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Electrochemistry
Na+ Cl-
Na Cl
Are you sure I can have that
electron?
I’m positive!
Useful Links
• This presentation: www.canisius.edu/~szczepas
• Past Years’ Exams + Answers– Google: ACS Chemistry Olympiad– http://www.acs.org/content/acs/en/education/
students/highschool/olympiad/pastexams.html
Typical Olympiad Topics• Oxidation Numbers• Balancing Redox Reactions• Galvanic Cell Architecture• Standard Cell Potentials• Non Standard Conditions• Electrolysis
• Common Oxyanions• Redox Concepts
From 2012
Oxidation Numbers (States)• Keeping Track of Electrons Gained or Lost
4 Rules for Assigning Oxidation Numbers:
In order of Importance!
1) Atom in Elemental Form– Oxidation Number is Always Zero
Examples: Fe Atom
Ar Atom
H Atom in H2 Molecule
O Atom in O2 Molecule
P Atom in P4 Molecule
Oxidation Numbers (States)
2) Monatomic Ions– Oxidation Number = Charge– Examples: K+ Oxidation Number = +1
Mg2+ Oxidation Number = +2
Al3+ Oxidation Number = +3
N3- Oxidation Number = -3
S2- Oxidation Number = -2
FeCl2 ON(Fe) = +2
Oxidation Numbers are written with the sign before the number to distinguish them from Actual Charges
Oxidation Numbers (States)3) Nonmetals (Usually Negative Oxidation #’s, But Can Be Positive)
Fluorine– Oxidation Number is -1 In All Compounds
Hydrogen– Oxidation Number is +1 When Bonded to Non Metals– Oxidation Number is -1 When Bonded to Metals
Oxygen– Oxidation Number is Usually -2 In Molecular and Ionic Compounds– In Peroxides (O2
2-) Oxidation Number is -1 for Each O Atom
Other Halogens– Oxidation Number is -1 in Most Binary Compounds– Oxidation Number When Combined with Oxyanions Can Be Positive
Oxidation Numbers (States)4) Sum of the Oxidation Number of All Atoms in a Neutral
Compound is ZeroH2SO3
Oxidation Number of H = +1 (H Bonded to NonMetal) Oxidation Number of O = -2 Sum is Zero 0 = 2×ON(H) + 1×ON(S) + 3×ON(O)
0 = 2× (+1) + 1×ON(S) + 3× (-2) ON(S) = +4
Sum of the Oxidation Numbers in a Polyatomic Ion Equals the Charge of the Ion
H2AsO4-
-1 = 2×ON(H) + 1×ON(As) + 4×ON(O)ON(As) = +5
From 2012
From 2014
K = +1
H = +1
O = -2
N = ???
From 2014
Split to ions
K+ NH4+ H2O “hydrate”
AsO43-
2012
Balancing Redox Reactions by Half-Reactions
Reduction Half Reaction (Electrons Taken In)2H+(aq) + 2e- H2(g)
Another Example: Ag+(aq) + e- Ag(s)
Oxidation Half Reaction (Electrons Given Off)Zn(s) Zn2+(aq) + 2 e-
Another Example: C2O42- (aq) 2 CO2(g) + 2 e-
22Zn(s) 2H (aq) Zn (aq) H (g)
To Balance Electrons, Reductions and Oxidations MUST Occur Simultaneously
NO3-(aq) + 4 H+ + 3 e- NO(g) + 2 H2O(l)
C2O42- (aq) 2 CO2(g) + 2 e-
ID the Reducing Agent in the Unbalanced Reaction:ClO3
- + Br- Cl2 + Br2
Half-Reactions
Balancing Redox Rxns1. Divide total reaction into two half reactions.2. Balance each half
a. All elements besides H and Ob. Balance O by adding H2Oc. Balance H by adding H+
d. Balance residual charge by adding e-
3. Multiply each half to least common multiple of electrons4. Add half reactions and cancel5. Check if balanced
The above procedure uses acid or neutral conditions as the default.
*To convert from acid to base conditions after steps 1-5, add enough OH- to both sides to neutralize all the H+ to H2O, then cancel out any excess.
2012
From 2014
A sample of copper metal is dissolved in 6 M nitric acid contained in a round bottom flask. This reaction yields a blue solution and emits a colorless gas which is found to be nitric oxide.
