1. Electrochemistry

2. Oxidation – Reduction Reactions
Consider the reaction of Copper wire and AgNO3(aq)
AgNO3(aq)
Cu(s)
Ag(s)

3. Oxidation – Reduction Reactions
If you leave the reaction a long time the solution goes blue!
The blue is due to Cu2+(aq)

4. Oxidation-Reduction Reactions
So when we mix Ag+(aq) with Cu(s) we get Ag(s) and Cu2+(aq)
Ag+(aq) + 1e-  Ag(s)
Cu(s)  Cu2+(aq) + 2e-
The electrons gained by Ag+ must come from the Cu
Can’t have reduction without oxidation (redox)
Each Cu can reduce 2 Ag+
2Ag+(aq) + 2e-  2Ag(s)
2Ag+(aq) + 2e- + Cu(s) 2Ag(s)+ Cu2+(aq) + 2e-
gain electrons = reduction
lose electrons = oxidation

5. Redox Cu/Ag E Cu
electron flow
Ag+
Ag+

6. Redox Cu/Ag E Cu2+ ΔE = e.Ecell Ag Ag e = charge on an electron
Ecell = Voltage in a electrochemical cell
If we could separate the two reactions we could use the energy gained by the e to do work
Redox Cu/Ag E
Cu2+
ΔE = e.Ecell Ag Ag

7. The maximum amount of energy available to do work is a definition of the free energy ΔG
Redox Cu/Ag
cell voltage
Free energy for redox rxn E
Cu2+
# electrons in redox rxn
F = Faraday’s constant
= C/mol Ag Ag

8. ΔEcell: the cell potential
In a redox reaction electrons are transferred to a more stable state
Most of the free energy of the reaction is due to these electrons
Can we access this energy?
Yes, by conducting the two half reactions in separate cells
2Ag+(aq) + 2e-  2Ag(s)
Cu(s)  Cu2+(aq) + 2e-
This is called the Voltaic electrochemical cell
The cell potential can be measured in such a cell with a voltmeter

9. Where oxidation happens
Voltaic Cell Cu/Ag
Ecell= V
DGcell= -2F x V
cations to the cathode
electrons to the cathode
Anions to the anode
Where oxidation happens
Where reduction
happens
Cu(s)  Cu2+(aq) + 2e-
2Ag+(aq) + 2e-  2Ag(s)

10. Electrochemistry and 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 (galvanic cell)
nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

11. Where oxidation happens
Voltaic Cell as a Battery: Electric Current Flowing Indirectly Between Atoms
Electrons do work here
electrons to the cathode
Anions to the anode
cations to the cathode
Where oxidation happens
Where reduction
happens

12. Electrodes Anode (donates electrons to the cathode)
electrode where oxidation occurs
In a Galvanic cell it is the –ve terminal and in an electrolytic cell it is the +ve terminal
anions attracted to it
connected to positive end of battery in electrolytic cell
loses weight in electrolytic cell
Cathode (attracts electrons from the anode)
electrode where reduction occurs
In a Galvanic cell it is the +ve terminal and in an electrolytic cell it is the -ve terminal
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

13. Current and Voltage
the number of electrons that flow through the system per second is the current
unit = Ampere (A)
1 A of current = 1 Coulomb (C) of charge flowing by each second
1 A = 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 per Coulomb of charge is called the Voltage
unit = Volt (V)
1 V of force = 1 J of energy/Coulomb of charge
the voltage is what drives electrons through the external circuit
amount of force pushing the electrons through the wire is called the electromotive force, emf, (also units of V)
the cell potential under standard conditions is called the standard emf, E°cell
25°C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions

14. Cell Notation shorthand description of Voltaic cell
anode | electrolyte || electrolyte | cathode
oxidation half-cell on left (anode), reduction half-cell on the right (cathode)
single | = phase barrier
if multiple electrolytes in same phase, a comma is used rather than |
often use an inert electrode
double line || = salt bridge

15. Write out the cell notation for this cell
anode | electrolyte || electrolyte | cathode
Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

16. Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s)
Practice - Sketch and Label the Voltaic Cell Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.
Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s)
anode
cathode
Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V
Fe2+(aq) + 2 e−  Fe(s) Ered = V
The anode is where oxidation happens
Reverse the reaction for Fe2+(aq) and flip the sign to get Eox
Ecell = Eox + Ered
= 0.45V V = 0.32V
Fe(s)  Fe2+(aq) + 2 e− Eox = V
Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V

17. ox: Fe(s)  Fe2+(aq) + 2 e− Eox = +0.45 V
red: Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V
tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s) Ecell = V

18. Measuring the Tendency to Reduce: Standard Reduction Potential
when two half-cells are connected, one side will oxidize and one side will reduce, but which will do what?
clearly the electrons will flow so that the half-reaction with the stronger tendency to reduce will reduce, making the other half-cell reaction oxidize
we cannot measure the absolute tendency of a half-reaction to reduce, we can only measure it relative to another half-reaction
To measure the tendency to reduce we measure the E (the voltage) in a voltaic cell where one of the electrodes is a standard hydrogen electrode (see right)
We assign a potential difference = 0.0 V to the following reaction
2H+(aq) +2e-  H2(g) Eored = 0.0V
E°cell = Eoox + Eored = Eoox V = Eoox
In so doing scientists can obtain standard reduction potentials as follows

20. The measured voltage can then be placed in a table like this one, where we write the voltage for the reduction reaction

21. Half-Cell Potentials ΔE°cell = E°oxidation + E°reduction
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
ΔE°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 (voltage is the energy per unit charge), even if you need to multiply the half-reactions to balance the equation
ΔGocell=-nFΔE°cell
Since ΔGocell < 0 to be spontaneous ΔE°cell > 0 for a redox reaction to be spontaneous if ΔE°cell < 0 the reaction will not be spontaneous

22. red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
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 Eo for each half-reaction from the standard reduction table. If you need the oxidation potential of one Eoox=-Eored and sum to get Eocell
Eoox = −Eored = V
Eored = V
Eocell = (+1.66 V) + (+0.96 V) = V
The ½ rxn with the largest reduction potential is the reduction reaction in the cell
The ½ rxn with the smallest reduction potential is reversed and sign flipped
reverse

23. red: Mg2+(aq) + 2 e−  Mg(s)
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) Ered = V
ox: Fe(S)  Fe2+(aq) + 2 e− Eox =+0.45V
Ecell = Eox + Ered = 0.45V – 2.37V = -1.92V
Since Ecell < 0 the cell reaction is not spontaneous as written
reverse

24. Keeping track of electrons
Redox reactions involve the transfer of electrons from the donor to the acceptor but sometimes it is hard to tell eg (2H2O22H2O + O2)
The donor loses electrons and is oxidized
The acceptor acquires electrons and is reduced
To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has
We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom

25. Assigning Oxidation Numbers
The sum of the oxidation numbers Q of all the atoms in a compound/ion up to the charge on the compound/ion. This means ….
free elements have an oxidation state = 0
QNa = 0 and QCl = 0 in 2 Na(s) + Cl2(g)
monatomic ions have an oxidation state equal to their charge
QNa = +1 and QCl = -1 in NaCl
The sum of the oxidation numbers of all the atoms in a neutral compound is 0
the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion
(a) Group I metals have an oxidation state of +1 in all their compounds
QNa = +1 in NaCl
(b) Group II metals have an oxidation state of +2 in all their compounds
QMg = +2 in MgCl2
F has an oxidation number QF = -1
H has an oxidation number QH = +1
O has an oxidation number QO = -2

26. Determine the oxidation states of the following
S in SO42-(aq)
Na in NaF(aq)
Fe in FeCl3(aq)
S in S2O32-(aq)
Cr in CrO42-(aq)

27. 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
rule 4
rule 1
rule 5
rule 4
CH O2 → CO2 + 2 H2O –
oxidation
reduction

28. Oxidation–Reduction 2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
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

29. 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

30. 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
oxidation
reduction

31. Common Oxidizing Agents

32. Common Reducing Agents

33. Balancing Redox Reactions: Aqueous Solution
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 H+1 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

34. 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:

35. Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
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)
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
then O by adding H2O
ox: 2 I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s) + 2 H2O(l)

