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Electrochemistry
Remember from CHM151…
A redox reaction in one in which electrons are transferred…
For example: Reduction Oxidation
0 0 +1 -2 2H (g) + O (g) → 2H O(g) Lose G ain 2 2 2 ox Electrons Electrons red O xidation R eduction Since hydrogen went from a 0 to +1 oxidation We can determine which element each hydrogen, lost an electron is oxidized or reduced by assigning oxidation numbers. Since oxygen went from a 0 to -2 reduction each oxygen, gained two electrons
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Balancing Redox Reactions – Half-Reaction Method In CHM151, we encountered equations such as:
3+ 2+ Notice that this is the net-ionic equation. That is, I removed Zn(s) + Al (aq) → Zn (aq) + Al(s) - any spectator ions (such as NO3 ). Without adding the spectator ion, the chemical equation might look like it is 3+ total charge on this 2+ total charge on this balanced. But is it? side… side… No…because the charges on each side are not balanced. Remember that the whole point of a redox reaction is the transfer of electrons. We can therefore break into two separate “half-reactions”…
Reduction Half-Reaction Al3+(aq) → Al(s) gain electrons Al3+(aq) + 3é → Al(s)
Oxidation Half-Reaction Zn(s) → Zn2+(aq) lose electrons Zn(s) → Zn2+(aq) + 2é
Total electrons transferred must be the same.
3+ 2Al (aq) + 2é → Al(s) x3 2Al3+(aq) + 6é → 2Al(s) 3Zn(s) + 2Al3+(aq) → 3Zn2+(aq) + 2Al(s)
Zn(s) → Zn2+(aq) + 3é x2 3Zn(s) → 3Zn2+(aq) + 6é Charges on both sides are now balanced
Balancing Redox Reactions – Half-Reaction Method
Let’s do a little more complicated one…between Ag and F2 gas
+ - + - Ag + F2 → Ag + F Ag + F2 → Ag + 2F Well, it looks balanced…what’s wrong?
0 -2
Charge imbalance
- - Reduction Half-Reaction F2 → 2F gain electrons F2+ 2é → 2F
Oxidation Half-Reaction Ag → Ag+ lose electrons Ag → Ag+ + é
Total electrons transferred must be the same.
F + 2é → 2F- - 2 F2+ 2é → 2F 2Ag + F → 2Ag+ + 2F- + + 2 Ag → Ag + é x2 2Ag → 2Ag + 2é
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Balancing Redox Reactions – Half-Reaction Method
Step 1: Balance all elements that are not H or O by using coefficients
Step 2: Balance O by using H2O
Step 3: Balance H by using H+
Step 4: If there is a charge imbalance, use electrons to fix the imbalance
Step 5: Make sure both half-reactions have the same number of electrons
Step 6: Recombine both half-reactions and mathematically cancel and species present on both sides + Step 7: If the reactions occurs in basic media, change all H to H2O by adding OH-. You must add the same amount of OH- to both sides of the equation
Balancing Redox Reactions – Half-Reaction Method
2- 3+ - Cr2O7 (aq) + HNO2(aq) → Cr (aq) + NO3 (aq) (acidic)
2- 3+ Reduction Half-Reaction Cr2O7 (aq) → Cr (aq)
- Oxidation Half-Reaction HNO2(aq) → NO3 (aq)
2- 3+ Cr2O7 (aq) → 2Cr (aq)
