Electrochemistry

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Electrochemistry
Chapter 17
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Contents

• Galvanic cells
• Standard reduction potentials
• Cell potential, electrical work, and free energy
• Dependence of cell potential on concentration
• Batteries
• Corrosion
• Electrolysis
• Commercial electrolytic processes

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Oxidation reduction reactions
17.1 Galvanic Cells

• Oxidation reduction reactions
• involve a transfer of electrons.
• Oxidation Involves Loss of electrons
• Increase in the oxidation number
• Reduction Involves Gain electrons
• Decrease in the oxidation number

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8HMnO4- 5Fe2 Mn2 5Fe3 4H2O
Example

• If we break the reactions into half reactions.
• 8HMnO4-5e- Mn2 4H2O (Red)
• 5(Fe2 Fe3 e- ) (Ox)
• Electrons are transferred directly.
• This process takes place without doing useful
  work

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• When the compartments of the two beakers are
  connected as shown the reaction starts
• Current flows for an instant then stops
• No flow of electrons in the wire, Why?
• Current stops immediately because charge builds
  up.

H MnO4-
Fe2
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Galvanic Cell
Solutions must be connected so ions can flow to
keep the net charge in each compartment zero
Salt Bridge allows ions to flow without extensive
mixing in order to keep net charge zero.
Electrons flow through the wire from reductant to
oxidant
Oxidant
reductant
H MnO4-
Fe2
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Porous Disk
H MnO4-
Fe2
8
e-
e-
e-
e-
Anode
Cathode
Fe2
e-
e-
Reducing Agent
Oxidizing Agent MnO4-
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_______ __________
_______ __________
Electrochemical Cells
Spontaneous redox reaction
19.2
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• Thus a Galvanic cell is a device in which a
  chemical energy is changed to electrical energy
• The electrochemical reactions occur at the
  interface between electrode and solution where
  the electron transfer occurs
• Anode the electrode compartment at which
  oxidation occurs
• Cathode the electrode compartment at which
  reduction occurs

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Cell Potential

• Oxidizing agent pulls the electrons
• Reducing agent pushes the electrons
• The total push or pull (driving force) is
  called the cell potential, Ecell
• Also called the electromotive force (emf)
• Unit is the volt(V)
• 1 joule of work/coulomb of charge
• Measured with a voltmeter

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Measuring the cell potential

• Can we measure the total cell potential??
• A galvanic cell is made where one of the two
  electrodes is a reference electrode whose
  potential is known.
• Standard hydrogen electrode (H 1M
• and the H2 (g) is at 1 atm) is used as a
• reference electrode and its potential was
  assigned to be zero at 25 0C.

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Standard Hydrogen Electrode

• This is the reference all other oxidations are
  compared to
• Eº 0
• (º) indicates standard states of 25ºC, 1 atm, 1 M
  solutions.

1 atm
H2
H Cl-
1 M HCl
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0.76
H2
Cathode
Anode
H Cl-
Zn2 SO4-2
1 M HCl
1 M ZnSO4
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Standard Electrode Potentials
Zn (s) Zn2 (1 M) H (1 M) H2 (1 atm) Pt
(s)
Anode (oxidation)
Cathode (reduction)
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17.2 Standard Reduction Potentials, E?

• The E? values corresponding to reduction half-
• reactions with all solutes at 1M and all
  gases at 1
• atm.
• E? can be measured by making a galvanic cell
  in
• which one of the two electrodes is the
  Standard
• Hydrogen electrode, SHE, whose E? 0 V
• The total potential of this cell can be
  measured
• experimentally
• However, the individual electrode potential can
  
• not be measured experimentally.
• Why?

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• If the cathode compartment of the cell is SHE,
  then the half reaction would be
• 2H 2e? ? H2 (g) Eo 0V
• And the anode compartment is Zn metal in Zn2,
  (1 M) then the half reaction would be
• Zn ? Zn2 2e?
• The total cell potential measured experimentally
  was found to be 0.76 V
• Thus, 0.76 V was obtained as a result of this
  calculation
• Eº cell EºZn Zn2 Eº H H2
• 0.76 V 0.76 V 0 V

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Standard Reduction Potentials

• The E? values corresponding to reduction half-
• reactions with all solutes at 1M and all
  gases at
• 1 atm. can be determined by making them half
  
