Chapter 18 Electrochemistry

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Chapter 18Electrochemistry
Chemistry A Molecular Approach, 1st Ed.Nivaldo
Tro
Roy Kennedy Massachusetts Bay Community
College Wellesley Hills, MA
2007, Prentice Hall
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Redox Reaction

• one or more elements change oxidation number
• all single displacement, and combustion,
• some synthesis and decomposition
• always have both oxidation and reduction
• split reaction into oxidation half-reaction and a
  reduction half-reaction
• aka electron transfer reactions
• half-reactions include electrons
• oxidizing agent is reactant molecule that causes
  oxidation
• contains element reduced
• reducing agent is reactant molecule that causes
  reduction
• contains the element oxidized

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Oxidation Reduction

• oxidation is the process that occurs when
• oxidation number of an element increases
• element loses electrons
• compound adds oxygen
• compound loses hydrogen
• half-reaction has electrons as products
• reduction is the process that occurs when
• oxidation number of an element decreases
• element gains electrons
• compound loses oxygen
• compound gains hydrogen
• half-reactions have electrons as reactants

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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
• 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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Oxidation and Reduction

• oxidation occurs when an atoms oxidation state
  increases during a reaction
• reduction occurs when an atoms oxidation state
  decreases during a reaction

CH4 2 O2 ? CO2 2 H2O -4 1
0 4 2 1 -2
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OxidationReduction

• oxidation and reduction must occur simultaneously
  
• if an atom loses electrons another atom must take
  them
• the reactant that reduces an element in another
  reactant is called the reducing agent
• the reducing agent contains the element that is
  oxidized
• the reactant that oxidizes an element in another
  reactant is called the oxidizing agent
• the oxidizing agent contains the element that is
  reduced

2 Na(s) Cl2(g) ? 2 NaCl(s) Na is oxidized, Cl
is reduced Na is the reducing agent, Cl2 is the
oxidizing agent
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Identify the Oxidizing and Reducing Agents in
Each of the Following

• 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
• MnO2 4 HBr MnBr2 Br2 2 H2O

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Identify the Oxidizing and Reducing Agents in
Each of the Following
ox ag
red ag

• 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
• MnO2 4 HBr MnBr2 Br2 2 H2O

1 -2 5 -2 1 0
2 -2 1 -2
red ag
ox ag
4 -2 1 -1 2 -1 0
1 -2
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Common Oxidizing Agents
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Common Reducing Agents
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Balancing Redox Reactions

• assign oxidation numbers
• determine element oxidized and element reduced
• write ox. red. half-reactions, including
  electrons
• ox. electrons on right, red. electrons on left of
  arrow
• balance half-reactions by mass
• first balance elements other than H and O
• add H2O where need O
• add H1 where need H
• neutralize H with OH- in base
• balance half-reactions by charge
• balance charge by adjusting electrons
• balance electrons between half-reactions
• add half-reactions
• check

