ELECTROCHEMISTRY Chapter 20

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ELECTROCHEMISTRYChapter 20
2
TRANSFER REACTIONS

• Atom/Group transfer
• HCl(aq) H2O(liq) ? Cl-(aq) H3O(aq)
• Electron transfer
• Cu(s) 2 Ag(aq) ? Cu2(aq) 2 Ag(s)

3
Electron Transfer Reactions

• Electron transfer reactions are
  oxidation-reduction or redox reactions.
• Redox reactions can result in the generation of
  an electric current or be caused by imposing an
  electric current.
• Therefore, this field of chemistry is often
  called ELECTROCHEMISTRY.

4
Why Study Electrochemistry?

• Batteries
• Corrosion
• Industrial production of chemicals such as
  Cl2, NaOH, F2 and Al
• Biological redox reactions

The heme group
5
Review of Terminology for Redox Reactions

• OXIDATIONloss of electron(s) by a species
  increase in oxidation number.
• REDUCTIONgain of electron(s) decrease in
  oxidation number.
• OXIDIZING AGENTelectron acceptor it is reduced.
• REDUCING AGENTelectron donor it is oxidized.

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OXIDATION-REDUCTION REACTIONS

• Direct Redox Reaction
• Oxidizing and reducing agents in direct contact.
• Cu(s) 2 Ag(aq) ? Cu2(aq) 2 Ag(s)

7
OXIDATION-REDUCTION REACTIONS

• Indirect Redox Reaction
• A battery functions by transferring electrons
  through an external wire from the reducing agent
  to the oxidizing agent.

8
Electrochemical Cells

• An apparatus that allows a redox reaction to
  occur by transferring electrons through an
  external connector.
• Product favored reaction ? voltaic or galvanic
  cell - ? produces electric current
• Reactant favored reaction ? electrolytic cell ?
  electric current used to cause a not spontaneous
  chemical change.

Batteries are voltaic cells
9
Electrochemistry
Alessandro Volta, 1745-1827, Italian scientist
and inventor.
Luigi Galvani, 1737-1798, Italian scientist and
inventor.
10
Balancing Equations for Redox Reactions

• Some redox reactions have equations that must be
  balanced by special techniques.
• MnO4-(aq) 5 Fe2(aq) 8 H(aq)
  ? Mn2 (aq) 5 Fe3(aq) 4
  H2O(liq)

Mn 7
Fe 2
Fe 3
Mn 2
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Cu Ag ?give? Cu2 Ag
Balancing Equations
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Balancing Equations

• Step 1 Divide the reaction into
  half- reactions, one for oxidation and the
  other for reduction.
• Ox Cu ? Cu2
• Red Ag ? Ag
• Step 2 Balance each for mass. Already done in
  this case.
• Step 3 Balance each half-reaction for charge
  by adding electrons.
• Ox Cu ? Cu2 2e-
• Red Ag e- ? Ag

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Balancing Equations

• Step 4 Multiply each half-reaction by a factor
  so that the reducing agent supplies as many
  electrons as the oxidizing agent requires.
• Reducing agent Cu ? Cu2 2e-
• Oxidizing agent 2 Ag 2 e- ? 2 Ag
• Step 5 Add half-reactions to give the overall
  equation.
• Cu 2 Ag ? Cu2 2Ag e-s are
  cancelled
• The equation is now balanced for both charge and
  mass.

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Reduction of VO2 with Zn
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Balancing Equations

• Balance the following in acid solution
• VO2 Zn ? VO2 Zn2
• Step 1 Write the half-reactions
• Ox Zn ? Zn2
• Red VO2 ? VO2
• Step 2 Balance each half-reaction for mass.
• Ox Zn ? Zn2
• Red

VO2 ? VO2 H2O
2 H
Add H2O on O-deficient side and add H on other
side for H-balance.
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Balancing Equations