36. 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- } x 3
red: {MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) } x 2
ox: 6 I-(aq)  3 I2(aq) + 6 e-
red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)

37. 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

38. Your turn
Your Turn

39. MnO4-(aq) + HSO3-(aq)  Mn2+(aq) + SO42-(aq)
Balance the following reaction in acid
MnO4-(aq) + HSO3-(aq)  Mn2+(aq) + SO42-(aq)
In the above reaction what is the oxidation number of Mn in MnO4-(aq)?
What is the oxidation number of Mn in Mn2+(aq)
What is the oxidation number of S in HSO3-(aq)?
What is the oxidation number of S in SO42-(aq)?
In the above reaction is MnO4-(aq) the oxidizing or reducing agent?
20. When permanganate is reacted with hydrogen sulfite in base instead of acid we get
MnO4-(aq) + HSO3-(aq)  MnO2(s)+ SO42-(aq)
Balance the above equation  

40. The Relationship between ΔGo and Eocell
Remember ΔGo is the maximum work the system can do
Eocell is the cell potential energy (standard emf) per unit of charge
Since the potential energy is the maximum amount of work that can be done on the surrounding we can write
The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write
Δ 𝐺 𝑐𝑒𝑙𝑙 =Δ 𝐺 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄
−𝑛𝐹 𝐸 𝑐𝑒𝑙𝑙 =−𝑛𝐹 𝐸 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄
𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄= 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑉 𝑛 𝑙𝑜𝑔𝑄
𝐸 𝑐𝑒𝑙𝑙 𝑜 = 𝑉 𝑛 𝑙𝑜𝑔𝐾
Divide b.s. by -nF
At equilibrium E = 0

41. E° at Nonstandard Conditions

42. Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
You are given a Zn/Cu Voltaic cell. Zn(s)|1M ZnSO4(aq)||IM CuSO4|Cu(s)
The net reaction in this cell is
Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
1. Which metal is the anode?
a. Cu
b. Zn
2. Which metal is the cathode?
3. The measured DEocell = 1.10V at 298K
4. What is DGo for this reaction?
5. Is the reaction spontaneous?
6. What is K for this reaction when equilibrium is reached?
7. Does equilibrium favor reactants or products? 
8. What is the DEcell for the following cell Zn(s)|2M ZnSO4(aq)||0.1M CuSO4|Cu(s) ? 
9. What is the free energy DG for this reaction at 298K?

43. Voltaic Cells as Batteries
Use chemical reactions to provide a source of energized electrons that can power electronics
Batteries in everything, phones, laptops, tablets, watches, cars
Actively seeking batteries that are light, powerful, and retain their effectiveness after many recharge cycles
Molten metal batteries developed for storing electricity in power stations (infinitely re-chargable)
Huge area of scientific and commercial interest
Batteries are being made ever smaller for energy efficient processor applications
Recent breakthroughs using gold nanowires that are gel coated make batteries that will last longer than the devices they are in
Nanotechnology, even biotechnology using viruses!
A benign bacteria virus serves as a template in water to grow manganese oxide nanowires that, when combined with palladium, increases the energy potential of a lithium battery

44. Voltaic Cells as Batteries Lead Storage Battery
Anode: Pb
Pb  Pb2+ + 2e-
Pb(s) + SO42-(aq)  PbSO4(s) + 2 e-
Cathode: Pb coated with PbO2
Pb4+ +2e- Pb2+
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- PbSO4(s) + 2 H2O(l)
Electrolyte: 30% H2SO4
cell voltage = 2.09 V
6 in series makes 12V
rechargeable, heavy

45. Voltaic Cells as Batteries Lithium Ion Battery
anode = graphite impregnated with Li ions
cathode = Li - transition metal oxide
Electrolyte is an organic solvent like dimethylcarbonate, with a Li salt like LiPF6
rechargeable, long life, very light, more environmentally friendly, greater energy density
High surface area electrodes
Tiny ions that move fast between electrodes

46. Voltaic Cells as Batteries
Why do we use Li-ion batteries in our computers and phones and not say Pb Acid Batteries?
We want to make batteries as portable (ie small and light) as possible

