells
- All based on spontaneous chemical reactions
- ?G must always be negative
- Voltage of voltaic cell is always positive (EMF)
- Subtract smaller reduction potential from larger
one - Same as EMF cathode anode
- Under standard conditions, voltage of cell is
same as total voltage of redox reaction - Standard emf of standard cell potential (E0cell)
- Under nonstandard conditions, cell voltage
computed by using Nernst equation - As galvanic cell operates
- Redox reaction of cell approaches equilibrium
- Capacity to deliver useful electrical energy
decreases - At equilibrium, cell ceases to function (dead
battery)
14Standard Reduction Potentials
- Cell potentials can be measured
- Half-cell potential cannot
15Measuring Potential
- Galvanic cell under standard conditions made
using arbitrary standard hydrogen electrode (SHE)
and test half-cell w/different half reaction - Assigned standard electrode potential of exactly
0.00 V - Half reaction always written as reduction
- Eo values corresponding to reduction
half-reactions with all solutes standard
reduction potentials-Eocell ?
Under ideal conditions where ideal behavior is
assumed
16Standard Reduction Potentials
- All solutions are 1M, gases at 1 atm, T 25oC
- Write oxidation/reduction half-reactions for cell
- Look of reduction potential (Eoreduction) for
reduction half-reaction in table - Half reaction w/higher reduction potential
- Look of reduction potential for reverse of
oxidation half-reaction and reverse sign
(Eooxidation -Eoreduction) - Half reaction w/lower reduction potential/sign
reversed - Add potentials to get overall standard cell
potential - Two half reactions are balanced for electrons
exchanged but value of each Eo remains unchanged
(intensive property-does not depend on how many
times reaction occurs) - If sum positive, reaction is spontaneous/runs on
own (always positive for electrochemical cells) - If sum negative, energy supplied to make reaction
go
17- Each half-reaction associated w/signed numerical
value - More positive it is, greater oxidizing power of
redox half-reaction - More negative it is, greater reducing power of
reverse redox half-reaction - Larger difference between Ered values, larger
Ecell
18(1 atm)
19Line Notation
- Anode written first on left/cathode on right
- Reactants written 1st on each side
- Vertical bar is boundary between two phases
- If both substances in same phase, separated by
comma, not vertical bar - Double line represents salt bridge or porous disk
- When platinum electrode present, placed at left
and/or right end of cell diagram
20Using the table of standard reductions provided
write the equation for the reaction between the
following two half cells, and determine its
voltage
- Au (s) Au 3 (aq) Cu 2 (aq) Cu (s)
- Gold is higher on table and written as found on
table - Au 3 (aq) 3 e - ?Au (s) Eo 1.50 V
- Copper is lower so it's written as oxidation
(sign reversed) - Cu (s) ? Cu 2 (aq) 2 e- Eo -0.34 V
- Equation balanced for exchange of electrons (each
needs 6) - 2 Au 3 (aq) 6e - ? 2 Au (s) E 1.50 V
- 3 Cu (s) ?3 Cu 2 (aq) 6e E -0.34 V
- Size of values for Eo of reactions not
changed - Add two half reactions and E0 values
- 2 Au 3 (aq) 3 Cu (s) ? 2 Au (s) 3 Cu 2 (aq)
- E 1.50 V ( - 0.34 V ) 1.16 V
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22Place the following in order of increasing
strength as oxidizing agents
-0.44 0.954 2.87 0.22
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24Describe a galvanic cell based on the two
half-reactions below
25Homework
- Read 17.1-17.2, pp.827-837
- Q pp. 867-869, 14 a/b/c/f/i/k, 16 a/b/d/f, 26a,
28 (a only), 30b, 32 (b only), 34a, 36b
26Cell Potential, Electrical Work, and Free Energy
27Complete Description of a Galvanic Cell
- Electrical energy delivered by galvanic cell
equal to quantity of useful work obtained as
result of cell operation - Work, w, measured in relation to amount of
charge, q, transferred between anode/cathode of