• Step 3 Balance half-reactions for charge.
• Ox Zn ? Zn2 2e- Zn looses 2 e-
• 5
  4 V gains one e-
• Red e- 2 H VO2 ? VO2 H2O
• Step 4 Multiply by an appropriate factor.
• Ox Zn ? Zn2 2e-
• Red 2e- 4 H 2 VO2 ? 2 VO2
  2 H2O
• Step 5 Add balanced half-reactions.
• Zn 4 H 2 VO2 ? Zn2 2
  VO2 2 H2O

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Tips on Balancing Equations

• Never add O2, O atoms, or O2- to balance oxygen.
• Balance O with OH- or H2O.
• Never add H2 or H atoms to balance hydrogen.
• Balance H with H and H2O in acid
• or OH-/H2O in base.

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Tips on Balancing Equations
Equations that include oxoanions like SO42-,
NO3-, ClO- , CrO42-, and MnO4-, also fall into
this category.
Be sure to write the correct charges on all the
ions.

• Check your work at the end to make sure mass and
  charge are balanced.
• PRACTICE!!!!!!!!!!!

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VOLTAIC CELLS

• Use a chemical rxn to produce an electric
  current. This is the conversion of chemical
  energy into electrical energy.
• 1st Law of thermodynamics.

The ZnZn2 and CuCu2 Cell
Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
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CHEMICAL CHANGE ? ELECTRIC CURRENT
We place a Zn(s) strip in a solutions tha
contains Cu2 ions. With time, Cu plates out
onto Zn metal strip, and Zn strip disappears.

• Oxidation Zn(s) ? Zn2(aq) 2e-
• Reduction Cu2(aq) 2e- ? Cu(s)
• --------------------------------------------------
  ------
• Cu2(aq) Zn(s) ? Zn2(aq) Cu(s)

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CHEMICAL CHANGE ?ELECTRIC CURRENT
Zn(s) ? Zn2(aq) 2e-

• To obtain a useful current, we separate the
  oxidizing and reducing agents so that electron
  transfer occurs thru an external wire.

This is accomplished in a GALVANIC or VOLTAIC
cell. A group of such cells is called a battery.
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Zn ? Zn2 2e-
Cu2 2e- ? Cu
Oxidation Anode Negative
Reduction Cathode Positive
? Anions Cations ?

• Electrons travel thru external wire.
• Salt bridge (such as NaNO3) allows anions and
  cations to move between electrode compartments.

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Electrochemical Cell
Electrons move from anode to cathode in the
wire. Anions cations move thru the salt bridge
to complete the circuit by cancelling charge
excess.
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Terms Used for Voltaic Cells
Figure 20.6
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CELL POTENTIAL, E
Zn and Zn2, anode
Cu and Cu2, cathode

• Electrons are driven from anode to cathode by
  an electromotive force or emf.
• For Zn/Cu cell, this is indicated by a voltage of
  1.10 V at 25 C and when Zn2 Cu2 1.0 M
  (or another case where the two concentrations are
  exactly the same).

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CELL POTENTIAL, E

• For Zn/Cu cell, the potential is 1.10 V at 25 C
  when Zn2 and Cu2 1.0 M or other equal M
• This is the STANDARD CELL POTENTIAL, Ecello
• a quantitative measure of the tendency of
  reactants to form products when all are in their
  standard states at 25 C.

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Calculating Cell Voltage

• Balanced half-reactions can be added together to
  get the overall, balanced equation.

Zn(s) ? Zn2(aq) 2e- Cu2(aq) 2e- ?
Cu(s) --------------------------------------------
Cu2(aq) Zn(s) ? Zn2(aq) Cu(s)
If we know Eo for each half-reaction, we add them
to get Eo for overall reaction.
-need Eo for Zn Cu half-cells
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Zn/Zn2 half-cell hooked up to a SHE (Standard
Hydrogen Electrode). Eo for the cell 0.76 V
Supplier of electrons
Acceptor of electrons
2 H 2e- ? H2 Reduction Cathode
Zn ? Zn2 2e- Oxidation Anode
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Overall reaction is reduction of H by Zn
metal. Zn(s) 2H(aq) ? Zn2(aq) H2(g) Eo
0.76 V Therefore, Eo for Zn ? Zn2 (aq) 2e-
is 0.76 V Because 2H 2e- ? H2(g) E 0.00
V Reference Zn is a (better) (poorer)? reducing
agent than H2.
(better)
30
Cu/Cu2 half-cell hooked up to a SHE. Eo for the
cell 0.34 V

