Chapter 18 Electrochemistry

1
Chapter 18Electrochemistry
2
GOALS
Balancing redox reactions Voltaic
cells Electrochemical potentials Electrolysis
the calculations!!
Review oxidation states oxidation/reduction oxidi
zing/reducing agent ch. 17
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Why Study Electrochemistry?

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

The heme group
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Electron Transfer Reactions

• Electron transfer reactions are
  oxidation-reduction or redox reactions (i.e.
  changes in oxidation states).
• Redox reactions can result in the generation of
  an electric current (battery), or, may be caused
  by applying an electric current (electroplating).
  
• Therefore, this field of chemistry is often
  called ELECTROCHEMISTRY.

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ELECTRON TRANSFER REACTIONS

• 0 0 1
  1-
• 2Na(s) Cl2(g) ? 2NaCl(s)
• 1 0
  2 0
  
• 2HCl(aq) Zn(s) ? ZnCl2(aq) H2(g)
• 0
  0
  
• Cu(s) 2 Ag(aq) ? Cu2(aq) 2 Ag(s)
• 0
  0 3 2-
• 2Fe(s) xH2O(l) 1½O2(g) ? Fe2O3.xH2O(s)
• 4 2-
  0 0
  
• CO2(g) H2O(l) energy ? (CH2O)n O2(g)

6
Review of Terminology for Redox Reactions

• OXIDATIONloss of electron(s) by a species
  increase in oxidation number e- to the right of
  arrow.
• Na ? Na e-
• REDUCTIONgain of electron(s) decrease in
  oxidation number e- to left of arrow.
• ½Cl2(g) e- ? Cl-
• OXIDIZING AGENTelectron acceptor it is reduced
  ½ Cl2(g) e- ? Cl-
• REDUCING AGENTelectron donor it is oxidized
• Na ? Na e-

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Electrochemical Cells

• Apparatus for generating an electric current
  through the use of a product favored reaction
  (spontaneous) voltaic or galvanic cell.
• An electrolytic cell is used to carry out
  electrolysis (an electric current is used to
  bring about a nonspontaneous chemical reaction).

Batteries are voltaic cells
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Electrochemistry
Alessandro Volta, 1745-1827, Italian scientist
and inventor.
Luigi Galvani, 1737-1798, Italian scientist and
inventor.
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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)
Fe 2
Mn 7
Fe 3
Mn 2
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Rules for Assigning Oxidation States

• rules are in order of priority
• free elements have an oxidation state 0
• Na 0 and Cl2 0 in 2 Na(s) Cl2(g)
• monatomic ions have an oxidation state equal to
  their charge
• Na 1 and Cl -1 in NaCl
• (a) the sum of the oxidation states of all the
  atoms in a compound is 0
• Na 1 and Cl -1 in NaCl, (1) (-1) 0

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Rules for Assigning Oxidation States

• (b) the sum of the oxidation states of all the
  atoms in a polyatomic ion equals the charge on
  the ion
• N 5 and O -2 in NO3, (5) 3(-2) -1
• (a) Group I metals have an oxidation state of 1
  in all their compounds
• Na 1 in NaCl
• (b) Group II metals have an oxidation
  state of 2 in all their compounds
• Mg 2 in MgCl2

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Rules for Assigning Oxidation States

• in their compounds, nonmetals have oxidation
  states according to the table below (grp - 8)
• nonmetals higher on the table take priority

Nonmetal Oxidation State Example
F -1 CF4
H 1 CH4
O -2 CO2
Group 7A -1 CCl4
Group 6A -2 CS2
Group 5A -3 NH3
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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
• The equation is now balanced for both charge and
  mass (the 2e- of the left are cancelled out with
  those on the right).

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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 2e-
• Red VO2 e- ? VO2 excess of 2
  on the right

Zn lost two electrons. They are written on the
right. V is 5 on the left and 4 on the right.
It gained one electron. That is written on the
left side.
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Balancing Equations

• Step 3 Balance half-reactions for charge
• Reaction is acidic, then we can use
  H.
• Ox Zn ? Zn2 2e-
• 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/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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More Practice - Balance the equations
below!I?(aq) MnO4?(aq) ? I2(aq) MnO2(s) in
basic solution
An alkaline (basic) solution of hypochlorite ions
reacts with solid chromium(III) hydroxide to
produce chromate and chloride ions.
ClO3- Cl- Cl2 (in acid)
Cr2O72- I- IO3- Cr3 (in acid)
MnO4- H2SO3 SO42- Mn2 (in acid)
Cr(OH)4- H2O2 CrO42- H2O (in basic
soln)
Zn NO3- Zn(OH)4- NH3 (in basic soln)
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VOLTAIC CELLS
- use a chemical rxn to produce an electric
current.
The ZnZn2 and CuCu2 Cell
Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
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A CHEMICAL CHANGE PRODUCES AN ELECTRIC
CURRENT
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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A CHEMICAL CHANGE PRODUCES AN ELECTRIC
CURRENT
Zn(s) ? Zn2(aq) 2e-

• To obtain a useful current, we separate the
  oxidizing and reducing agents so that electron
  transfer occurs through 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 through external wire.
• Salt bridge allows anions and cations to move
  between electrode compartments.