cell - This quantity, potential difference, E, is
defined as E w/q. - SI unit is joule per coulomb or volt (V)
- Cell potential (always positive for galvanic
cell) - Direction of electron flow (direction that yields
potential) - Designation of anode/cathode
- Nature of each electrode/ions present in
compartments - Chemically inert conductor (such as Pt) required
if only ions are present (no substance in
reaction is conducting solid)
28Electric work and cell potential
- Free energy change occurring during chemical
reaction is measure of maximum work that system
can perform - Potential (E) -Work (w) / Charge (q)
- So w -qE
- Work leaving system has negative charge
- Faraday (F) charge in coulombs per mole of
electrons (96,485 C/mol e-) - Then q nF and w -nFE
- n number of moles of electrons
29- ?G (Gibbs free energy) is measure of spontaneity
of process occurring at constant T/P - emf, E, of redox reaction also indicates whether
reaction is spontaneous - From thermodynamics we know that
- ?G ?U T?S ?(PV) and U heat w
- Therefore, at constant T/P ?G w
- Therefore ?G -nFE and at standard state ?G0
-nFE0 (relationship between emf/free energy
changes ) - ?Go Standard Gibbs free energy change (kJ/mol
or Joules) - n moles of electrons exchanged in reaction
(mol) - F Faradays constant, 96,485 coulombs/mole (1
mole of electrons has a charge of 96,485
coulombs) - Eo Standard reaction potential (V or
Joules/Coulomb)
30- Since both n/F are
- value of E leads to value of ?G, which
indicates spontaneous reaction - If ?G and E have opposite signs, E predicts
direction of reaction - If Eo is positive, ?Go is negative (lt 0)-reaction
spontaneous (has positive cell potentials) - If Eo is negative, ?Go is positive (gt 0)-reaction
is nonspontaneous (but is spontaneous in reverse
direction)
31Relationship between thermodynamics (push behind
electrons) and electrochemistry
- Relationship between reaction potential and free
energy for a redox reaction is given by - Emf potential difference (V) work (J)
- charge (C)
- Driving force (emf) is defined in terms of
potential difference (in V) between two points in
circuit - One coulomb is amount of charge that moves past
any given point in circuit when current of 1
ampere (amp) is supplied for one second (1 ampere
1 coulomb/sec) - Faradays law states that during electrolysis,
passage of 1 faraday through circuit brings about
oxidation of one equivalent weight of substance
at one electrode (anode) and reduction of one
equivalent weight at other electrode (cathode)
32- emf is not converted to work with 100 efficiency
- Energy always lost as heat, but wmax useful for
calculating efficiency of conversion - wmax -qEmax
- Relationship to free energy (energy driving
reaction due to movement of charged particles
giving rise to potential difference) - wmax EG
- EG -qEmax -nFEmax
- EG -qEmax
- EG0 -nFE0
- When Ecell positive (spontaneous), EG will be
negative (spontaneous), so there is agreement - Standard cell potential, Eocell, measured and
standard electrode potential of test half-cell
determined by using Eocell Eocathode - Eoanode - Eocathode-standard reduction potential for
reaction occurring at cathode, represents
tendency to remove es from electrode surface - Eoanode-standard reduction potential for reaction
occurring at anode and represents its tendency to
remove es from anode
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35Dependence of Cell Potential on Concentration
- Cell voltages at nonstandard concentrations (not
1 M)
36Nernst Equation-way to relate E0 at standard
conditions and E, potential at any real condition
- Standard state impossible to achieve in reality
- As soon as wire hooked to two half-cells,
reaction proceeds and changes concentrations of
all reactants and products - Heating or cooling makes reaction deviate from
standard temperature
37Calculating Nernst Equation
- From thermodynamics, recall G Go RTlnQ