• cells where the other half is the SHE.
• E0 values for all species were determined as
• reduction half potentials and tabulated.
  For
• example
• Cu2 2e? ? Cu E? 0.34 V
• SO42? 4H 2e? ? H2SO3 H2O E? 0.20 V
• Li e- ? Li E? -3.05 V

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Some Standard Reduction Potentials

• Li e- ---gt Li -3.045 v
• Zn2 2 e- ---gt Zn -0.763v
• Fe2 2 e- ---gt Fe -0.44v
• 2 H(aq) 2 e- ---gt H2(g) 0.00v
• Cu2 2 e- ---gt Cu 0.337v
• O2(g) 4 H(aq) 4 e- ---gt 2
  H2O(l) 1.229v
• F2 2e- ---gt 2 F- 2.87v

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Standard Reduction Potentials at 25C
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• E0 is for the reaction as written
• The more positive E0 the greater the tendency for
  the substance to be reduced
• The more negative E0 the greater the tendency for
  the substance to be oxidized
• Under standard-state conditions, any species on
  the left of a given half-reaction will react
  spontaneously with a species that appears on the
  right of any half-reaction located below it in
  the table

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• The half-cell reactions are reversible
• The sign of E0 changes when the reaction is
  reversed
• Changing the stoichiometric coefficients of a
  half-cell reaction does not change the value of E0

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Can Sn reduce Zn2 under standard-state
conditions?
Look up the Eº values in in the table of
reduction potentials
Zn2 2 e- ---gt Zn(s) -0.763v
Sn2 2 e- ---gt Sn -0.143v

• How do we find the answer?
• Look up the Eº values in in the table of
  reduction potentials
• \Which reactions in the table will reduce
  Zn2(aq)?

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Standard cell potential

• Zn(s) Cu2 (aq) Zn2(aq) Cu(s)
• The total standard cell potential is the sum of
  the potential at each electrode.
• Eº cell EºZn Zn2 Eº Cu2 Cu
• We can look up reduction potentials in a table.
• One of the reactions must be reversed, in order
  to change its sign.

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Standard Cell Potential

• Determine the cell potential for a galvanic cell
  based on the redox reaction.
• Cu(s) Fe3(aq) Cu2(aq) Fe2(aq)
• Fe3(aq) e- Fe2(aq) Eº 0.77 V ?
• Cu2(aq)2e- Cu(s) Eº 0.34 V
• Cu(s) Cu2(aq)2e- Eº -0.34 V
• 2Fe3(aq) 2e- 2Fe2(aq) Eº 0.77 V ?
• Eo cell EoFe3 Eo Fe2 EoCuEo Cu2
• Eo cell 0.77 (-0.34) o.43 V

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• The total reaction
• Cu(s) Cu2(aq)2e- Eº -0.34 V
• 2Fe3(aq) 2e- 2Fe2(aq) Eº 0.77 V ?

Cu(s) 2Fe3(aq) Cu2 2Fe2
Eºcell 0.43 V
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Line Notation

• Solid½Aqueous½½Aqueous½solid
• Anode on the left½½Cathode on the right
• Single line different phases.
• Double line porous disk or salt bridge.
• Zn(s)½Zn2(aq)½½Cu2½Cu
• If all the substances on one side are aqueous, a
  platinum electrode is indicated.
• For the last reaction
• Cu(s)½Cu2(aq)½½Fe2(aq),Fe3(aq)½Pt(s)

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Complete description of a Galvanic Cell

• The reaction always runs spontaneously in the
  direction that produces a positive cell
  potential.
• Four parameters are needed for a complete
  description
• Cell Potential
• Direction of flow
• Designation of anode and cathode
• Nature of all the components- electrodes and ions

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Exercise

• Describe completely the galvanic cell based on
  the following half-reactions under standard
  conditions.
• MnO4- 8 H 5e- Mn2 4H2O Eº1.51
• Fe3 3e- Fe(s) Eº0.036V
• Write the total cell reaction
• Calculate Eo cell
• Define the cathode and anode
• Draw the line notation for this cell

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17.3 Cell potential, electrical work and
free energy

• The work accomplished when electrons are
  transferred through a wire depends on the push
  (thermodynamic driving force) behind the
  electrons
• The driving force (emf) is defined in terms of
  potential difference (in volts) between two
  points in the circuit
• emf potential difference (V)
• work (J) / Charge(C)

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• The work done by the system has a
• ve sign
• Potential produced as a result of doing a work
  should have a ve sign
• The cell potential, E, and the work, w, have
  opposite signs.
• Relationship between E and w can be expressed as
  follows
• E work done by system / charge
• ( )

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• Charge is measured in coulombs. Thus,
• -w qE
• Faraday 96,485 C/mol e-
• q nF moles of e- x charge/mole e-
• w -qE -nFE DG
• Thus, DG -nFE and
• DGo -nFEo

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Potential, Work, DG and spontaneity

• DGº -nFE º
• if E º gt 0, then DGº lt 0 spontaneous
• if E º lt 0, then DGº gt 0 nonspontaneous
• In fact, the reverse process is spontaneous.

