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Electrochemistry-study of interchange of
chemical and electricalenergy
Generating electric current from spontaneous chemical
reactions and use of current to produce chemicalchange
Labs
#31 The Thermodynamics of the Dissolution of Borax#33 Electrolytic Cells; Avogadros Number
Chemical Equations
Chapter 12
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Oxidation-Reduction Reactions
(Redox Reactions)
Oxidation: (LEO)
Loss of electrons-atoms or ions undergo increasein oxidation state (is oxidized)
Electrons given to another atom which is beingreduced (reducing agentor reductant)
Reduction: (goes GER)
Gain of electrons-atoms/ions undergo decrease inoxidation state (is reduced)
Takes electrons away from another atom which isbeing oxidized (oxidizing agentor oxidant)
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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 toprovide useful work
Reactants separated by salt bridge/porouspartition
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
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Electrochemical cells
device 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 electronsto anode
Species in solution act as reducing agenthttp://college.hmco.com/chemistry/shared/media/animations/anodereaction.html
http://college.hmco.com/chemistry/shared/media/animations/cathodereaction.html
http://college.hmco.com/chemistry/shared/media/animations/anodereaction.htmlhttp://college.hmco.com/chemistry/shared/media/animations/cathodereaction.htmlhttp://college.hmco.com/chemistry/shared/media/animations/cathodereaction.htmlhttp://college.hmco.com/chemistry/shared/media/animations/anodereaction.html -
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Galvanic Cell (voltaic cell)
Chemical energyelectrical energy
Harnesses energy of spontaneous redox reactions
Physically separate chemicals in 2 half-reactions
Electrons generated by oxidation half-reaction flowthrough electrical conductor before being used inreduction half-reaction
Flow diverted through meters, motors, light bulbs toperform useful work before reaching destination
Current (defined by physicists as flow of positivecharge)always in opposite direction from flow ofelectrons (always from anode to cathode)
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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
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Porous barrier separate two
compartments
Allows for migration of
positive/negative ions
between half-cells,
completing electriccircuit
Problem with porous
barriers
Inside barriers, ionicsolutions mix
Has effect on operation
of cell
http://college.hmco.com/chemistry/shared/media/animations/electrochemicalhalf.html
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Galvanic 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 toreplace - ions being used up at
cathode (during reduction) Anions always migrate toward anode
Cations toward cathode
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Electron Flow
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The ANODE... The CATHODE...
Supplies electronsto external
circuit (wire)
Accepts electronsfrom external
circuit (wire)
Is negativepole of battery Is positivepole of battery
Is site of OXIDATION Is site of REDUCTION
Is written on left- hand sideif
convention is followed
Is written on right-hand sideif
convention is followed
Is half-cell with lowestelectrode
potential
Is half-cell with highestelectrode
potential
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Cell 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 createdbetween two dissimilar metals
Greater tendency or potential of two half-reactions tooccur 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 throughknown resistance (heat is produced)
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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 = cathodeanode
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 Nernstequation 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)
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Standard Reduction
PotentialsCell potentials can be
measured
Half-cell potential cannot4/7/2014 14
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Measuring Potential
Galvanic cell understandard conditions madeusing arbitrary standardhydrogen electrode(SHE) and test half-cell
w/different half reaction Assigned standard electrodepotential of exactly 0.00 V
Half reaction alwayswritten as reduction
Eo
values corresponding toreduction half-reactions withall solutes standardreduction potentials-Eocell
Under ideal conditions whereideal behavior is assumed
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Standard 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 # electronsexchanged but value of each Eoremains unchanged(intensive property-does not depend on how manytimes 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
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Each half-reactionassociated w/signednumerical value
More positive it is,greater oxidizing powerof redox half-reaction
More negative it is,greater reducing powerof reverse redox half-reaction
Larger differencebetween Ered values,larger Ecell
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(1 atm)
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Line Notation
Anodewritten first on left/cathodeon right Reactants written 1ston each side
Vertical baris 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 leftand/or right end of cell diagram
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Using 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 Eoof reactions not changed ***
Add two half reactions and E0values
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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Place the following in order of increasing
strength as oxidizing agents:
-0.44 0.954 2.87 0.22
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Describe a galvanic cell based
on the two half-reactions below:
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Homework:
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
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Cell Potential,Electrical Work,
and Free Energy
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Complete 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)
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Electric work and cell potential
Free energy change occurring during chemicalreaction is measure of maximum work thatsystem 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
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G (Gibbs free energy) is measure of spontaneity of
process occurring at constant T/P
emf, E, of redox reaction also indicates whetherreaction is spontaneous
From thermodynamics we know that
G =UTS +(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)
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Since both n/F are + + value of E leads tovalue ofG,
which indicates spontaneous reaction IfG and E have opposite signs, E
predicts direction of reaction If Eois positive, Gois negative (< 0)-
reaction spontaneous (has positive cellpotentials)
If Eois negative, Gois positive (> 0)-reaction is nonspontaneous (but isspontaneous in reverse direction)
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Relationship 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)
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emf is not converted to work with 100% efficiency Energy always lost as heat, but wmaxuseful for calculating
efficiency of conversion wmax = -qEmax
Relationship to free energy (energy driving reactiondue to movement of charged particles giving rise topotential difference)
wmax= EG
EG = -qEmax= -nFEmax
EG = -qEmax
EG0= -nFE0 When Ecellpositive (spontaneous), EG will be negative
(spontaneous), so there is agreement
Standard cell potential, Eocell, measured and standardelectrode 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 electrodesurface
Eoanode-standard reduction potential for reaction occurring atanode and represents its tendency to remove es from anode
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Dependence of CellPotential on
ConcentrationCell voltages at
nonstandardconcentrations (not 1 M)
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Nernst Equation-
way to relate E0at 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 deviatefrom standard temperature
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From thermodynamics, recall G = Go+ RTlnQ
If we divide everything bynF
sinceG =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 + bBcC + dD
Expressed in terms of base 10 rather than ln (standardconditions 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)
Calculating Nernst Equation
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Consider the Daniell Cell at 25 C
Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) Find cell potential at following conditions
when [Cu2+] = 1.00 M, [Zn2+] = 1.0109M and when [Cu2+] = 0.10 M, [Zn2+] =0.90 M.