- If we divide everything by nF
- since ?G nFE, or E ?G/(nF)
- Ecell cell potential at non-standard conditions
- E0cell standard reduction potential
- R 8.314 J/molK (the gas constant)
- F 96485 coul/mol (Faraday's constant)
- T absolute temperature
- n number of moles of electrons transferred in
balanced equation - Q reaction quotient for reaction aA bB ? cC
dD - Expressed in terms of base 10 rather than ln
(standard conditions of 298K) - Can be used to find cell potential at any set of
conditions - Cells spontaneously discharge until they achieve
equilibrium (at equilibrium, cell is dead)
38Consider the Daniell Cell at 25 CZn(s)
Cu2(aq) Cu(s) Zn2(aq)
- Find cell potential at following conditions when
Cu2 1.00 M, Zn2 1.0109 M and when
Cu2 0.10 M, Zn2 0.90 M. - Recall that standard potential for Daniell cell
is Eo 1.10 V - Nernst equation used to find potentials at
nonstandard conditions
When Cu2 1.00 M, Zn2 1.0109 M
When Cu2 0.10 M, Zn2 0.90 M
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43Depiction of concentration cell
44- Cell in which current flows due only to
difference in concentration of ion in 2 different
compartments of cell - Le Châteliers principle used to determine effect
on potential - Shift to left reduces potential
- Shift to right increases potential
- If concentrations are different, stress is put on
system that will be equalized by electron flow to
allow reduction and oxidation to occur - Voltages typically small
- When concentrations in half-cells become equal,
E0cell 0 and system is at equilibrium
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46Calculate the EMF of the cell Zn(s) Zn2 (0.024
M) Zn2 (2.4 M) Zn(s)
- Zn2 (2.4 M) 2 e Zn Reduction
- Zn Zn2 (0.024 M) 2 e Oxidation
- Zn2 (2.4 M) Zn2 (0.024 M), DE 0.00
- Using Nernst equation
- (0.024) DE 0.00 - 0.0592/2 log (0.024/(2.4)
- (-0.296)(-2.0)
- 0.0592 V
47Show that voltage of electric cell is unaffected
by multiplying reaction equation by positive
number
- Mg Mg2 Ag Ag
- Mg 2 Ag Mg2 2 Ag
- DE DEo 0.0592/2 log Mg2/Ag2
- 2 Mg 4 Ag 2 Mg2 4 Ag
- DE DEo 0.0592/4 log Mg22/Ag4
- Simplified to the 1st equation, showing cell
potential DE not affected
48Calculation of Equilibrium Constants for Redox
Reactions
- The standard reaction potential is related to the
equilibrium constant - At equilibrium, Ecell 0 and Q K
- As cells discharge, concentration changes, Ecell
changes. - For a cell at concentrations and conditions other
than standard, a potential can be calculated
using the Nernst equation - If Eo is positive, then K gt 1 and forward
reaction favored - If Eo is negative, then K lt 1 and reverse
reaction is favored
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50The standard cell potential dE for the reaction
Fe Zn2 Zn Fe2 is -0.353 V. If a piece of
iron is placed in a 1 M Zn2 solution, what is
the equilibrium concentration of Fe2?
- Equilibrium constant K may be calculated using K
10(n DE)/0.0592 - 10-11.93
- 1.2x10-12
- Fe2/Zn2. Since Zn2 1 M, it is
evident that Fe2 1.2-12 M
51Homework
- Read 17.3-17.4, pp. 837-846
- Q pp. 869-871, 38, 40, 46, 48, 54, 60, 66, 70
52Batteries
- Portable, self-contained electrochemical power
source (DC) consisting of one or more voltaic
cells, connected in series - Greater voltages achieved by using multiple
voltaic cells in single battery (12V) - When connected in series, battery produces
voltage that is sum of emfs of individual cells - Higher emfs achieved by using multiple batteries
in series - Electrodes marked (cathode) and (anode)
53Lead-acid storage battery
- As battery discharged, uses up sulfate
ions/electrodes become coated w/lead sulfate - Reverse reactions regenerate sulfate ion in
solution/reduce amount of lead sulfate
contaminating electrode surfaces - Each pair produces 2 volts (6 pairs of
electrodes used in 12-volt car battery) - When jump starting car, connect ground cable on
dead car to metallic contact away from battery.