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Ex 18.3 Balance the equationI?(aq)
MnO4?(aq) ? I2(aq) MnO2(s) in basic solution
Assign Oxidation States I?(aq) MnO4?(aq) ? I2(aq) MnO2(s)
Separate into half-reactions ox red
Assign Oxidation States
Separate into half-reactions ox I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s)
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Ex 18.3 Balance the equationI?(aq)
MnO4?(aq) ? I2(aq) MnO2(s) in basic solution
Balance half-reactions by mass ox I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s)
Balance half-reactions by mass ox 2 I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s)
Balance half-reactions by mass then O by adding H2O ox 2 I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s) 2 H2O(l)
Balance half-reactions by mass then H by adding H ox 2 I?(aq) ? I2(aq) red 4 H(aq) MnO4?(aq) ? MnO2(s) 2 H2O(l)
Balance half-reactions by mass in base, neutralize the H with OH- ox 2 I?(aq) ? I2(aq) red 4 H(aq) MnO4?(aq) ? MnO2(s) 2 H2O(l) 4 H(aq) 4 OH?(aq) MnO4?(aq) ? MnO2(s) 2 H2O(l) 4 OH?(aq) 4 H2O(aq) MnO4?(aq) ? MnO2(s) 2 H2O(l) 4 OH?(aq) MnO4?(aq) 2 H2O(l) ? MnO2(s) 4 OH?(aq)
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Ex 18.3 Balance the equationI?(aq)
MnO4?(aq) ? I2(aq) MnO2(s) in basic solution
Balance Half-reactions by charge ox 2 I?(aq) ? I2(aq) 2 e? red MnO4?(aq) 2 H2O(l) 3 e? ? MnO2(s) 4 OH?(aq)
Balance electrons between half-reactions ox 2 I?(aq) ? I2(aq) 2 e? x3 red MnO4?(aq) 2 H2O(l) 3 e? ? MnO2(s) 4 OH?(aq) x2 ox 6 I?(aq) ? 3 I2(aq) 6 e? red 2 MnO4?(aq) 4 H2O(l) 6 e? ? 2 MnO2(s) 8 OH?(aq)
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Ex 18.3 Balance the equationI?(aq)
MnO4?(aq) ? I2(aq) MnO2(s) in basic solution
Add the Half-reactions ox 6 I?(aq) ? 3 I2(aq) 6 e? red 2 MnO4?(aq) 4 H2O(l) 6 e? ? 2 MnO2(s) 8 OH?(aq) tot 6 I?(aq) 2 MnO4?(aq) 4 H2O(l) ? 3 I2(aq) 2 MnO2(s) 8 OH?(aq)
Check
Reactant Count Element Product Count
6 I 6
2 Mn 2
12 O 12
8 H 8
2? charge 2?
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Practice - Balance the Equation H2O2 KI
H2SO4 K2SO4 I2 H2O
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Practice - Balance the Equation H2O2 KI
H2SO4 K2SO4 I2 H2O
1 -1 1 -1 1 6 -2 1 6 -2
0 1 -2
oxidation
reduction
ox 2 I-1 I2 2e-1 red H2O2 2e-1 2 H 2
H2O tot 2 I-1 H2O2 2 H I2 2 H2O
1 H2O2 2 KI H2SO4 K2SO4 1 I2 2 H2O
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Practice - Balance the EquationClO3-1 Cl-1
Cl2 (in acid)
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Practice - Balance the EquationClO3-1 Cl-1
Cl2 (in acid)
5 -2 -1 0
oxidation
reduction
ox 2 Cl-1 Cl2 2 e-1 x5 red 2 ClO3-1 10
e-1 12 H Cl2 6 H2O x1 tot 10 Cl-1 2
ClO3-1 12 H 6 Cl2 6 H2O
1 ClO3-1 5 Cl-1 6 H1 3 Cl2 3 H2O
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Electrical Current

• when we talk about the current of a liquid in a
  stream, we are discussing the amount of water
  that passes by in a given period of time
• when we discuss electric current, we are
  discussing the amount of electric charge that
  passes a point in a given period of time
• whether as electrons flowing through a wire or
  ions flowing through a solution

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Redox Reactions Current

• redox reactions involve the transfer of electrons
  from one substance to another
• therefore, redox reactions have the potential to
  generate an electric current
• in order to use that current, we need to separate
  the place where oxidation is occurring from the
  place that reduction is occurring

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Electric Current Flowing Directly Between Atoms
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Electric Current Flowing Indirectly Between Atoms
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Electrochemical Cells

• electrochemistry is the study of redox reactions
  that produce or require an electric current
• the conversion between chemical energy and
  electrical energy is carried out in an
  electrochemical cell
• spontaneous redox reactions take place in a
  voltaic cell
• aka galvanic cells
• nonspontaneous redox reactions can be made to
  occur in an electrolytic cell by the addition of
  electrical energy

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Electrochemical Cells

• oxidation and reduction reactions kept separate
• half-cells
• electron flow through a wire along with ion flow
  through a solution constitutes an electric
  circuit
• requires a conductive solid (metal or graphite)
  electrode to allow the transfer of electrons
• through external circuit
• ion exchange between the two halves of the system
• electrolyte