47. H2 Fuel Cell
like batteries in which reactants are constantly being added
so it never runs down!
Anode and Cathode both Pt coated metal (may change with cheaper technology)
Electrolyte is OH– solution
Anode Reaction:
2 H2 + 4 OH– → 4 H2O(l) + 4 e-
Cathode Reaction:
O2 + 4 H2O + 4 e- → 4 OH–
Toyota Mirai available in CA $499/month 312 miles per tank

48. Electrolysis
electrolysis is the process of using electricity to make non-spontaneous redox reactions happen
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
Make halogen gasses from halide salts

50. 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 connects to the anode
the - terminal of the battery connects to the cathode
cations attracted to the cathode, anions to the anode
Oxidation at the anode, reduction at the cathode
some electrolysis reactions require more voltage than Etot, called the overvoltage

51. Applications of electrolysis: 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
Gold plated terminals in electronics (Au is the most conductive of all metals)

52. 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

53. Electrolysis of NaCl(l)

54. 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

55. 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

56. Electrolysis of NaI(aq): Inert Electrodes
Write down ions/molecules/metals present and don’t forget water
Here Na+, I-, H2O are what is present (the electrodes here are inert)
Write down the reduction reactions involving these species
I2(aq)+2e-  2I-(aq) Ered = 0.54V
O2(g) + 4e-1 + 4H+1(aq) 2 H2O(l) Ered = V
2 H2O + 2 e-1  H2 + 2 OH-1 Ered = 0.41V
Na+(aq) + e-  Na(s) Ered = -2.71V
Rearrange the equations so the reactants are on the left (reverse the sign of E if you flip the equation – these will be the oxidation reactions
possible oxidations (anode –ve terminal)
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible oxidations
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductions (cathode +ve terminal)
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2(g) + 2 OH-1E° = -0.41V
possible reductions
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v
brown liquid
overall reaction
2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq)
bubbles
Ecell = -0.54V – 0.41V = -0.95V
The battery needs to at least be 1V to make this happen

57. Electrolysis in aqueous solutions: Overpotentials
There are (potentially) competing processes in the electrolysis of an aqueous solution consider now NaCl(aq) with C electrodes
Cathode
Anode
We choose those reactions that are thermodynamically favored
But…chlorine is evolved at the anode!

58. Overpotentials
Thermodynamics true at equilibrium but when the process has a current flowing kinetics plays a role
Overpotential represents the additional voltage that must be applied to drive the process
In the NaCl(aq) solution the overpotential for evolution of oxygen is greater than that for chlorine, and so chlorine is evolved preferentially
The limiting process in electrolysis is usually diffusion of the ions in the electrolyte
Overpotential will depend on the electrolyte and electrode
Driving the cell at the least current will give rise to the smallest overpotential

59. Faraday’s Law
the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current (A), and the length of time (s) the cell runs
charge = current x time
Q = I x t

60. Calculate the mass of Au that can be plated in 25 min using 5
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:
t(s), amp
charge (C)
mol e−
mol Au
g Au
Concept Plan:
Relationships:
Solve:
Check:
units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

61. Corrosion
corrosion is the spontaneous oxidation of a metal by chemicals in the environment
since many materials we use are active metals (metals that like to make ions), corrosion can be a very big problem

62. Rusting rust is hydrated iron(III) oxide moisture must be present
water is a reactant
required for flow between cathode and anode
electrolytes (sea-water) promote rusting
enhances current flow
acids (acid rain) promote rusting
lower pH = lower E°red

63. O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Rusting
Anode (inside the droplet):
Fe(s)  Fe2+(aq) + 2e-
Cathode (at the surface of the droplet:
O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Within the droplet
Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s)
Rust is then quickly produced by the oxidation of the precipitate at the edge of the droplet.
4Fe(OH)2(s) + O2(g)  2Fe2O3 •H2O(s) + 2H2O(l)

64. Preventing Corrosion Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V
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 steel (Zinc coating)
Zn  Zn2+ + 2e V
Fe  Fe2+ + 2e V

65. Sacrificial Anode Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V
Galvanic anode on the hull of a ship