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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.

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Need Calculate 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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Standard Reduction Potential

• We can measure ALL other half-cell potentials
  relative to another half-reaction.
• We select as a standard half-reaction, the
  reduction of H to H2 under standard conditions
  (1atm, 1 M _at_ 25 oC) and which we assign a
  potential difference 0 V
• Standard Hydrogen Electrode, SHE

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Zn/Zn2 half-cell hooked up to a SHE. Eo for the
cell 0.76 V
Supplier of electrons
Acceptor of electrons
2 H 2e- ? H2 Reduction Cathode
Zn ? Zn2 2e- Oxidation Anode
Zn is a better reducing agent than H2
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Cu/Cu2 half-cell hooked up to a SHE. Eo for the
cell 0.34 V
Positive
Negative
Acceptor of electrons
Supplier of electrons
H2 ? 2 H 2e- Oxidation Anode
Cu2 2e- ? Cu Reduction Cathode
H2 is now a better reducing agent than Cu!
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Zn/Cu Electrochemical Cell
Anode negative source of electrons
Cathode positive sink for electrons

• oxid Zn(s) ? Zn2(aq) 2e- Eo 0.76 V
• red 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
When a reaction is reversed, the sign of E is
reversed!
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(No Transcript)
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reducing agents
oxidizing agents
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Using Standard Potentials, Eo
Which is the best oxidizing agentO2 (1.23 V)
H2O2 (1.77 V) or Cl2 (1.36 V)?
H2O2 (1.77 V)

• Which is the best reducing agent
• Hg (0.79 V), Al (-1.66 V), or Sn (-0.14 V)?

Al (-1.66 V)
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Using Standard Potentials, Eo
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)
Cr2O72-

• 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)

Sn2
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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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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 (ANODE)
• oxidizing agent at northwest corner (CATHODE)

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

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

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Using Standard Potentials, Eo

• 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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Eo for a Voltaic Cell
Fe(s) Cd2(aq) ? Cd(s) Fe2(aq)
Cd ? Cd2 2e- or Cd2 2e- ? Cd (-0.40 V)
Fe ? Fe2 2e- or Fe2 2e- ? Fe (-0.44 V)
Which way does the reaction proceed? In which
direction is it spontaneous?
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Eo for a Voltaic Cell

• From the table,
• 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 (reverse the
smaller, more
negative, then add) cathode Cd2(aq)
2e- ? Cd(s) -0.40 V (red) - anode
Fe(s) ? Fe2(aq) 2e- 0.44 V
(oxid) Overall Fe(s) Cd2(aq) ? Cd(s)
Fe2(aq) 0.04 V
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More 0n Cell Voltage
When two half-rxns (written as reduction) are
joined in an electrochemical cell, the one with
the larger half-cell potential occurs in the
forward direction, and the one with the smaller
potential occurs in the reverse direction.
Cd2 2e- ? Cd (-0.40 V)
larger
Fe2 2e- ? Fe (-0.44 V)
smaller (reverse this rxn)
Cd2 2e- ? Cd (-0.40 V)
larger Fe ? Fe2 2e- (0.44 V)
overall Cd2 Fe ? Cd Fe2 (0.04 V)
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More 0n Cell Voltage
When two half-rxns (written as reduction) are
joined in an electrochemical cell, the one with
the larger half-cell potential occurs in the
forward direction, and the one with the smaller
potential occurs in the reverse direction.
Ni2 2e- ? Ni (-0.23 V)
larger
Mn2 2e- ? Mn (-1.18 V)
smaller (reverse this rxn)
Ni2 2e- ? Ni (-0.23 V)
Mn ? Mn2 2e-
(1.18 V) overall Ni2 Mn ? Ni
Mn2 (0.95 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 net rxn. occurs in the opposite direction
(favors I- H2O).
I2 2 OH- H2 ? 2I- H2O (1.363 V)!!!
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Calculate E?cell for the reaction at 25?CAl(s)
NO3-(aq) 4 H(aq) ? Al3(aq) NO(g) 2 H2O(l)
(This is the reaction of Al with nitric acid)
Separate the reaction into the oxidation and reduction half-reactions
Find the E? for each half-reaction and sum to get E?cell
E?ox -E?red 1.66 V
ox Al(s) ? Al3(aq) 3 e-
red NO3-(aq) 4 H(aq) 3 e- ? NO(g) 2
H2O(l)
Ered 0.96 V
E?ox -(E?red) 1.66 V
E?red 0.96 V
E?cell (1.66 V) (0.96 V) 2.62 V
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For the reaction Al(s) NO3-(aq) 4 H(aq) ?
Al3(aq) NO(g) 2 H2O(l)
ox Al(s) ? Al3(aq) 3 e-
E?ox -E?red 1.66 V
E?red 0.96 V
red NO3-(aq) 4 H(aq) 3 e- ? NO(g) 2
H2O(l)
E?cell (1.66 V) (0.96 V) 2.62 V
The symbol of the cell is Al(s) Al3(aq)
NO3-(aq), 4 H(aq) , NO(g) Pt
This is the symbol for the salt bridge
This is the symbol for the electrode-solxn contact
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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