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Electrodes

• Anode
• electrode where oxidation occurs
• anions attracted to it
• connected to positive end of battery in
  electrolytic cell
• loses weight in electrolytic cell
• Cathode
• electrode where reduction occurs
• cations attracted to it
• connected to negative end of battery in
  electrolytic cell
• gains weight in electrolytic cell
• electrode where plating takes place in
  electroplating

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Voltaic Cell
the salt bridge is required to complete the
circuit and maintain charge balance
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Current and Voltage

• the number of electrons that flow through the
  system per second is the current
• unit Ampere
• 1 A of current 1 Coulomb of charge flowing by
  each second
• 1 A 6.242 x 1018 electrons/second
• Electrode surface area dictates the number of
  electrons that can flow
• the difference in potential energy between the
  reactants and products is the potential
  difference
• unit Volt
• 1 V of force 1 J of energy/Coulomb of charge
• the voltage needed to drive electrons through the
  external circuit
• amount of force pushing the electrons through the
  wire is called the electromotive force, emf

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Cell Potential

• the difference in potential energy between the
  anode the cathode in a voltaic cell is called the
  cell potential
• the cell potential depends on the relative ease
  with which the oxidizing agent is reduced at the
  cathode and the reducing agent is oxidized at the
  anode
• the cell potential under standard conditions is
  called the standard emf, Ecell
• 25C, 1 atm for gases, 1 M concentration of
  solution
• sum of the cell potentials for the half-reactions

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Cell Notation

• shorthand description of Voltaic cell
• electrode electrolyte electrolyte
  electrode
• oxidation half-cell on left, reduction half-cell
  on the right
• single phase barrier
• if multiple electrolytes in same phase, a comma
  is used rather than
• often use an inert electrode
• double line salt bridge

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Fe(s) Fe2(aq) MnO4?(aq), Mn2(aq), H(aq)
Pt(s)
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Standard Reduction Potential

• a half-reaction with a strong tendency to occur
  has a large half-cell potential
• when two half-cells are connected, the electrons
  will flow so that the half-reaction with the
  stronger tendency will occur
• we cannot measure the absolute tendency of a
  half-reaction, we can only measure it relative to
  another half-reaction
• we select as a standard half-reaction the
  reduction of H to H2 under standard conditions,
  which we assign a potential difference 0 v
• standard hydrogen electrode, SHE

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Half-Cell Potentials

• SHE reduction potential is defined to be exactly
  0 v
• half-reactions with a stronger tendency toward
  reduction than the SHE have a value for Ered
• half-reactions with a stronger tendency toward
  oxidation than the SHE have a ? value for Ered
• Ecell Eoxidation Ereduction
• Eoxidation ?Ereduction
• when adding E values for the half-cells, do not
  multiply the half-cell E values, even if you
  need to multiply the half-reactions to balance
  the equation

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Ex 18.4 Calculate E?cell for the reaction at
25?CAl(s) NO3-(aq) 4 H(aq) ? Al3(aq)
NO(g) 2 H2O(l)
Separate the reaction into the oxidation and reduction half-reactions ox Al(s) ? Al3(aq) 3 e- red NO3-(aq) 4 H(aq) 3 e- ? NO(g) 2 H2O(l)
find the E? for each half-reaction and sum to get E?cell 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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Ex 18.4a Predict if the following reaction is
spontaneous under standard conditionsFe(s)
Mg2(aq) ? Fe2(aq) Mg(s)
Separate the reaction into the oxidation and reduction half-reactions ox Fe(s) ? Fe2(aq) 2 e- red Mg2(aq) 2 e- ? Mg(s)
look up the relative positions of the reduction half-reactions red Mg2(aq) 2 e- ? Mg(s) red Fe2(aq) 2 e- ? Fe(s) since Mg2 reduction is below Fe2 reduction, the reaction is NOT spontaneous as written
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the reaction is spontaneous in the reverse direction Mg(s) Fe2(aq) ? Mg2(aq) Fe(s) ox Mg(s) ? Mg2(aq) 2 e- red Fe2(aq) 2 e- ? Fe(s)
sketch the cell and label the parts oxidation occurs at the anode electrons flow from anode to cathode
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Practice - Sketch and Label the Voltaic
CellFe(s) ½ Fe2(aq) ½½ Pb2(aq) ½ Pb(s) , Write
the Half-Reactions and Overall Reaction, and
Determine the Cell Potential under Standard
Conditions.
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ox Fe(s) ? Fe2(aq) 2 e- E? 0.45 V
red Pb2(aq) 2 e- ? Pb(s) E? -0.13 V
tot Pb2(aq) Fe(s) ? Fe2(aq) Pb(s) E?
0.32 V
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Predicting Whether a Metal Will Dissolve in an
Acid