Recallthat 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+] = 1.0109M
When [Cu2+] = 0.10 M, [Zn2+] = 0.90 M
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Depiction of concentration cell
E E
n
Q 0.0592
log
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Cell in which current flows due only to difference
in concentration of ion in 2 different
compartments of cell Le Chteliers 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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Calculate 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
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Show that voltage of electric cell is
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Show 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 = DEo0.0592/2 log [Mg2+]/[Ag+]2
2 Mg + 4 Ag+= 2 Mg2++ 4 Ag
DE = DEo0.0592/4 log [Mg2+]2/[Ag+]4
Simplified to the 1stequation, showing cell
potential DE not affected
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Calculation of Equilibrium
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Calculation of EquilibriumConstants for Redox
Reactions The standard reaction potential is related to theequilibrium 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 Eois positive, then K > 1 and forward reaction favored
If Eois negative, then K < 1 and reverse reaction is favored
log( ).
K nE
0 0592at 25 C
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The standard cell potential dE for the reaction
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pFe + 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 Kmay be calculated
using K= 10(nDE)/0.0592
= 10-11.93
= 1.2x10-12
= [Fe2+]/[Zn2+]. Since [Zn2+] = 1 M, it isevident that [Fe2+] = 1.2-12 M
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Homework:
Read 17.3-17.4, pp. 837-846
Q pp. 869-871, #38, 40, 46, 48, 54, 60, 66,
70
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Batteries 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)
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Lead-acid storage battery
As battery discharged, uses up sulfateions/electrodes become coated w/lead sulfate
Reverse reactions regenerate sulfate ion insolution/reduce amount of lead sulfate
contaminating electrode surfaces Each pair produces ~2 volts (6 pairs ofelectrodes used in 12-volt car battery)
When jump starting car, connect ground cable
on dead car to metallic contact away frombattery. Otherwise, could explode
Causes electrolysis of water/production ofH2/O2which could ignite
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C D C ll B tt
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Common Dry Cell Battery
(acid version)
Zinc-anode
Carbon rod in
contact with moist
paste-cathode
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Alkaline 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
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Mercury Battery
Fuel Cells
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Fuel Cells Galvanic cells where
reactants continuouslysupplied
Energy normally lost asheat is captured andused to produce anelectric current
Redox reaction Hydrogen oxidized at
anode and oxygenreduced at cathode toform water and electricity
2x as effective as gas, oilor coal-poweredgenerators in convertingchemical energy intoelectricity
Type of Battery Advantages Disadvantages
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Lead acid Rechargeable
Long life
12 V
Heavy
Contain acid
Weather issues
Alkaline/Dry cell InexpensiveNo toxic metals used
Lots of power (1.5 V down to 1.2 V)
Heavier
Lithium (solid state) Lightweight
Higher capacity (3.6-3.9 V)
Rechargeable
Longer lastingLess likely to leak/explode
Expensive
Fuel cells No rechargingNo harmful pollutants
high initial cost
Fuel not readily
available
Nickel-Cadmium (Ni-Cd) Fast/simple charge (rechargeable)High # charges
1.2 V continuously
Toxic metalsExpensive
Mercury More constant voltage (1.35 V)
Longer life
Lighter
More expensive
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Corrosionoxidation of metal
Oxidation of most metals by oxygen isspontaneous redox reactions
Many metals develop thin coating ofmetal oxide on outside that preventsfurther oxidation
Some metals, such as copper, gold,silver and platinum (noble metals), arerelatively difficult to oxidize
C i f I
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Corrosion of Iron Anodic regions
Regions of steel alloy where iron is more easilyoxidized
FeFe2+ + 2e-
Cathodic regions
Areas resistant to oxidation
Electrons flow from anodic regions & react
w/oxygen
O2+ 2H
2O + 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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Prevention of Corrosion
Coating w/metal that forms oxide coat to protect metalthat 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
Cathodic Protection
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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
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Electrolysis
Decomposition of substanceby electric current
Galvanic Cell Electrolytic Cell
Cathode + (reduction) - (reduction)Anode - (oxidation) + (oxidation)
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Electrolytic cells
Nonspontaneous reactions
Electrical energy required to induce reaction Two electrodes immersed in electrically conductive sample
Electrical voltage (>1.10 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
(a) Standard galvanic cell based on spontaneous
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( ) g
reaction
Zn + Cu2+Zn2++ Cu
(b) Standard electrolytic cell: Power source forcesopposite reaction
Cu + Zn2+Cu2++ Zn
What voltage is necessary to force the
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What 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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Calculating Quantity of Substance
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Calculating 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 thatwas produced/consumed at electrode
Convert moles of substance to desired units of
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A 40.0 amp current flowed through molten iron(III) chloride for 10.0 hours
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(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 ofelectrons 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).