Write a balanced equation for this reaction.
23Cu(s) NO (aq) Cu (aq) NO(g) Unbalanced
Zn(s) → Zn2+(aq) + 2 e-
Zn2+
e-
e-
Cu2+
Cu2+(aq) + 2 e- → Cu(s)
Cl-
Cl-
Electrode Electrode
Oxidation Reduction
Voltaic (Galvanic) Cell
to the cathode
to the anode
2008 Local 39. Which occurs at the anode of any voltaic cell?
I. A metal electrode dissolves.
SO32- + H2O → SO4
2- + 2 H+ + 2 e-
II. A substance undergoes oxidation.
Fe(s) → Fe2+ + 2 e-
III. Positive ions are deposited from the solution.
(A) I only
(B) II only
(C) I and II only
(D) I and III only
Cell Potential• Voltaic Cell
Spontaneous Redox Reaction (Ecell>0) Used to Perform Electrical Work
Similar to a Waterfall (Water Falls from High to Low Potential Energy) Electrons Flow Spontaneously from High to Low Electric Potential
Use Cell Potential (Cell EMF) (Ecell)• Volt Difference in Potential Energy per Electrical
Charge (1V = 1J/C) (e- charge = 1.60x10-19C)
• Potential Difference Between 2 Electrodes
Standard Cell Potential
• Use Standard Reduction Potentials for the Reduction and Oxidation Half-Reactions
• Note: No Multiplying Reduction Potential By Stoichiometry
• Voltaic (Galvanic) Cell: Positive Ecell
• Electrolytic Cell: Negative Ecell
0 0 0cell red red(anode)(cathode)E =E -E
T = 25°C
Standard State Gas (P = 1atm) Species in Solution (1 M Concentration)
Most easily oxidized
2008 Local 41. What is the standard cell potential for the voltaic cell: Cr | Cr3+ || Pb2+ | Pb ?
E0red / V
Pb2+ + 2 e- → Pb -0.13
Cr3+ + 3 e- → Cr -0.74
(A) 1.09
(B) 0.61
(C)-0.61
(D)-1.09
Half Reaction E0 (V)Zn2+(aq) + 2e- Zn(s) -0.763Cr3+(aq) + e- Cr2+(aq) -0.408Tl+(aq) + e- Tl(s) -0.336Cu2+(aq) + e- Cu+(aq) +0.161Fe3+(aq) + e- Fe2+(aq) +0.769
Use the Standard Reduction Potentials to Find the Standard Cell Potential, E0
cell, for the Reaction:
Zn(s) + 2 Tl+(aq) Zn2+(aq) + 2 Tl(s) 81Tl: Thallium
2redZn(s) Zn (aq) 2e E 0.763V (Anode/Oxidation)
red2Tl (aq) 2e Tl(s) E 0.336V (Cathode/Reduction)
0 0 0cell red red(anode)(cathode)E =E -E -0.336V-(-0.763V) 0.427V
Calculate the E0rxn based on the standard reduction potentials above.
Which reaction(s) is(are) spontaneous?
Half Reaction E0 (V)Zn2+(aq) + 2e- Zn(s) -0.763Cr3+(aq) + e- Cr2+(aq) -0.408Tl+(aq) + e- Tl(s) -0.336Cu2+(aq) + e- Cu+(aq) +0.161Fe3+(aq) + e- Fe2+(aq) +0.769
2+ 3+ 3+ 2+Cr (aq) + Fe (aq) Cr (aq) + Fe (aq)
2+ 2+ + 3+Cu (aq) + Fe (aq) Cu (aq) + Fe (aq)
2+ 3+ -red
3+ - 2+red
0 0 0cell red red(anode)(cathode)
Cr (aq) Cr (aq) + e (oxidation, E = -0.408V)
Fe (aq) +e Fe (aq) (reduction, E = +0.769V)
E =E -E 0.769V-(-0.408V) 1.177V
2+ 3+ -red
2+ - +red
0 0 0cell red red(anode)(cathode)
Fe (aq) Fe (aq) + e (oxidation, E = +0.769V)
Cu (aq) + e Cu (aq) (red uction, E = +0.161V)
E =E -E 0.161V-(0.769V) 0.608V
G n F E
0
0
0
0
0
ΔG = ΔG +RTln Q
-nFE = -nFE +RTln Q
-nFE +RTln Q-nFE =
-nF -nFRT
E = E - ln QnF0.0592 V
E = E - log Qn
Ecell Non-Standard Conditions
Nernst Equation
2012
An electrochemical cell is constructed with a piece of copper wire in a 1.00 M solution of Cu(NO3)2 and a piece of chromium wire in a 1.00 M solution of Cr(NO3)3. (1) Write a balanced equation for the spontaneous reaction that occurs in this cell and calculate the potential it produces. (2) Sketch a diagram for this cell. Label the anode. Show the direction of electron flow in the external circuit. Show the direction of movement of nitrate ions. Explain. (3) The cell is allowed to operate until the [Cu2+] = 0.10 M. Find the [Cr3+] (4) Calculate the cell potential at these concentrations.