• Eo 0.34 V

Positive
Negative
Acceptor of electrons
Supplier of electrons
Cu2 2e- ? Cu Reduction Cathode
H2 ? 2 H 2e- Oxidation Anode
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Cu/Cu2 and H2/H Cell

• Overall reaction is reduction of Cu2 by H2 gas.
• Cu2 (aq) H2(g) ? Cu(s) 2 H(aq)
• Measured Eo 0.34 V
• Therefore, Eo for Cu2 2e- ? Cu is 0.34 V,
• because H2(g) ? 2H 2e- E 0.00 V Ref.
  electrode

H2 is now a better reducing agent than Cu! (Zn is
a better reducing agent than H2).
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Zn/Cu Electrochemical Cell
Anode, negative, source of electrons
Cathode, positive, sink for electrons

• Zn(s) ? Zn2(aq) 2e- Eo 0.76 V
• Cu2(aq) 2e- ? Cu(s) Eo 0.34 V
• --------------------------------------------------
  -------------
• Cu2(aq) Zn(s) ? Zn2(aq) Cu(s)
• Eo (calcd) 1.10 V

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Uses of Eo Values

• Organize half-reactions by relative ability to
  act as oxidizing/reducing agents. Half-rxns are
  written as reduction rxns!

Cu2(aq) 2e- ? Cu(s) Eo 0.34 V Zn2(aq)
2e- ? Zn(s) Eo 0.76 V
Note that when a reaction is reversed the sign of
E is reversed! (-) to (), () to
(-)
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Uses of Eo Values

• Organize half-reactions by relative ability to
  act as oxidizing/reducing agents
• Table 20.1
• Use this to predict direction of redox reactions
  and overall cell potentials.

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? oxidants
(p. 967)
?Reference electrode
reductants ?
36
Potential Ladder for Reduction Half-Reactions
Figure 20.14
Best oxidizing agents
Best reducing agents
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Using Standard Potentials, EoTable 20.1
Which is the best oxidizing agent look at
reductionO2 (1.23 V) H2O2 (1.77 V) or Cl2 (1.36
V)? All are reduction E, and H2O2 2H 2e-
? 2H2O has the most positive one. Then,
H2O2 (1.77 V) is the best oxidizing agent

• Which is the best reducing agent look at
  oxidation
• Hg (0.79 V), Al (-1.66 V), or Sn (-0.14 V)?
• Hg(l) ? Hg2 2e- E -0.79 V
• Al(s) ? Al3 3e- E 1.66 V
• Sn(s) ? Sn2 2e- E 0.14 V

Al (1.66 V)
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Using Standard Potentials, EoTable 20.1
Which substance is the best oxidizing
agent?Cr2O72- 6e- 14H ? 2Cr3 7H2O (1.33
V) O2 4e- 4H ? 2H2O (1.23 V) Fe3 e-
? Fe2 (0.77 V) All are reductions, then
the most positive E is for
Cr2O72- , the best oxidizing agt.

• Which element/ion is the best reducing agent?
• Fe3 e- ? Fe2 (0.77 V)
• I2 2e- ? 2I- (0.54 V)
• Sn4 2e- ? Sn2 (0.15 V)
• We have to reverse the E -0.77 V, -0.54 V,
  -0.15 V

Sn2 is the best
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TABLE OF STANDARD REDUCTION POTENTIALS













2


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Standard Redox Potentials, Eo

• Any substance on the right will reduce any
  substance HIGHER than it on the LEFT.
• Zn can reduce H and Cu2.
• H2 can reduce Cu2 but not Zn2
• Cu cannot reduce H or Zn2.