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Spontaneity of Redox Reactions
DG -nFEcell
n number of moles of electrons in reaction
96,500 C/mol
DG0 -RT ln K
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Spontaneity of Redox Reactions
If you know one, you can calculate the other If
you know K, you can calculate ?Eº and ?Gº If you
know ?Eº, you can calculate ?Gº
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Spontaneity of Redox Reactions
Relationships among ?G º, K, and Eºcell
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Calculate DG0 for the following reaction at 250C.
2Al3(aq) 3Mg(s) 2Al(s) 3Mg2(aq)
3 (Mg Mg2 2e-)
Oxidation
n ?
Reduction
0
0
___ X (96,500 J/V mol) X ___ V
DG0 _______ kJ/mol
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17.4 Dependence of Cell Potential on Concentration

• Qualitatively we can predict direction of change
  in E from LeChâtelier pinciple
• 2Al(s) 3Mn2(aq) 2Al3(aq)
  3Mn(s) Eo cell 0.48 V
• Predict if Ecell will be greater or less than
  Eºcell for the following cases
• if Al3 1.5 M and Mn2 1.0 M
• if Al3 1.0 M and Mn2 1.5M
• An increase in conc. of reactants would favor
  forward reaction thus increasing the driving
  force for electrons i.e. Ecell becomes gt Eo cell

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Concentration Cell both compartments contain
same components but at different concentrations
Half cell potential are not identical Because
the Ag Conc. On both sides are not same
Eright gt Eleft

• To make them equal, Ag
• On both sides should same
• Electrons move from left to
• right

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The Nernst Equation Effect of Concentration on
Cell Emf
DG DG0 RT ln Q
DG -nFE
-nFE -nFE0 RT ln Q
Nernst equation
At 298K
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The Nernst Equation

• As reactions proceed concentrations of products
  increase and reactants decrease.
• When equilibrium is reached
• Q K Ecell 0
• and ?G 0 (the cell no longer has
• the
  ability to do work)

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Predicting spontaneity using Nernst equation

• Qualitatively we can predict the direction of
  change in E from Lechatelier principle
• Find Q
• Calculate E
• E gt 0 the reaction is spontaneous to the
• right
• E lt 0 the reaction is spontaneous to the
• left

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Will the following reaction occur spontaneously
at 250C if Fe2 0.60 M and Cd2 0.010 M?
Fe2 (aq) Cd (s) Fe (s) Cd2 (aq)
Oxidation
n 2
Reduction
2e- Fe2 2Fe
E ____________
E ___ 0
________________
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Exercise- p. 843

• Determine the cell potential at 25oC for the
  following cell, given that
• 2Al(s) 3Mn2(aq) 2Al3(aq) 3Mn(s)
• Mn2 0.50 M Al31.50 M E0 cell 0.4
• Always we have to figure out n from the balanced
  equation
• 2(Al(s) Al3(aq) 3e-)
• 3(Mn2(aq) 2e- Mn(s))

n 6
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Calculation of Equilibrium Constants for redox
reactions
At equilibrium, Ecell 0 and Q K.
Then,
at 25 oC
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What is the equilibrium constant for the
following reaction at 250C? Fe2 (aq) 2Ag (s)
Fe (s) 2Ag (aq)
Oxidation
n ___
Reduction
K ________________
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17.5 Batteries
Batteries are Galvanic Cells

• Lead-Storage Battery
• A 12 V car battery consists of 6 cathode/anode
  pairs each producing 2 V.
• Cathode PbO2 on a metal grid in sulfuric acid
• PbO2(s) SO42-(aq) 4H(aq) 2e- ?
• 
  PbSO4(s) 2H2O(l)
• Anode Pb
• Pb(s) SO42-(aq) ? PbSO4(s) 2e-