Half Reaction E0 (V) Cr3+(aq) + 3 e- Cr(s) -0.744 Cu2+(aq) + 2 e- Cu(s) +0.340
An electrochemical cell is constructed with a piece of copper wire in a 1.00 M solution of Cu(NO3)2 and a piece of chromium wire in a 1.00 M solution of Cr(NO3)3. (1) Write a balanced equation for the spontaneous reaction that occurs in this cell and calculate the potential it produces. (2) Sketch a diagram for this cell. Label the anode. Show the direction of electron flow in the external circuit. Show the direction of movement of nitrate ions. Explain.
Half Reaction E0 (V) Cr3+(aq) + 3 e- Cr(s) -0.744 Cu2+(aq) + 2 e- Cu(s) +0.340
2 33Cu (aq) 2Cr(s) 2Cr (aq) 3Cu(s) E 1.084V
An electrochemical cell is constructed with a piece of copper wire in a 1.00 M solution of Cu(NO3)2 and a piece of chromium wire in a 1.00 M solution of Cr(NO3)3. (3) The cell is allowed to operate until the [Cu2+] = 0.10 M. Find the [Cr3+] (4) Calculate the cell potential at these concentrations.
Half Reaction E0 (V) Cr3+(aq) + 3 e- Cr(s) -0.744 Cu2+(aq) + 2 e- Cu(s) +0.340
2+
3+2+ 3+
2+
3+
Assume Vol 1.00L
Change in Moles Cu =1.00 mol(initial) - 0.10 mol(final) = 0.90 mol
2 mol Cr0.90 mol Cu 0.60 mol Cr
3 mol Cu
Change in Moles Cr =1.00 mol(initial) + 0.60 mol(produced by oxidat
3+
ion) = 1.60 mol
Concentration Cr = 1.60M
23+
032+
2
3
Cr0.0592 VE = E - log
n Cu
1.600.0592 VE = 1.084V- log 1.050V
6 0.10
How many moles of electrons must pass through a cell to produce 5.00 kg of Aluminum from Al2O3?
3 1 65.00 10 556
26.9815 2
mol Al eg mol e
g Al
Al2O3 + 6 e- 2 Al(s) + 3 O2-
Using F = 96485 C/mol e-, and A=C/s, 2. How long will this take using a current of 33.5 A?
696485 1 1556 1.60 10 18.5
33.5
C Amol e s days
Cmol e As
1. Calculate the number of moles of electrons needed.
Should also know that W = J/s.
2008 Local 42. During the electrolysis of AgNO3, what would happen to the mass of silver metal deposited if the current is doubled and the electrolysis time is decreased to ½ of its initial value?
(A) It would stay the same.
(B) It would increase to twice its initial value.
(C) It would decrease to ¼ of its initial value.
(D) It would decrease to ½ of its initial value.
2012 Local #42
How to memorize the negative ions:
______-ateBO3
3- CO32- NO3
-
SiO32- PO4
3- SO42- ClO3
-
AsO43- SeO4
2- BrO3-
IO3-
Rule: Most common ion consisting of one nonmetal atom, 2 or 3 oxygen atoms, and a negative charge.Element name may be truncated, and followed by suffix “ate”.
Example: NO3- is nitrate, PO4
3- is phosphate, ClO3- is chlorate
Note, not all “-ate”s have a corresponding “-ite”!