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H2(g) H(aq) Cu2(aq) Cu(s)
Cathode Positive
Anode Negative
Electrons lt----------
H2 ? 2 H 2 e- (0.0 V) or 2 H 2e- ? H2
(0.0 V)
Cu2 2e- ? Cu (0.34 V) Or Cu ? Cu2 2 e-
(-0.34 V)
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H2(g) H(aq) Cu2(aq) Cu(s)
Cathode Positive
Anode Negative
Electrons lt----------
Cu2 2e- ? Cu, 0.34 V
H2 ? 2 H 2 e-
The sign of the electrode in Table 20.1 is the
polarity when hooked to the H/H2 half-cell.
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Standard Redox Potentials, Eo
Ox. agent
Red. agent
Any substance on the right will reduce any
substance higher than it on the left.

• Northwest-southeast rule product-favored
  reactions occur between
• reducing agent at southeast corner
• oxidizing agent at northwest corner

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Standard Redox Potentials, Eo
Ox. agent
Red. agent
Expt 29 Zn will reduce Ni2, Cu2 Ni will
reduce Cu2.

• Northwest-southeast rule product-favored
  reactions occur between
• reducing agent at southeast corner
• oxidizing agent at northwest corner

45
Using Standard Potentials, EoTable 20.1

• In which direction do the following reactions go?
• Cu(s) 2 Ag(aq) ? Cu2(aq) 2 Ag(s) 0.46
  V
• Cu2(aq) Zn(s) ? Cu(s) Zn2(aq) 1.10 V
• Go to the right as written
• Fe2(aq) Cd(s) ? Fe(s) Cd2(aq)
• Goes LEFT, opposite to direction written
• What is Eonet for this reverse reaction?

-0.04 V
0.04 V
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Standard Redox Potentials, Eo
CATHODE
ANODE

• Northwest-southeast rule
• reducing agent _at_ southeast corner
  ANODE
• oxidizing agent _at_ northwest corner
  CATHODE

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Standard Redox Potentials, Eo
Enet distance from top half-reaction
(cathode) to bottom half-reaction (anode) Enet
Ecathode - Eanode
oxidizing agent
Reducing agent
Eonet for Cu 2Ag ? Cu2 2Ag 0.46 V
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Eo for a Voltaic Cell
Cd ? Cd2 2e- or Cd2 2e- ? Cd (-0.40 V)
Fe ? Fe2 2e- or Fe2 2e- ? Fe (-0.44 V)
All substances are present. Which way does the
reaction proceed?
49
Eo for a Voltaic Cell

• From the table, you see
• Fe is a better reducing agent than Cd (anode)
• Cd2 is a better oxidizing agent than Fe2
  (cathode)

Eo Ecathode - Eanode (-0.40 V) - (-0.44
V) 0.04 V Overall reaction Fe(s)
Cd2(aq) ? Cd(s) Fe2(aq)
50
Eo for a Voltaic Cell

• From the table, you see
• Fe is a better reducing agent than Cd (-0.44 V
  anode)
• Cd2 is a better oxidizing agent than Fe2
  (-0.40 V cathode)

Eo Ecathode - Eanode cathode Cd2(aq)
2e- ? Cd(s) -0.40 V - anode Fe(s) ?
Fe2(aq) 2e- 0.44 V Overall Fe(s)
Cd2(aq) ? Cd(s) Fe2(aq) 0.04 V
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More 0n Cell Voltage

• Assume I- ion can reduce water.