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Lead storage battery
Anode
Pb (s) SO2- (aq) PbSO4 (s) 2e-
4
Cathode
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• Lead-Storage Battery
• The overall electrochemical reaction is
• PbO2(s) Pb(s) 2SO42-(aq) 4H(aq) ?
• 2PbSO4(s)
  2H2O(l)
• for which
• E?cell E?red(cathode) - E?red(anode)
• (1.685 V) - (-0.356 V)
• 2.041 V.
• H2SO4 is consumed while the battery is
  discharging
• H2SO4 is 1.28g/ml and must be kept
• Water is depleted thus the battery should be
  topped off always

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Dry cell Batteries
Anode
Cathode
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• Dry Cell Battery
• Anode Zn cap
• Zn(s) ? Zn2(aq) 2e-
• Cathode MnO2, NH4Cl and C paste
• 2NH4(aq) 2MnO2(s) 2e- ? Mn2O3(s) 2NH3(aq)
  2H2O(l)
• Total reaction
• Zn NH4 MnO2 Zn2 NH3 H2O
• This cell produces a potential of about 1.5 V.
• The graphite rod in the center is an inert
  cathode.

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Alkaline Cell Battery

• For an alkaline battery, NH4Cl is replaced with
  KOH.
• Anode oxidation of Zn
• Zn(s) 2OH- ? ZnO H2O 2e-
• Cathode reduction of MnO2.
• 2MnO2 H2O 2e- ? Mn2O3 2OH-
• Total reaction
• Zn(s) 2 MnO2(s) ---gt ZnO(s) Mn2O3(s)
• It lasts longer because Zn anode corrodes
• less rapidly than under acidic conditions.

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Alkaline Battery
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Nickel-Cadmium (Ni-Cad) Battery

• Anode Cd(s) 2OH- Cd(OH)2
  2e-
• Cathode NiO2 2H2O 2e-
  Ni(OH)2 2OH-
• NiO2 Cd 2H2O Cd(OH)2 Ni(OH)2
  
• NiCad 1.25 v/cell
• The products adhere to the
• electrodes thus the battery can
• be recharged indefinite number
• of times.

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Fuel Cells
A fuel cell is a galvanic cell that requires a
continuous supply of reactants to keep functioning
2H2 (g) 4OH- (aq) 4H2O (l) 4e-
Anode
Cathode
O2 (g) 2H2O (l) 4e- 4OH- (aq)
2H2 (g) O2 (g) 2H2O (l)
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17.6 Corrosion

• Rusting - spontaneous oxidation of metals.
• Most metals used for structural purposes have
  reduction potentials that are less positive than
  O2 . (They are readily oxidized by O2)
• Fe2 2e- Fe Eº - 0.44 V
• O2 2H2O 4e- 4OH- Eº 0.40 V
• When a cell is formed from these two half
  reactions a cell with ve potential will be
  obtained
• Au, Pt, Cu, Ag are difficult to be oxidized
  (noble metals)
• Most metals are readily oxidized by O2 however,
  this process develops a thin oxide coating that
  protect the internal atoms from being further
  oxidized.
• Al that has Eo -1,7V is easily oxidized. Thus,
  it is used for making the body of the airplane.

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Electrochemical corrosion of iron
Salt speeds up process by increasing conductivity
Water
Cathodic area
Rust
Anodic area
e-
Iron dissolves forming a pit
Fe Fe2 2e- Anodic reaction
O2 2H2O 4e- 4OH- cathodic reaction
Fe2 (aq) O2(g) (4-2n) H2O(l)
2F2O3(s).nH2O (s) 8H(aq)
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• Fe on the steel surface is oxidized (anodic
  regions)
• Fe Fe2 2e- Eº- 0.44 V
• e-s released flow through the steel to the areas
  that have O2 and moisture (cathodic regions).
  Oxygen is reduced
• O2 2H2O 4e- 4OH- Eº 0.40 V
• Thus, in the cathodic region Fe2 will react
  with O2
• The total reaction is
• Fe2 (aq) O2(g) (4-2n) H2O(l)
• 2F2O3(s).nH2O
  (s) 8H(aq)
• Thus, iron is dissolved to form pits in steel
• Moisture must be present to act as the salt
  bridge
• Steel does not rust in the dry air
• Salts accelerates the process due to the increase
  in conductivity on the surface

