Chapter 18Electrochemistry
2007, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
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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 reductionsplit 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 increaseselement loses electronscompound adds oxygencompound loses hydrogenhalf-reaction has electrons as products
• reduction is the process that occurs whenoxidation number of an element decreaseselement gains electronscompound loses oxygencompound gains hydrogenhalf-reactions have electrons as reactants
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Rules for Assigning Oxidation States
• rules are in order of priority1. free elements have an oxidation state = 0
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equal to their charge
Na = +1 and Cl = -1 in NaCl
3. (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 States3. (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
4. (a) Group I metals have an oxidation state of +1 in all their compounds
Na = +1 in NaCl
4. (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 States5. 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 atom’s oxidation state increases during a reaction
• reduction occurs when an atom’s oxidation state decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2
oxidationreduction
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Oxidation–Reduction• 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 Na+Cl–(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+ 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
3 H2S + 2 NO3– + 2 H+ S + 2 NO + 4 H2O
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
+1 -2 +5 -2 +1 0 +2 -2 +1 -2
ox agred ag
+4 -2 +1 -1 +2 -1 0 +1 -2
oxidationreduction
oxidation
reduction
red agox ag
11
Common Oxidizing AgentsOxidizing Agent Product when Reduced
O2 O-2
H2O2 H2O
F2, Cl2, Br2, I2 F-1
, Cl-1
, Br-1
, I-1
ClO3-1
(BrO3-1
, IO3-1
) Cl-1
, (Br-1
, I-1
)
H2SO4 (conc) SO2 or S or H2S
SO3-2
S2O3-2
, or S or H2S
HNO3 (conc) or NO3-1
NO2, or NO, or N2O, or N2, or NH3
MnO4-1
(base) MnO2
MnO4-1
(acid) Mn+2
CrO4-2
(base) Cr(OH)3
Cr2O7-2
(acid) Cr+3
12
Common Reducing AgentsReducing Agent Product when Oxidized
H2 H+1
H2O2 O2
I-1
I2
NH3, N2H4 N2
S-2
, H2S S
SO3-2
SO4-2
NO2-1
NO3-1
C (as coke or charcoal) CO or CO2
Fe+2
(acid) Fe+3
Cr+2
Cr+3
Sn+2
Sn+4
metals metal ions
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Balancing Redox Reactions1) assign oxidation numbers
a) determine element oxidized and element reduced
2) write ox. & red. half-reactions, including electronsa) ox. electrons on right, red. electrons on left of arrow
3) balance half-reactions by massa) first balance elements other than H and Ob) add H2O where need Oc) add H+1 where need Hd) neutralize H+ with OH- in base
4) balance half-reactions by chargea) balance charge by adjusting electrons
5) balance electrons between half-reactions6) add half-reactions7) check
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Ex 18.3 – Balance the equation:I
(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 equation:I
(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 equation:I
(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 equation:I
(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 ReactantCount Element
ProductCount
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
oxidationreduction
ox: 2 I-1 I2 + 2e-1
red: H2O2 + 2e-1 + 2 H+ 2 H2Otot 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
oxidationreduction
ox: 2 Cl-1 Cl2 + 2 e-1 } x5red: 2 ClO3
-1 + 10 e-1 + 12 H+ Cl2 + 6 H2O} x1tot 10 Cl-1 + 2 ClO3
-1 + 12 H+ 6 Cl2 + 6 H2O
1 ClO3-1 + 5 Cl-1 + 6 H+1 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 timewhether 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 cellaka 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 systemelectrolyte
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Electrodes• Anode
electrode where oxidation occursanions attracted to itconnected to positive end of battery in electrolytic
cellloses weight in electrolytic cell
• Cathodeelectrode where reduction occurscations attracted to itconnected to negative end of battery in electrolytic
cellgains 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 currentunit = Ampere1 A of current = 1 Coulomb of charge flowing by each second1 A = 6.242 x 1018 electrons/secondElectrode surface area dictates the number of electrons that can
flow• the difference in potential energy between the reactants
and products is the potential differenceunit = Volt1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the external circuitamount 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, E°cell
25°C, 1 atm for gases, 1 M concentration of solutionsum 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 vstandard 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 E°red
• half-reactions with a stronger tendency toward oxidation than the SHE have a value for E°red
• E°cell = E°oxidation + E°reduction
E°oxidation = E°reduction
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 Ecell for the reaction at 25CAl(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 Ecell
Eox = −Ered = +1.66 v
Ered = +0.96 v
Ecell = (+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 conditions
Fe(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) Fe2+(aq) Pb2+(aq) 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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E°cell, G° and K
• for a spontaneous reaction one the proceeds in the forward direction with the
chemicals in their standard statesG° < 1 (negative)E° > 1 (positive)K > 1
• G° = −RTlnK = −nFE°cell
n is the number of electronsF = Faraday’s Constant = 96,485 C/mol e−
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Example 18.6- Calculate G° for the reactionI2(s) + 2 Br−