2 H2O 2e- ? H2 2 OH- Cathode 2
I- ? I2 2e- Anode --------------------
----------------------------- 2 I- 2 H2O ?
I2 2 OH- H2
Assuming reaction occurs as written, Enet
Ecathode - Eanode (from values in table)
(-0.828 V) - (0.535 V) -1.363 V Minus Enet
means rxn. occurs in opposite direction
(product-favored).
I2 2 OH- H2 ? 2I- H2O (1.363 V)!!!
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E at Nonstandard Conditions

• The NERNST EQUATION
• E potential under nonstandard conditions
• n no. of electrons exchanged
• ln natural log
• If P and R 1 mol/L, then E E
• If R gt P, then E is LARGER than E
• If R lt P, then E is smaller than E

53
Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)Zn is anode
(-0.76 V) in 0.40 M Zn2(aq) and Cu is cathode
(0.34 V) in 4.8 ? 10-3 M Cu2(aq) .
Q. 28 Calculate the cell potential.
Solution (need standard cell potential, Eocell)
Eºcell Eºcathode Eºanode (0.34 V) (0.76
V) 1.10 V
Substituting
54
Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)Zn is anode
(-0.76 V) and Cu is cathode (0.34 V) Eocell
1.10 V
Solution
E 1.10 V (0.01285)(4.42) 1.04 V
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BATTERIESPrimary, Secondary, and Fuel Cells
56
Dry Cell Battery
Primary battery uses redox reactions that
cannot be restored by recharge.

• Anode (-)
• Zn ? Zn2 2e-
• Cathode ()
• 2 NH4 2e- ? 2 NH3 H2

57
Alkaline Battery

• Nearly same reactions as in common dry cell, but
  under basic conditions.

Anode (-) Zn 2 OH- ? ZnO H2O
2e- Cathode () 2 MnO2 H2O 2e- ?
Mn2O3 2 OH-
58
Lead Storage Battery

• Secondary battery
• Uses redox reactions that can be reversed.
• Can be restored by recharging

59
Lead Storage Battery

• Anode (-) Eo 0.36 V
• Pb HSO4- ?? PbSO4 H 2e-
• Cathode () Eo 1.68 V
• PbO2 HSO4- 3 H 2e- ? PbSO4 2
  H2O

60
Ni-Cad Battery

• Anode (-)
• Cd 2 OH- ? Cd(OH)2 2e-
• Cathode ()
• NiO(OH) H2O e- ? Ni(OH)2 OH-

61
Fuel Cells H2 as a Fuel

• Fuel cell - reactants are supplied continuously
  from an external source.
• Cars can use electricity generated by H2/O2 fuel
  cells.
• H2 carried in tanks or generated from
  hydrocarbons.

62
Storing H2 as a Fuel
One way to store H2 is to adsorb the gas onto a
metal or metal alloy. (Energy, p. 290)
63
HydrogenAir Fuel Cell
Anode 2H2(g) ? 4H(aq) 4e-
Cathode O2(g) 2H2O(liq) 4e- ? 4OH- (aq)
----------------------------------
Net O2(g) 2H2(g) ? 2H2O(liq)
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Electrolysis

• Using electrical energy to produce chemical
  change.
• Sn2(aq) 2 Cl-(aq) ? Sn(s) Cl2(g)

Electrolysis of water electroplating refining
metals production of chemicals.
65
Electrolysis Electric Energy ? Chemical Change
  Electrolysis of molten NaCl.   Here a battery
pumps electrons from Cl- to Na.   NOTE
Polarity of electrodes is reversed from batteries.
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Electrolysis of Molten NaCl

• Anode ()
• 2 Cl- ? Cl2(g) 2e-
• Cathode (-)
• Na e- ? Na

Eo for cell (in water) Ec - Ea - 2.71 V
(1.36 V) - 4.07 V (in water) External
energy needed because Eo is (-).
67
Electrolysis of Aqueous NaOH
Electric Energy ? Chemical Change

• Anode ()
• 4 OH- ? O2(g) 2 H2O 4e-
• Cathode (-)
• 4 H2O 4e- ? 2 H2 4 OH-
• Eo for cell -1.23 V

Anode
Cathode
H2O is more easily reduced than Na!!
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Electrolysis of Aqueous NaCl

• Anode ()
• 2 Cl- ?
• Cl2(g) 2e-
• Cathode (-)
• 2 H2O 2e- ?
• H2 2 OH-
• Eo for cell -2.19 V
• Note that H2O is more easily reduced than
  Na.