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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) .
Calculate the cell potential.
Solution (need standard cell potential, Eocell)
Eºcell Eºcathode Eºanode (0.34 V) (0.76
V) 0.34 0.76 1.10 V
Substituting
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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 n 2 (Cu2 2e- Cu0)
Solution
E 1.10 V (0.01285)(4.42) 1.04 V
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2Fe3(aq) 3Mg(s) ? 2Fe(s) 3Mg2(aq) Fe3
1.0 ? 10-3 M Mg2 2.5 M Calculate the
cell potential.
Solution (need standard cell potential, Eocell)
Eºcell Eºcathode Eºanode (-0.036 V) (2.37
V) -0.036 2.37 2.33 V
Substituting
What is n???
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2Fe3(aq) 3Mg(s) ? 2Fe(s) 3Mg2(aq) Fe3
1.0 ? 10-3 M Mg2 2.5 M Calculate the
cell potential.
Solution (need standard cell potential, Eocell)
Eºcell 2.33 V n 6
Substituting
Ans 2.26 V
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BATTERIESPrimary, Secondary, and Fuel Cells
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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

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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-
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Lead Storage Battery

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

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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

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Ni-Cad Battery

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

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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.

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Storing H2 as a Fuel
One way to store H2 is to adsorb the gas onto a
metal or metal alloy.
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HydrogenAir (O2) 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.
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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 ()
• 2Cl-(l) ? Cl2(g) 2e- (-1.36 V)
• Cathode (-)
• Na(l) e- ? Na (-2.71 V)

Eo for cell (in melted NaCl) Ec Ea -
2.71 V (-1.36 V) - 4.07 V (in melted
NaCl) rxn is nonspontaneous External electrical
energy needed because Eo is (-).
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Electrolysis of Aqueous NaOH
NaOH H2O ? Na(aq) OH-(aq)
Electric Energy ? Chemical Change
Anode
Cathode

• Anode () E -0.40 V
• 4 OH- ? O2(g) 2 H2O 4e-
• Cathode (-)
• 4 H2O 4e- ? 2 H2 4 OH-
• Eo for cell -1.23 V

H2O is more easily reduced than Na!! (Eo
-2.71 V)
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Electrolysis of Aqueous NaCl
NaCl H2O ? Na(aq) Cl-(aq)

• Anode ()
• 2 Cl- ?
• Cl2(g) 2e- E -1.36 V
• Cathode (-)
• 2 H2O 2e- ?
• H2 2 OH-
• Eo for cell -2.19 V
• Note that H2O is more
• Easily reduced than Na(E -2.71 V)
• 2H2O(l) ?
• O2(g) 4H 4e-

Also, Cl- is oxidized in preference to H2O
because of kinetics (overvoltage) Ered lt -1.23
V, may be down to -2.00 V
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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.

68
Electrolysis of Aqueous CuCl2
CuCl2 H2O ? Cu2(aq) 2Cl-(aq)

• 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 (check redox potentials).

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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
• ?Go gt 0, reaction is nonspontaneous
• Reagents are favored

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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

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Ecell, DG and K!!

• for a spontaneous reaction
• DG lt 0 (negative)
• E gt 0 (positive)
• K gt 1 (large)

When Ecell 0 (no net rxn), reactants and
products are at equilibriumand Q K
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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-?

74
Quantitative Aspects of Electrochemistry

• But , how is charge related to moles of
  electrons?

96,500 C/mol e- 1 Faraday
75
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

76
Quantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag(aq) solution for 15.0
min. What mass of Ag metal is deposited?
Ag e- ? Ag(s)

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

(c) Calc. quantity of Ag
77
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 there is 454
  g of Pb, how long will the battery last?
• Solution
• a) 454 g Pb 2.19 mol Pb
• Calculate moles of e-
• each Pb atom is loosing 2e-

c) Calculate charge 4.38 mol e- 96,500 C/mol
e- 423,000 C
78
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