(aq) → Br2(l) + 2 I−(aq)
since G° is +, the reaction is not spontaneous in the forward direction under standard conditions
Answer:
Solve:
Concept Plan:
Relationships:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−
(aq)
G, (J)
Given:
Find:
E°ox, E°red E°cell G°redoxcell EEE
cellFEG n
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v
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
cellFEG n
J 101.1G
55.0485,96 mol 2G
5
C
J
mol
C
ee
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125.11 102.310
5.11V 0592.0
mol 2V 34.0log
K
eK
Example 18.7- Calculate at 25°C for the reactionCu(s) + 2 H+
(aq) → H2(g) + Cu2+(aq)
since < 1, the position of equilibrium lies far to the left under standard conditions
Answer:
Solve:
Concept Plan:
Relationships:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+
(aq)
Given:
Find:
E°ox, E°red E°cell redoxcell EEE K
nlog
V 0592.0Ecell
ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 v
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
Kn
logV 0592.0
Ecell
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Nonstandard Conditions - the Nernst Equation
• G = G° + RT ln Q
• E = E° - (0.0592/n) log Q at 25°C
• 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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V 41.1E
.0]1[.0]2[
]010.0[log
6
V 0592.0V 34.1E
][H][MnO
]Cu[log
V 0592.0EE
cell
83
3
cell
834
32
cellcell
n
Example 18.8- Calculate Ecellat 25°C for the reaction3 Cu(s) + 2 MnO4
−(aq) + 8 H+
(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
units are correct, Ecell > E°cell as expected because [MnO4−]
> 1 M and [Cu2+] < 1 M
Check:
Solve:
Concept Plan:
Relationships:
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:
E°ox, E°red E°cell Ecell
redoxcell EEE Q
nlog
V 0592.0EE cellcell
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
Qn
logV 0592.0
EE cellcell
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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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when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow
Concentration Cell
when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)
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 + NH4Clor MgBr2
• anode = Zn (or Mg)Zn(s) Zn2+(aq) + 2 e-
• cathode = graphite rod
• MnO2 is reduced2 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 reduced2 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 = PbPb(s) + SO4
2-(aq) PbSO4(s) + 2 e-
• cathode = Pb coated with PbO2
• PbO2 is reducedPbO2(s) + 4 H+(aq) + SO4
2-(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 reducedNiO2(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 H+1
M∙H(s) + OH-1(aq) M(s) + H2O(l) + e-1 E° = 0.89 v
• cathode = Ni coated with NiO2
• NiO2 is reducedNiO2(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 addedso 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 reactionmust 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 (−) chargecathode draws electrons and has a (+) charge
• in electrolytic cellselectrons are drawn off the anode, so it must have a place to
release the electrons, the + terminal of the batteryelectrons 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 compoundsgenerate 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 cathodeleast negative or most positive E°red
• 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 E°ox
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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 @ stand. cond.E° = -0.41 v @ pH 7
• possible anode reactionsoxidation of anion to elementoxidation of H2O to O2
2 H2O O2 + 4e-1 + 4H+1 E° = -1.23 v @ stand. cond.E° = -0.82 v @ pH 7
oxidation of electrodeparticularly Cugraphite doesn’t oxidize
• half-reactions that lead to least negative Etot will occur unless overvoltage changes the conditions
Tro, Chemistry: A Molecular Approach 71
Electrolysis of NaI(aq)
with Inert Electrodespossible oxidations2 I-1 I2 + 2 e-1 E° = −0.54 v2 H2O O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductionsNa+1 + 1e-1 Na0 E° = −2.71 v2 H2O + 2 e-1 H2 + 2 OH-1 E° = −0.41 v
possible oxidations2 I-1 I2 + 2 e-1 E° = −0.54 v2 H2O O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductionsNa+1 + 1e-1 Na0 E° = −2.71 v2 H2O + 2 e-1 H2 + 2 OH-1 E° = −0.41 v
overall reaction2 I−
(aq) + 2 H2O(l) I2(aq) + H2(g) + 2 OH-1(aq)
Tro, Chemistry: A Molecular Approach 72
Faraday’s 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 runscharge that flows through the cell = current x time
Tro, Chemistry: A Molecular Approach 73
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)
units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−
Check:
Solve:
Concept Plan:
Relationships:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Given:
Find:
s 1
C 5.5
Au g 6.5
Au mol 1
g 196.97
mol 3
Au mol 1
C 96,485
mol 1
s 1
C 5.5
min 1
s 60min 25
e
e
t(s), amp charge (C) mol e− mol Au g Au
C 6,4859
mol 1 ee mol 3
Au mol 1
Au mol 1
g 196.97
Tro, Chemistry: A Molecular Approach 74
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
Tro, Chemistry: A Molecular Approach 75
Rusting• rust is hydrated iron(III) oxide• moisture must be present
water is a reactantrequired for flow between cathode and anode
• electrolytes promote rustingenhances current flow
• acids promote rustinglower pH = lower E°red
Tro, Chemistry: A Molecular Approach 76
Tro, Chemistry: A Molecular Approach 77
Preventing Corrosion• one way to reduce or slow corrosion is to coat
the metal surface to keep it from contacting corrosive chemicals in the environmentpaintsome 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 cheapsacrificial electrode
galvanized nails
Tro, Chemistry: A Molecular Approach 78
Sacrificial Anode
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