Also, Cl- is oxidized in preference to H2O
because of kinetics.
69
Electrolysis of Aqueous NaI

• Anode () 2 I- ? I2(g) 2e-
• Cathode (-) 2 H2O 2e- ? H2 2 OH-
• Eo for cell -1.36 V

70
Electrolysis of Aqueous CuCl2

• Anode ()
• 2 Cl- ? Cl2(g) 2e-
• Cathode (-)
• Cu2 2e- ? Cu
• Eo for cell -1.02 V
• Note that Cu is more easily reduced than either
  H2O or Na.

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Eo and Thermodynamics

• Eo is related to ?Go, the free energy change for
  the reaction.
• ?G proportional to nE
• ?Go -nFEo
• where F Faraday constant 9.6485 x
  104 J/Vmol of e-
• (or 9.6485 ? 104 coulombs/mol)
• and n is the number of moles of electrons
  transferred.

72
Calculate ?Go for the reaction,
Zn2(aq) Ni(s) ? Zn(s) Ni2(aq)
Solution use ?Go -nFE no. of electrons, n
2 F 9.6485 ? 104 C
need Eocell
Zn2(aq) 2e- ? Zn(s), -0.763 V
cathode
Ni(s) ? Ni2(aq) 2e-, 0.25 V
anode
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Eocell Ecathode - Eanode
Eocell -0.763 (-0.25 V) -0.51 V
?Go (-2 ? 96485 J/V ?-0.51 V) ? 1 kJ/1000 J
? 98 kJ
74
Eo and ?Go

• ?Go - n F Eo
• For a product-favored reaction
• Reactants ? Products
• ?Go lt 0 and so Eo gt 0
• Eo is positive
• For a reactant-favored reaction
• Reactants ? Products
• ?Go gt 0 and so Eo lt 0
• Eo is negative

75
Quantitative Aspects of Electrochemistry

• Consider electrolysis of aqueous silver ion.
• Ag (aq) e- ? Ag(s)
• 1 mol e- ? 1 mol Ag
• If we could measure the moles of e-, we could
  know the quantity of Ag formed.
• But do we measure moles of e-?

76
Quantitative Aspects of Electrochemistry

• But how is charge related to moles of electrons?

96,500 C/mol e- 1 Faraday
77
Quantitative Aspects of Electrochemistry

• 1.50 amps flow thru a Ag(aq) solution for 15.0
  min. What mass of Ag metal is deposited?
• Solution
• (a) Calc. charge
• Charge (C) current (A) x time (t)
• (1.5 amps)(15.0 min)(60 s/min) 1350 C

78
Quantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag(aq) solution for 15.0
min. What mass of Ag metal is deposited?

• Solution
• (a) Charge 1350 C
• (b) Calculate moles of e- used

(c) Calc. quantity of Ag
79
Quantitative Aspects of Electrochemistry

• The anode reaction in a lead storage battery is
• Pb(s) HSO4-(aq) ? PbSO4(s) H(aq) 2e-
• If a battery delivers 1.50 amp, and you have 454
  g of Pb, how long will the battery last?
• Solution
• a) 454 g Pb 2.19 mol Pb
• b) Calculate moles of e-

c) Calculate charge 4.38 mol e- 96,500 C/mol
e- 423,000 C
80
Quantitative Aspects of Electrochemistry

• The anode reaction in a lead storage battery is
• Pb(s) HSO4-(aq) ? PbSO4(s) H(aq) 2e-
• If a battery delivers 1.50 amp, and you have 454
  g of Pb, how long will the battery last?
• Solution
• a) 454 g Pb 2.19 mol Pb
• b) Mol of e- 4.38 mol
• c) Charge 423,000 C

d) Calculate time
About 78 hours
81
Useful Equations
Eocell Ecathode - Eanode
Ecathode and Eanode as written for reduction in
table.
?Go - n F Eo