• acids dissolve in metals if the reduction of the
  metal ion is easier than the reduction of H(aq)
• metals whose ion reduction reaction lies below H
  reduction on the table will dissolve in acid

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Ecell, DG and K

• for a spontaneous reaction
• one the proceeds in the forward direction with
  the chemicals in their standard states
• DG lt 1 (negative)
• E gt 1 (positive)
• K gt 1
• DG -RTlnK -nFEcell
• n is the number of electrons
• F Faradays Constant 96,485 C/mol e-

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Example 18.6- Calculate DG for the
reactionI2(s) 2 Br-(aq) ? Br2(l) 2 I-(aq)
I2(s) 2 Br-(aq) ? Br2(l) 2 I-(aq) DG, (J)
Given Find
Concept Plan Relationships
ox 2 Br-(aq) ? Br2(l) 2 e- E -1.09 v
Solve
red I2(l) 2 e- ? 2 I-(aq) E 0.54 v
tot I2(l) 2Br-(aq) ? 2I-(aq) Br2(l) E
-0.55 v
since DG is , the reaction is not spontaneous
in the forward direction under standard conditions
Answer
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Example 18.7- Calculate K at 25C for the
reactionCu(s) 2 H(aq) ? H2(g) Cu2(aq)
Cu(s) 2 H(aq) ? H2(g) Cu2(aq) K
Given Find
Concept Plan Relationships
ox Cu(s) ? Cu2(aq) 2 e- E -0.34 v
Solve
red 2 H(aq) 2 e- ? H2(aq) E 0.00 v
tot Cu(s) 2H(aq) ? Cu2(aq) H2(g) E
-0.34 v
since K lt 1, the position of equilibrium lies far
to the left under standard conditions
Answer
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Nonstandard Conditions - the Nernst Equation

• DG DG RT ln Q
• E E - (0.0592/n) log Q at 25C
• when Q K, E 0
• use to calculate E when concentrations not 1 M

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E? at Nonstandard Conditions
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Example 18.8- Calculate Ecell at 25C for the
reaction3 Cu(s) 2 MnO4-(aq) 8 H(aq) ? 2
MnO2(s) Cu2(aq) 4 H2O(l)
3 Cu(s) 2 MnO4-(aq) 8 H(aq) ? 2 MnO2(s)
Cu2(aq) 4 H2O(l) Cu2 0.010 M, MnO4-
2.0 M, H 1.0 M Ecell
Given Find
Concept Plan Relationships
Solve
ox Cu(s) ? Cu2(aq) 2 e- x3 E -0.34 v
red MnO4-(aq) 4 H(aq) 3 e- ? MnO2(s) 2
H2O(l) x2 E 1.68 v
tot 3 Cu(s) 2 MnO4-(aq) 8 H(aq) ? 2 MnO2(s)
Cu2(aq) 4 H2O(l)) E 1.34 v
units are correct, Ecell gt Ecell as expected
because MnO4- gt 1 M and Cu2 lt 1 M
Check
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Concentration Cells

• it is possible to get a spontaneous reaction when
  the oxidation and reduction reactions are the
  same, as long as the electrolyte concentrations
  are different
• the difference in energy is due to the entropic
  difference in the solutions
• the more concentrated solution has lower entropy
  than the less concentrated
• electrons will flow from the electrode in the
  less concentrated solution to the electrode in
  the more concentrated solution
• oxidation of the electrode in the less
  concentrated solution will increase the ion
  concentration in the solution the less
  concentrated solution has the anode
• reduction of the solution ions at the electrode
  in the more concentrated solution reduces the ion
  concentration the more concentrated solution
  has the cathode