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Calculating 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/consumedin moles
Write balanced half-reaction involved
Calculate number of moles of electrons required Convert moles of electrons into coulombs
Calculate time required
How long must a 20.0 amp current flow
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g p
through a solution of ZnSO4in order to
produce 25.00 g of Zn metal
Convert mass of Zn produced into moles using molar massof 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 usingstoichiometry 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
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Calculating Current Required
To determine amount of current necessary to
produce known quantity of substance in
given amount of time
Find quantity of substance produced/orconsumed 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 required4/7/2014
74
What current is required to produce 400.0 L of
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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
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How many grams of copper can be reduced by
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applying a 3.00 A current for 16.2 min to a
solution containing Cu2+ions?
TimecurrentCoulombsmoles e-moles Cug 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
Electrolysis can be used to separate mixture of
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Electrolysis can be used to separate mixture of
ions, if reduction potentials are fairly far apart
Remember, metal ion with highestreduction potential is easiest to reduce
Predict order of reduction and which of
following ions will reduce first at cathode ofelectrolytic cell: Ag+, Zn2+, IO3
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Electroplating
Deposit neutral metal atoms onelectrode by reducing metal
ions in solution
One metal coated with another
Presence of active electrodethat takes part in electrolysis
reaction
Anode-piece of plating metal
Cathode-object to be plated Plating solution is NiSO4because
SO42-ion does not participate in
plating reaction
How long must a current of 5.00 A be
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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
El t l i f W thttp://college.hmco.com/chemistry/shared/media/animations/electrolysisofwater.html
http://college.hmco.com/chemistry/shared/media/animations/electrolysisofwater.htmlhttp://college.hmco.com/chemistry/shared/media/animations/electrolysisofwater.html -
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Electrolysis of Water 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
What volume of H2(g) and O2(g) is produced by
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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 2:1 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 H20.167 L O2
El t l i f lt lt
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Electrolysis of molten salts
NaCl Good conductor as
liquid
Melting salt freesions
Makes it electricallyconductive
Ions of oppositecharge migrate tothese electrodesand react
Electrolysis of aqueous solutions
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Electrolysis 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 (2H2OO2+ 4H+ + 4e-)
To determine which will occur at anode
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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 aqueouselectrolysis, 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 AlBr3will all produce Br2at anode
Commercial
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Commercial
ElectrolyticProcesses
Since metals are easily oxidized,most found as ores, mixtures of
ionic compounds. Au, Ag, and Ptare more difficult to oxidize, so often
found as pure metals.
Abundance of elements on earth:
1st Oxygen
2nd Silicon
3rd Aluminum (very active metal so difficult
and expensive originally to purify)
Production of aluminum from molten-saltl t l i
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electrolysis
(purification of aluminum from bauxite ore)
Hall-Heroult process
Electrorefining (purifying) metals
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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
Metal Plating
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Metal Plating
Electroplating thin layers of decorative metal
on less expensive metal (silver and gold ontoiron, chromium on to car parts for decoration
and resistance to corrosion)
Electrolysis of concentrated aqueous sodium chloride
l i (b i ) d h d d h d id
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solutions (brine) produces hydrogen and hydroxide
ions at cathode and chlorine gas at anode
If electrodes separated byporous membrane, H2,
NaOH, and Cl2 produced
If solution stirred, chlorine
gas reacts with sodiumhydroxide to form sodium
hypochlorite (NaOCl)
solution (bleach)
Electrolysis of moltenNaCl produces sodium
metal and chlorine gas
(Downs cell)
H k
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Homework:
Read 17.5-17.8, pp. 846-866Q 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:
http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xml
http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xml
http://www.cengage.com/chemistry/book_content/0547125321 zumdahl/ace/launch ace html?folder path=/chemistry
http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xml