1. Chapter 18 Electrochemistry
Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 18 Electrochemistry

2. Redox Reaction one or more elements change oxidation number
all single displacement, and combustion,
some synthesis and decomposition
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3. Redox Reaction always have both oxidation and reduction
split reaction into oxidation half-reaction and a reduction half-reaction
aka e- transfer reactions
half-reactions include e-
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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4. Oxidation & Reduction oxidation: ox number of an element increases
element loses e-
compound adds O
compound loses H
half-reaction has e- as products
reduction:
ox number of an element decreases
element gains e-
compound loses O
compound gains H
half-reactions have e- as reactants
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5. 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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6. 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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7. 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
CCl4
Group 6A
CS2
Group 5A -3
NH3
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8. 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
CH O2 → CO2 + 2 H2O –
oxidation
reduction
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9. Oxidation–Reduction oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them
reactant that reduces an element in another reactant = reducing agent
the reducing agent contains the element that is oxidized
reactant that oxidizes element in another reactant = 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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10. Identify the Oxidizing and Reducing Agents in Each of the Following
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
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11. Identify the Oxidizing and Reducing Agents in Each of the Following
red ag
ox ag
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
reduction
oxidation
ox ag
red ag
reduction
oxidation
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12. Common Oxidizing Agents

13. Common Reducing Agents

14. Balancing Redox Reactions
1. assign oxidation numbers
determine element oxidized and element reduced
2. write ox. & red. half-reactions, including e-
ox. electrons on right, red. electrons on left of arrow
3. balance half-reactions by mass
first balance elements other than H and O
add H2O where need O
add H+ where need H
neutralize H+ with OH- in base
4. balance half-reactions by charge
balance charge by adjusting e-
5. balance e- between half-reactions
6. add half-reactions
7. check
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15. Assign Oxidation States
Ex 18.3 – Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Assign Oxidation States
Separate into half-reactions
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
Assign Oxidation States
I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
Separate into half-reactions
ox:
red:
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16. Balance half-reactions by mass
Ex 18.3 – Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
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
ox: 2 I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
Balance half-reactions by mass
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
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)
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)
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17. Balance Half-reactions by charge ox: 2 I(aq)  I2(aq) + 2 e
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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18. Add the Half-reactions ox: 6 I(aq)  3 I2(aq) + 6 e
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
Reactant
Count
Element
Product 6 I 2 Mn 12 O 8 H 8
charge
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19. Practice - Balance the Equation H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O
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20. Practice - Balance the Equation H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O
oxidation
reduction
ox: 2 I- ® I2 + 2e-
red: H2O2 + 2e- + 2 H+ ® 2 H2O
tot 2 I- + H2O2 + 2 H+ ® I2 + 2 H2O
H2O2 + 2 KI + H2SO4 ® K2SO4 + I2 + 2 H2O
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21. Practice - Balance the Equation ClO3- + Cl- ® Cl2 (in acid)
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22. Practice - Balance the Equation ClO3- + Cl- ® Cl2 (in acid)
oxidation
reduction
ox: 2 Cl- ® Cl2 + 2 e-1 } x5
red: 2 ClO e H+ ® Cl2 + 6 H2O} x1
tot 10 Cl- + 2 ClO H+ ® 6 Cl2 + 6 H2O
ClO Cl H+ ® 3 Cl2 + 3 H2O
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23. Electrical Current
current of a liquid in a stream, = amount of water that passes by in a given period of time
electric current = amount of electric charge that passes a point in a given period of time
whether as e- flowing through a wire or ions flowing through a solution
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24. Redox Reactions & Current
redox reactions involve the transfer of e- 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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25. Electric Current Flowing Directly Between Atoms
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26. Electric Current Flowing Indirectly Between Atoms
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27. 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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28. Electrochemical Cells
redox reactions kept separate
half-cells
e- flow in a wire along and ion flow in solution constitutes an electric circuit
requires a conductive metal or graphite electrode to allow the transfer of e-
through external circuit
ion exchange between the two halves of the system
electrolyte
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29. 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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30. Voltaic Cell
the salt bridge is required to complete the circuit and maintain charge balance
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31. Current and Voltage
# e- that flow through the system per second is the current
unit = Ampere
1 A of current = 1 Coulomb of charge per second
1 A = x 1018 e-/sec.
Electrode surface area dictates the number of e- that can flow
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32. Current and Voltage
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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33. Cell Potential
the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential
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 solution
Ecell = EOX + ERED
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34. 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, double line || = salt bridge
if multiple electrolytes in same phase, a comma is used rather than |
often use an inert electrode
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35. Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)
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36. Standard Reduction Potential
a half-reaction with a strong tendency to occur has a large + half-cell potential
two half-cells are connected, e- 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
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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37. Tro, Chemistry: A Molecular Approach

38. 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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40. Tro, Chemistry: A Molecular Approach