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Concentration Cell
Cu(s)? Cu2(aq) (0.010 M) ?? Cu2(aq) (2.0 M)?
Cu(s)
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LeClanche Acidic Dry Cell

• electrolyte in paste form
• ZnCl2 NH4Cl
• or MgBr2
• anode Zn (or Mg)
• Zn(s) Zn2(aq) 2 e-
• cathode graphite rod
• MnO2 is reduced
• 2 MnO2(s) 2 NH4(aq) 2 H2O(l) 2 e-
• 2 NH4OH(aq) 2 Mn(O)OH(s)
• cell voltage 1.5 v
• expensive, nonrechargeable, heavy, easily corroded

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Alkaline Dry Cell

• same basic cell as acidic dry cell, except
  electrolyte is alkaline KOH paste
• anode Zn (or Mg)
• Zn(s) Zn2(aq) 2 e-
• cathode brass rod
• MnO2 is reduced
• 2 MnO2(s) 2 NH4(aq) 2 H2O(l) 2 e-
• 2 NH4OH(aq) 2 Mn(O)OH(s)
• cell voltage 1.54 v
• longer shelf life than acidic dry cells and
  rechargeable, little corrosion of zinc

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Lead Storage Battery

• 6 cells in series
• electrolyte 30 H2SO4
• anode Pb
• Pb(s) SO42-(aq) PbSO4(s) 2 e-
• cathode Pb coated with PbO2
• PbO2 is reduced
• PbO2(s) 4 H(aq) SO42-(aq) 2 e-
• PbSO4(s) 2 H2O(l)
• cell voltage 2.09 v
• rechargeable, heavy

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NiCad Battery

• electrolyte is concentrated KOH solution
• anode Cd
• Cd(s) 2 OH-1(aq) Cd(OH)2(s) 2 e-1 E0 0.81
  v
• cathode Ni coated with NiO2
• NiO2 is reduced
• NiO2(s) 2 H2O(l) 2 e-1 Ni(OH)2(s) 2OH-1
  E0 0.49 v
• cell voltage 1.30 v
• rechargeable, long life, light however
  recharging incorrectly can lead to battery
  breakdown

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Ni-MH Battery

• electrolyte is concentrated KOH solution
• anode metal alloy with dissolved hydrogen
• oxidation of H from H0 to H1
• MH(s) OH-1(aq) M(s) H2O(l) e-1 E 0.89
  v
• cathode Ni coated with NiO2
• NiO2 is reduced
• NiO2(s) 2 H2O(l) 2 e-1 Ni(OH)2(s) 2OH-1
  E0 0.49 v
• cell voltage 1.30 v
• rechargeable, long life, light, more
  environmentally friendly than NiCad, greater
  energy density than NiCad

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Lithium Ion Battery

• electrolyte is concentrated KOH solution
• anode graphite impregnated with Li ions
• cathode Li - transition metal oxide
• reduction of transition metal
• work on Li ion migration from anode to cathode
  causing a corresponding migration of electrons
  from anode to cathode
• rechargeable, long life, very light, more
  environmentally friendly, greater energy density

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Fuel Cells

• like batteries in which reactants are constantly
  being added
• so it never runs down!
• Anode and Cathode both Pt coated metal
• Electrolyte is OH solution
• Anode Reaction
• 2 H2 4 OH
• ? 4 H2O(l) 4 e-
• Cathode Reaction
• O2 4 H2O 4 e-
• ? 4 OH

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Electrolytic Cell

• uses electrical energy to overcome the energy
  barrier and cause a non-spontaneous reaction
• must be DC source
• the terminal of the battery anode
• the - terminal of the battery cathode
• cations attracted to the cathode, anions to the
  anode
• cations pick up electrons from the cathode and
  are reduced, anions release electrons to the
  anode and are oxidized
• some electrolysis reactions require more voltage
  than Etot, called the overvoltage

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electroplating
In electroplating, the work piece is the
cathode. Cations are reduced at cathode and plate
to the surface of the work piece. The anode is
made of the plate metal. The anode oxidizes and
replaces the metal cations in the solution
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Electrochemical Cells