Otherwise, could explode - Causes electrolysis of water/production of H2/O2
which could ignite
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55Common Dry Cell Battery(acid version)
- Zinc-anode
- Carbon rod in contact with moist paste-cathode
56Alkaline dry cell
- NH4Cl replaced w/KOH or NaOH
- Last longer than acid cells because zinc (anode)
corrodes more slowly in basic environment - Cathode (graphite rod) inserted into paste made
of manganese dioxide, water and potassium
hydroxide - Zn(s) 2OH-(aq) ? ZnO(s) H2O 2e- (anode)
- 2MnO2(s) H2O 2e- ? Mn2O3(s) 2OH-(aq)
(cathode) - Total voltage is 1.54 volts
- Not rechargeable
57Mercury Battery
58- Fuel Cells
- Galvanic cells where reactants continuously
supplied - Energy normally lost as heat is captured and used
to produce an electric current - Redox reaction
- Hydrogen oxidized at anode and oxygen reduced at
cathode to form water and electricity - 2x as effective as gas, oil or coal-powered
generators in converting chemical energy into
electricity
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60Corrosionoxidation of metal
- Oxidation of most metals by oxygen is spontaneous
redox reactions - Many metals develop thin coating of metal oxide
on outside that prevents further oxidation - Some metals, such as copper, gold, silver and
platinum (noble metals), are relatively difficult
to oxidize
61Corrosion of Iron
- Anodic regions
- Regions of steel alloy where iron is more easily
oxidized - Fe ? Fe2 2e-
- Cathodic regions
- Areas resistant to oxidation
- Electrons flow from anodic regions react
w/oxygen - O2 2H2O 4e- ? ?4OH-
- Presence of water essential to iron corrosion
- Presence of salt accelerates corrosion by
increasing electron conduction from anodic to
cathodic regions
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63Prevention of Corrosion
- Coating w/metal that forms oxide coat to protect
metal that would not develop protective coat - Galvanizing
- Place sacrificial of more easily oxidized metal
on top of metal to protect - Zinc over iron
- Alloying
- Addition of metals that change steels reduction
potential. - Nickel and chromium alloyed to iron
64- Cathodic Protection
- Connection of easily oxidized metals (an anode)
to less easily oxidized metals keeps less from
experiencing corrosion - Anode corrodes-must be replaced periodically
- Magnesium as anode to iron pipe
- Titanium as anode to steel ships hull
65Electrolysis
- Decomposition of substance by electric current
66Electrolytic cells
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lpt3.html
- Nonspontaneous reactions
- Electrical energy required to induce reaction
- Two electrodes immersed in electrically
conductive sample - Electrical voltage (gt1.10 V) applied to them
- Voltage increased until electrons flow in
opposite direction (electrolytic) - At cathode-reduction occurs (RED CAT)
- At anode-oxidation occurs (AN OX)
- Electrical energy is converted into chemical
energy - Electrolytic cells are used for electroplating
67- (a) Standard galvanic cell based on spontaneous
reaction - Zn Cu2 ? Zn2 Cu
- (b) Standard electrolytic cell Power source
forces opposite reaction - Cu Zn2 ? Cu2 Zn
68What voltage is necessary to force the following
electrolysis reaction to occur?
- 2I-(aq) Cu2(aq)? I2(s) Cu(s)
- Which process would occur at the anode? Cathode?
Assuming the iodine oxidation takes place at a
platinum electrode, what is the direction of
electron flow in this cell?