41. Separate the reaction into the oxidation and reduction half-reactions
Ex 18.4 – Calculate Ecell for the reaction at 25C Al(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 = v
Ered = v
Ecell = (+1.66 v) + (+0.96 v) = v
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42. Separate the reaction into the oxidation and reduction half-reactions
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: Fe2+(aq) + 2 e−  Fe(s)
since Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written
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43. 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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44. Practice - Sketch and Label the Voltaic Cell Fe(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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45. 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 = V
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46. 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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47. E°cell, DG° and K for a spontaneous reaction
one the proceeds in the forward direction with the chemicals in their standard states
DG° < 1 (negative)
E° > 1 (positive)
K > 1
DG° = −RTlnK = −nFE°cell
n is the number of electrons
F = Faraday’s Constant = 96,485 C/mol e−
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48. Example 18.6- Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given:
Find:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
DG°, (J)
Concept Plan:
Relationships:
E°ox, E°red
E°cell
DG°
Solve:
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v
red: I2(l) + 2 e− → 2 I−(aq) E° = v
tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 v
Answer:
since DG° is +, the reaction is not spontaneous in the forward direction under standard conditions
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49. E°cell, DG° and K DG° = −RTlnK = −nFE°cell E°cell = RT x ln K nF
R = J/mol.K
lnK = 2.303log K
F = 96,485 C/mol e-
E°cell = logK n
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50. Example 18.7- Calculate K at 25°C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given:
Find:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K
Concept Plan:
Relationships:
E°ox, E°red
E°cell K
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 v
red: 2 H+(aq) + 2 e− → H2(aq) E° = v
tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 v
Answer:
since K < 1, the position of equilibrium lies far to the left under standard conditions
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51. Nonstandard Conditions - the Nernst Equation
Relationship between Ecell (nonstandard) and E°cell (standard)
DG = DG° + RT ln Q
Subs. DG° = -nFE°cell into above eqn.
-nFEcell = -nFE°cell + -RTlnQ (divide by -nF)
Ecell = E°cell - (RT/nF) log Q
R = J/mol.K, (RT/nF)lnQ = (0.0592/n)logQ
Ecell = E° - (0.0592/n) logQ
Called the Nernst equation
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52. Nonstandard Conditions - the Nernst Equation
Ecell = E°cell - (0.0592/n) log Q at 25°C
1. when Q = 1 (std. conditions) Ecell = E°cell
2. At equilibrium,Q = K,
Ecell = E°cell - (0.0592/n) log K and (0.0592/n) log K = E°cell
Ecell = 0
Potential reaches zero as concentrations approach equilibrium
Used to calculate E when concentrations not 1 M
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53. E at Nonstandard Conditions
Reactant conc. > standard conditions
Product conc. < standard conditions … reaction shifts right
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54. Example Calculate Ecell at 25°C for the reaction 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
Given:
Find:
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
[Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Concept Plan:
Relationships:
E°ox, E°red
E°cell
Ecell
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° = v
tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)) E° = v
Check:
units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M
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55. 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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56. Concentration Cell Cu(s) Cu2+(aq) (1 M)  Cu2+(aq) (1 M) Cu(s)
when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow
Cu2+(aq) + 2e- → Cu(s) V
Cu(s) → Cu2+(aq) + 2e V
Cu2+(aq) + Cu(s) → Cu(s) + Cu2+(aq)
E°cell = E°red + E°ox = 0V
Cu(s) Cu2+(aq) (1 M)  Cu2+(aq) (1 M) Cu(s)
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57. Concentration Cell
when the cell concentrations are different, e- flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)
Cell potential Ecell calculated using Nernst eqn.
Ecell = E° - (0.0592/n) logQ
Ecell = E° - (0.0592/n) log([OX]/[RED])
= 0.068V
Cu(s) Cu2+(aq) (0.010 M)  Cu2+(aq) (2.0 M) Cu(s)
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58. 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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59. 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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60. PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-
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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61. NiCad Battery electrolyte is concentrated KOH solution anode = Cd
Cd(s) + 2 OH-(aq) ® Cd(OH)2(s) + 2 e- E0 = 0.81 v
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e- ® Ni(OH)2(s) + 2OH- 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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62. Ni-MH Battery electrolyte is concentrated KOH solution
anode = metal alloy with dissolved hydrogen
oxidation of H from H0 to H+
M∙H(s) + OH-(aq) ® M(s) + H2O(l) + e- E° = 0.89 v
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e- ® Ni(OH)2(s) + 2OH- 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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63. 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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64. Tro, Chemistry: A Molecular Approach

65. 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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66. 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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67. Tro, Chemistry: A Molecular Approach

68. 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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69. 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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70. 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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71. Electrolysis of Water
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72. 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
Tro, Chemistry: A Molecular Approach

73. Electrolysis of NaCl(l)
Tro, Chemistry: A Molecular Approach

74. 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 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
Tro, Chemistry: A Molecular Approach

75. 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° = stand. cond.
E° = pH 7
possible anode reactions
oxidation of anion to element
oxidation of H2O to O2
2 H2O ® O2 + 4e-1 + 4H+1 E° = stand. cond.
E° = pH 7
oxidation of electrode
particularly Cu
graphite doesn’t oxidize
half-reactions that lead to least negative Etot will occur
unless overvoltage changes the conditions
Tro, Chemistry: A Molecular Approach

76. Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations
2 I- ® I2 + 2 e- E° = −0.54 v
2 H2O ® O2 + 4e- + 4H+1 E° = −0.82 v
possible oxidations
2 I- ® I2 + 2 e- E° = −0.54 v
2 H2O ® O2 + 4e- + 4H+ E° = −0.82 v
possible reductions
Na+ + e- ® Na0 E° = −2.71 v
2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = −0.41 v
possible reductions
Na+ + e- ® Na0 E° = −2.71 v
2 H2O + 2 e- ® H2 + 2 OH- E° = −0.41 v
overall reaction
2 I−(aq) + 2 H2O(l) ® I2(aq) + H2(g) + 2 OH-(aq)
Tro, Chemistry: A Molecular Approach

77. 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 runs
charge that flows through the cell = current x time
Tro, Chemistry: A Molecular Approach

78. Example 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)
Given:
Find:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Concept Plan:
Relationships:
t(s), amp
charge (C)
mol e−
mol Au
g Au
Solve:
Check:
units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−
Tro, Chemistry: A Molecular Approach

79. 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

80. 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 E°red
Tro, Chemistry: A Molecular Approach

81. Tro, Chemistry: A Molecular Approach

82. 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
Tro, Chemistry: A Molecular Approach

83. Sacrificial Anode
Tro, Chemistry: A Molecular Approach