______-ideIrregular
* N3- O2- F-
P3- S2- Cl-
As3- Se2- Br-
I-
Rule: Ion consisting of one nonmetal atom with a negative charge has truncated element name with suffix “ide”.Negative charge is how many steps from right edge of periodic table (noble gas group).
Example: N3- is nitride, O2- is oxide, F- is fluoride
* Carbide is actually C22-. This does not follow the naming rules above, and you do not need to know this ion.
Also:H-
______-iteNO2
-
PO33- SO3
2- ClO2-
AsO33- SeO3
2- BrO2-
IO2-
Rule: Less common ion consisting of one nonmetal atom, 2 or 3 oxygen atoms, and a negative charge.Element name may be truncated, and followed by suffix “ide”.Same charge, but one less oxygen from the “-ate”s.
Example: NO2- is nitrite, PO3
3- is phosphite, ClO2- is chlorite
Note, not all “-ate”s have a corresponding “-ite”!
per-______-ate* O2
2-
ClO4-
BrO4-
IO4-
Rule: Ion consisting of one halogen atom, 4 oxygen atoms, and a negative charge.One more oxygen atom than the “-ate”s. Highest possible nonmetal oxidation state in the halogen group.
Example: ClO4- is perchlorate
* Adding one more oxygen to oxide gives O22-. The prefix rule follows with its own suffix for the name peroxide.
Also:MnO4
-
hypo-______-ite
ClO-
BrO-
IO-
Rule: Most common ion consisting of one halogen atom, one oxygen atom, and a negative charge.Example: ClO- is hypochlorite
hydrogen -______-ideOH-
HP2- HS-
Rule: Ion consisting of one H+ added to the “-ide” ion with 2- or greater charge.Example: HS- is hydrogen sulfide, OH- is hydroxide (a contraction of hydrogen oxide). Note that there is typically a space between “hydrogen” and the rest of the name.
hydrogen ______-ateHCO3
-
HSiO3- HPO4
2- HSO4-
HAsO42- HSeO4
-
Rule: Ion consisting of one H+ added to the “-ate” ion with 2- or greater charge.Example: HCO3
- is hydrogen carbonate, HPO42- is hydrogen phosphate
Important Note: H2PO4- is dihydrogen phosphate
Corresponding AcidsHF
HNO2
H3BO3 H2CO3 HNO3
H2S HCl
HClO
H3PO3 H2SO3 HClO2
H2SiO3 H3PO4 H2SO4 HClO3
HClO4
H2Se HBr
HBrO
H3AsO3 H2SeO3 HBrO2
H3AsO4 H2SeO4 HBrO3
HBrO4
Other visual redox stuff:
22Zn(s) 2H (aq) Zn (aq) H (g)
ZnZn
ZnZn+
Zn
Zn
Zn
ZnZn
Zn
ZnZn Zn
Zn
Zn
Zn
Zn
Zn
Zn
Zn+e-
e-
ZnZn
Zn+
Zn
Zn
Zn
ZnZn
ZnZn Zn
Zn
Zn
Zn
Zn
Zn
Zn
Zn+
HH
Oxidation-Reduction Reactions
22Zn(s) 2 H (aq) Zn (aq) H (g)
0 +1 +2 0
Use Oxidation # to ID Oxidized and Reduced Species
Zn is Oxidized (Reducing Agent) to Zn2+
H+ is Reduced (Oxidizing Agent) to H2
VO3- VO2
+
CrO2- CrO4
2-
SO3 SO42-
NO3 NO2-
Identify Half-Rxn (Ox or Red)
Neither: Lewis Acid/Base
Neither: Lewis Acid/Base
Oxidation
Reduction
+5 +5
+3 +6
+6 +6
+6 +3
2008 Local 38. For a stoichiometric mixture of reactants, which statement best describes the changes that occur when this reaction goes to completion?
0 +5 +2 +5 +4
Zn + 4 HNO3 → Zn(NO3)2 + 2 NO2 + 2 H2O
(A) All of the zinc is oxidized and some of the nitrogen is reduced.
(B) All of the zinc is oxidized and all of the nitrogen is reduced.
(C)Some of the zinc is oxidized and all of the nitrogen is reduced.
(D)Some of the zinc is oxidized and some of the nitrogen is reduced.
Cue-Cu+
OH
H
OHH
O H
H
O
H
H
O
H
H
Cl-H+
e-
e-