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Preventing of Corrosion

• Coating to keep out air and water.
• Galvanizing - Putting on a zinc coat
• Fe Fe2 2e- Eoox 0.44V
• Zn Zn2 2e- Eoox 0.76 V
• Zn has a more positive oxidation potential than
  Fe, so it is more easily oxidized.
• Any oxidation dissolves Zn rather than Fe
• Alloying is also used to prevent corrosion.
  stainless steel contains Cr and Ni that make make
  steel as a noble metal
• Cathodic Protection - Attaching large pieces of
  an active metal like magnesium by wire to the
  pipeline that get oxidized instead. By time Mg
  must be replaced since it dissolves by time

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Cathodic Protection of an Underground Pipe
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Cathodic Protection of an Iron Storage Tank
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17.7 Electrolysis

• Running a galvanic cell backwards.
• Put a voltage bigger than the galvanic potential
  and reverse the direction of the redox reaction.
• Electrolysis Forcing a current through a cell to
  produce a chemical change for which the cell
  potential is negative.
• That is causing a nonspontaneous reaction to
  occur
• It is used for electroplating.

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1.10
e-
e-
Galvanic cell based on spontaneous reaction Zn
Cu2 Zn2 Cu
Zn
Cu
1.0 M Cu2
1.0 M Zn2
Cathode
Anode
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Electrolytic cell
e-
A battery gt1.10V
e-
Zn2 Cu Zn Cu2
Zn
Cu
1.0 M Cu2
1.0 M Zn2
Anode
Cathode
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Electrolytic Cell
Galvanic Cell
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Calculating plating

• How much chemical change occurs with the flow of
  a given current for a specified time?
• Determine quantity of electrical charge in
  coulombs
• Measure current, I (in amperes) per a period of
  time
• 1 amp 1 coulomb of charge per second
• coulomb of charge amps X seconds
• q I x t
• q/nF moles of metal
• Mass of plated metal can then be calculated

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Exercise

• Calculate mass of Cu that is plated out when a
  current of 10.0 amps is passed for 30.0 min
  through a solution of Cu2

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Excercise

• How long must 5.00 amp current be applied to
  produce 10.5 g of Ag from Ag?

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Electroplating

• How many grams of chromium can be plated from a
  Cr6 solution in 45 minutes at a 25 amp current?

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• How many grams of chromium can be plated from a
  Cr6 solution in 45 minutes at a 25 amp current?
• (45 min)
• g Cr ------------

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• (45 min)(60 sec)
• g Cr ---------------------
• (1 min)

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• How many grams of chromium can be plated from
  a Cr6 solution in 45 minutes at a 25 amp
  current?
• (45) (60 sec) (25 amp)
• g Cr ---------------------------
• (1)

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• (45)(60 sec)(25 amp)(1 C)
• g Cr -----------------------------
• (1) (1 amp sec)

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• Faradays constant
• (45)(25)(60)(1 C)(1 mol e-)
• g Cr ----------------------------------
• (1)(1)(96,500 C)

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• (45)(60)(25)(1)(1 mol e-)(1 mol Cr)
• g Cr -----------------------------------------
  --------
• (1)(1)(96,500) (6 mol e-)

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• (45)(60)(25)(1)(1 mol e-)(52 g Cr)
• g Cr ------------------------------------------
  -
• (1)(1)(96,500) (1 mol Cr)

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Electroplating

• How many grams of chromium can be plated from
  a Cr6 solution in 45 minutes at a 25 amp
  current?
• (45)(60)(25)(1)(1 mol e-)(52 g Cr)
• g Cr ------------------------------------------
  -
• (1)(1)(96,500)(6 mol e-)
• 58 g Cr

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Michael Faraday Lecturing at the Royal
Institution Before Prince Albert and Others (1855)
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Electrolysis of Water
83
The Electrolysis of Water Produces Hydrogen Gas
at the Cathode (on the Right) and Oxygen Gas at
the Anode (on the Left)
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Electrolysis of water
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Other uses of electrolysis

• Separating mixtures of ions.
• More positive reduction potential means the
  reduction reaction proceeds forward.
• We want the reverse.
• Most negative reduction potential is easiest to
  plate out of solution.

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17.8 Commercial electrolytic processes
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A Schematic Diagram of an Electrolytic Cell for
Producing Aluminum by the Hall-Heroult Process.
To reduce mp of Al From 2000 to 1000
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The Downs Cell for the Electrolysis of Molten
Sodium Chloride
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The Mercury Cell for Production of Chlorine and
Sodium Hydroxide
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Metal Plating
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