• in all electrochemical cells, oxidation occurs at
  the anode, reduction occurs at the cathode
• in voltaic cells,
• anode is the source of electrons and has a (-)
  charge
• cathode draws electrons and has a () charge
• in electrolytic cells
• electrons are drawn off the anode, so it must
  have a place to release the electrons, the
  terminal of the battery
• electrons are forced toward the anode, so it must
  have a source of electrons, the - terminal of the
  battery

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Electrolysis

• electrolysis is the process of using electricity
  to break a compound apart
• electrolysis is done in an electrolytic cell
• electrolytic cells can be used to separate
  elements from their compounds
• generate H2 from water for fuel cells
• recover metals from their ores

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Electrolysis of Water
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Electrolysis of Pure Compounds

• must be in molten (liquid) state
• electrodes normally graphite
• cations are reduced at the cathode to metal
  element
• anions oxidized at anode to nonmetal element

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Electrolysis of NaCl(l)
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Mixtures of Ions

• when more than one cation is present, the cation
  that is easiest to reduce will be reduced first
  at the cathode
• least negative or most positive Ered
• when more than one anion is present, the anion
  that is easiest to oxidize will be oxidized first
  at the anode
• least negative or most positive Eox

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Electrolysis of Aqueous Solutions

• Complicated by more than one possible oxidation
  and reduction
• possible cathode reactions
• reduction of cation to metal
• reduction of water to H2
• 2 H2O 2 e-1 H2 2 OH-1 E -0.83 v _at_
  stand. cond.
• E -0.41 v _at_ pH 7
• possible anode reactions
• oxidation of anion to element
• oxidation of H2O to O2
• 2 H2O O2 4e-1 4H1 E -1.23 v _at_ stand.
  cond.
• E -0.82 v _at_ pH 7
• oxidation of electrode
• particularly Cu
• graphite doesnt oxidize
• half-reactions that lead to least negative Etot
  will occur
• unless overvoltage changes the conditions

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Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations 2 I-1 I2 2 e-1 E
-0.54 v 2 H2O O2 4e-1 4H1 E -0.82 v
possible oxidations 2 I-1 I2 2 e-1 E
-0.54 v 2 H2O O2 4e-1 4H1 E -0.82 v
possible reductions Na1 1e-1 Na0 E -2.71
v 2 H2O 2 e-1 H2 2 OH-1 E -0.41 v
possible reductions Na1 1e-1 Na0 E -2.71
v 2 H2O 2 e-1 H2 2 OH-1 E -0.41 v
overall reaction 2 I-(aq) 2 H2O(l) I2(aq)
H2(g) 2 OH-1(aq)
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Faradays Law

• the amount of metal deposited during electrolysis
  is directly proportional to the charge on the
  cation, the current, and the length of time the
  cell runs
• charge that flows through the cell current x
  time

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Example 18.10- Calculate the mass of Au that can
be plated in 25 min using 5.5 A for the
half-reaction Au3(aq) 3 e- ? Au(s)
3 mol e- 1 mol Au, current 5.5 amps, time
25 min mass Au, g
Given Find
Concept Plan Relationships
Solve
units are correct, answer is reasonable since 10
A running for 1 hr 1/3 mol e-
Check
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Corrosion

• corrosion is the spontaneous oxidation of a metal
  by chemicals in the environment
• since many materials we use are active metals,
  corrosion can be a very big problem

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Rusting

• rust is hydrated iron(III) oxide
• moisture must be present
• water is a reactant
• required for flow between cathode and anode
• electrolytes promote rusting
• enhances current flow
• acids promote rusting
• lower pH lower Ered

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Preventing Corrosion

• one way to reduce or slow corrosion is to coat
  the metal surface to keep it from contacting
  corrosive chemicals in the environment
• paint
• some metals, like Al, form an oxide that strongly
  attaches to the metal surface, preventing the
  rest from corroding
• another method to protect one metal is to attach
  it to a more reactive metal that is cheap
• sacrificial electrode
• galvanized nails

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Sacrificial Anode