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70Calculating Quantity of Substance Produced or
Consumed
- To determine quantity of substance either
produced or consumed during electrolysis given
time known current flowed - Write balanced half-reactions involved
- Calculate number of moles of electrons that were
transferred - Calculate number of moles of substance that was
produced/consumed at electrode - Convert moles of substance to desired units of
measure
71A 40.0 amp current flowed through molten
iron(III) chloride for 10.0 hours (36,000 s).
Determine the mass of iron and the volume of
chlorine gas (measured at 25oC and 1 atm) that is
produced during this time.
- Write half-reactions that take place at
anode/cathode - anode (oxidation) 2 Cl- Cl2(g) 2 e
- cathode (reduction) Fe3 3 e- Fe(s)
- Calculate number of moles of electrons
- Calculate moles of iron/chlorine produced using
number of moles of electrons calculated and
stoichiometries from balanced half-reactions. (3
moles electrons produce 1 mole of Fe/2 moles of
electrons produce 1 mole of chlorine gas) - Calculate mass of iron using molar mass and
calculate volume of chlorine gas using ideal gas
law (PV nRT).
72Calculating Time Required
- To determine quantity of time required to produce
known quantity of substance given amount of
current that flowed - Find quantity of substance produced/consumed in
moles - Write balanced half-reaction involved
- Calculate number of moles of electrons required
- Convert moles of electrons into coulombs
- Calculate time required
73How long must a 20.0 amp current flow through a
solution of ZnSO4 in order to produce 25.00 g of
Zn metal
- Convert mass of Zn produced into moles using
molar mass of Zn - Write the half-reaction for the production of Zn
at the cathode Zn2(aq) 2 e- Zn(s) - Calculate moles of e- required to produce moles
of Zn using stoichiometry of the balanced
half-reaction (2 moles of electrons produce 1
mole of zinc) - Convert moles of electrons into coulombs of
charge using Faraday's constant - Calculate time using current and coulombs of
charge
74Calculating Current Required
- To determine amount of current necessary to
produce known quantity of substance in given
amount of time - Find quantity of substance produced/or consumed
in moles - Write equation for half-reaction taking place
- Calculate number of moles of electrons required
- Convert moles of electrons into coulombs of
charge - Calculate current required
75What current is required to produce 400.0 L of
hydrogen gas, measured at STP, from the
electrolysis of water in 1 hour (3600 s)?
- Calculate number of moles of H2
- Write equation for half-reaction that takes
place. Hydrogen produced during reduction of
water at cathode. Equation for this
half-reaction is 4 e- 4 H2O(l) 2 H2(g) 4
OH-(aq) - Calculate number of moles of electrons (4 mole of
e- required to produce 2 moles of hydrogen gas,
or 2 moles of e-'s for every one mole of hydrogen
gas) - Convert moles of electrons into coulombs of
charge - Calculate current required
76How many grams of copper can be reduced by
applying a 3.00 A current for 16.2 min to a
solution containing Cu2 ions?
- Time ? current ? Coulombs ? moles e- ? moles Cu ?
g Cu - 16.2 min 60 sec 3 C 1 mol e- 1 mol Cu
63.54 g Cu - 1 min 1 sec 96,486 C 2
mol e- 1 mol Cu - 0.96 g Cu
77Electrolysis can be used to separate mixture of
ions, if reduction potentials are fairly far apart
- Remember, metal ion with highest reduction
potential is easiest to reduce - Predict order of reduction and which of following
ions will reduce first at cathode of electrolytic
cell Ag, Zn2, IO3-
78Electroplating
- Deposit neutral metal atoms on electrode by
reducing metal ions in solution - One metal coated with another
- Presence of active electrode that takes part in
electrolysis reaction - Anode-piece of plating metal
- Cathode-object to be plated
- Plating solution is NiSO4 because SO42- ion does
not participate in plating reaction
79How long must a current of 5.00 A be applied to a
solution of Ag to produce 10.5 g silver metal?
- 10.5 g Ag 1 mol Ag 1 mol e- 96,485 C 1
sec 1 min - 107.868 g Ag 1 mol Ag 1 mol e-
5 C 60 sec - 31.3 min
80Electrolysis of Water
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- Requires soluble salt/dilute acid to serve as
electrolyte - Anode
- 2H2O(l) ? O2(g) 4H(aq) 4e- Eoox -1.23 V
- Cathode
- 2H2O(l) ? H2(g) 2OH-(aq) Eored -0.83 V
- Overall reaction
- 6H2O(l) ? 2 H2(g) O2(g) 4H(aq) 4e-
Eocell -2.06 V - If in single container, H/OH- combine to yield 4
additional water molecules
81What volume of H2(g) and O2(g) is produced by
electrolyzing water at a current of 4.00 A for
12.0 minutes (assuming ideal conditions)?
- 2H2O(l) ? ?2H2(g) O2(g)
- Actual ratio is not exactly 21 for a variety of
reasons including oxygen solubility. - 12 min 60 sec 4 C 1 mol e- 1 mol H2
22.4 L - 1 min 1 sec 96,486 C 2 mol e-
1 mol H2 - 0.334 L H2 ? 0.167 L O2
82Electrolysis of molten salts
- NaCl
- Good conductor as liquid
- Melting salt frees ions
- Makes it electrically conductive
- Ions of opposite charge migrate to these
electrodes and react
83Electrolysis of aqueous solutions
- Aqueous solutions of salts are electrically
conductive and can be electrolyzed - For solutions, two possible reactions occur
(water can be both oxidized and reduced) - At cathode
- If metal ion is very active metal, water will be
reduced (2H2O 2e- ? H2 2OH-) - If metal ion is inactive or active metal, metal
ion will be reduced - At anode
- Oxidation of salts anion or (?)
- Oxidation of water (2H2O ? O2 4H 4e-)
84- To determine which will occur at anode
- If anion is polyatomic ion, it generally will not
be oxidized - SO42-/NO3-/ClO4- not oxidized in aqueous solution
- Cl-/Br-/l- will be oxidized in aqueous solution
- If anion in one salt is oxidized in aqueous
electrolysis, that same anion in any other salt
will also be oxidized - If solution of NaBr results in Br- being oxidized
to Br2, predict that solutions of KBr, CaBr2,
NH4Br and AlBr3 will all produce Br2 at anode
85Commercial Electrolytic Processes
- Abundance of elements on earth
- 1st Oxygen
- 2nd Silicon
- 3rd Aluminum (very active metal so difficult and
expensive originally to purify)
- Since metals are easily oxidized, most found as
ores, mixtures of ionic compounds. Au, Ag, and
Pt are more difficult to oxidize, so often found
as pure metals.
86Production of aluminum from molten-salt
electrolysis (purification of aluminum from
bauxite ore)
Hall-Heroult process
87- Electrorefining (purifying) metals
- Copper ore is refined by roasting
- Impure copper is anode
- Small strip pure copper is cathode
- During electrolysis, copper is oxidized to Cu2
at anode and then reduced to copper metal again
at cathode - Impurities such as silver and gold drop to bottom
as sludge which is then salvaged
88- Metal Plating
- Electroplating thin layers of decorative metal on
less expensive metal (silver and gold onto iron,
chromium on to car parts for decoration and
resistance to corrosion)
89Electrolysis of concentrated aqueous sodium
chloride solutions (brine) produces hydrogen and
hydroxide ions at cathode and chlorine gas at
anode
- If electrodes separated by porous membrane, H2,
NaOH, and Cl2 produced - If solution stirred, chlorine gas reacts with
sodium hydroxide to form sodium hypochlorite
(NaOCl) solution (bleach) - Electrolysis of molten NaCl produces sodium metal
and chlorine gas (Downs cell)
90Homework
- Read 17.5-17.8, pp. 846-866
- Q pp. 871-872, 74, 76, 78, 80, 84, 88 a (dont
forget water) - Do 1 additional exercise and 1 challenge problem
- Submit quizzes by email to me
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