2- 3+ - Cr2O7 (aq) → 2Cr (aq) + 7H2O(l) HNO2(aq) + H2O(l) → NO3 (aq)
2- + 3+ - + Cr2O7 (aq) + 14H (aq) → 2Cr (aq) + 7H2O(l) HNO2(aq) + H2O(l) → NO3 (aq) + 3H (aq)
2- + 3+ - + Cr2O7 (aq) + 14H (aq) + 6é → 2Cr (aq) + 7H2O(l) HNO2(aq) + H2O(l) → NO3 (aq) + 3H (aq) + 2é
- + 3(HNO2(aq) + H2O(l) → NO3 (aq) + 3H (aq) + 2é) 2- + 3+ Cr2O7 (aq) + 514H (aq) + 6é → 2Cr (aq) + 47H2O(l) - + - + 3HNO2(aq) + 3H2O(l) → 3NO3 (aq) + 9H (aq) + 6é 3HNO2(aq) + 3H2O(l) → 3NO3 (aq) + 9H (aq) + 6é
2- + 3+ - Cr2O7 (aq) + 5H (aq) 3HNO2(aq) → 2Cr (aq) + 4H2O(l) + 3NO3 (aq)
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Balancing Redox Reactions – Half-Reaction Method
- 2- - Cr(OH)3(s) + ClO3 ( aq) → CrO4 (aq) + Cl (aq) (basic)
- - Reduction Half-Reaction ClO3 ( aq) → Cl (aq)
Oxidation Half-Reaction 2- Cr(OH)3(s) → CrO4 (aq)
- - 2- ClO3 ( aq) → Cl (aq) Cr(OH)3(s) → CrO4 (aq)
- - 2- ClO3 ( aq) → Cl (aq) + 3H2O Cr(OH)3(s) +H2O → CrO4 (aq)
- + - 2- + ClO3 ( aq) + 6H → Cl (aq) + 3H2O Cr(OH)3(s) +H2O → CrO4 (aq) + 5H
- + - 2- + ClO3 ( aq) +6H + 6é → Cl (aq) + 3H2O Cr(OH)3(s) +H2O → CrO4 (aq) + 5H + 3é
2- + - + - 2(Cr(OH)3(s) +H2O → CrO4 (aq) + 5H + 3é) ClO3 ( aq) +6H + 6é → Cl (aq) + 3H2O 2- + 2Cr(OH)3(s) + 2H2O → 2CrO4 (aq) + 410H + 6é 2- + 2Cr(OH)3(s) + 2H2O → 2CrO4 (aq) + 10H + 6é
- - - 2- 4OH + ClO3 ( aq) + 2Cr(OH)3(s) → Cl (aq) + 2CrO4 (aq) + 4H2O + H2O - - - 2- 4OH + ClO3 ( aq) + 2Cr(OH)3(s) → Cl (aq) + 2CrO4 (aq) + 5H2O
Balancing Redox Reactions – Half-Reaction Method
Cd(s) + Ni2O3(s) → Cd(OH)2(s)+ Ni(OH)2(s) (basic)
Reduction Half-Reaction Ni2O3(s) → Ni(OH)2(s)
Oxidation Half-Reaction Cd(s) → Cd(OH)2(s)
Ni2O3(s) → 2Ni(OH)2(s)
Ni2O3(s) + H2O → 2Ni(OH)2(s) Cd(s) + 2H2O(l) → Cd(OH)2(s)
+ + Ni2O3(s) + H2O + 2H → 2Ni(OH)2(s) Cd(s) + 2H2O → Cd(OH)2(s) + 2H (aq)
+ + Ni2O3(s) + H2O + 2H + 2é → 2Ni(OH)2(s) Cd(s) + 2H2O → Cd(OH)2(s) + 2H (aq) + 2é
+ Ni2O3(s) + H2O + 2H + 2é → 2Ni(OH)2(s) + Cd(s) + 2H2O → Cd(OH)2(s) + 2H (aq) + 2é
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Redox Reactions
Zn(NO3)2 Cu(NO3)2
Zn + Cu2+ → Cu + Zn2+
Although electrons are transferred, we are unable to use these electrons. Is there a way we can harness these electrons??
Anatomy of a Voltaic/Galvanic Cell
e- e- (-) e- (+) e- Cathode Anode - + NO3 KNO3 K
Zn Salt Bridge Cu
- 2+ Zn2+ NO3 Cu 2+ - Cu NO3 - - 2+ - NO3 NO3 Zn NO3 - - - NO3 NO3 NO3
1M 1M
It is traditional to draw the anode on the left hand side, cathode on the other
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As the Voltaic/Galvanic Cell runs…
e- e- (-) e- (+) e- Cathode Anode - + NO3 KNO3 K Cu Zn Salt Bridge
- NO - Cu2+ Zn2+ NO3 3 2+ Zn2+ - NO - Cu NO3 3 - + - 2+ - NO3K NO3 Zn NO3 - - - K+ NO3 NO3 NO3
>1M?M <1M?M
It is traditional to draw the anode on the left hand side, cathode on the other
Anatomy of a Voltaic/Galvanic Cell
What about a voltaic cell made from this REDOX reaction? + - 2Ag(s) + F2(g) → 2Ag (aq) + 2F (aq)
How do we use a gas?? (-) (+) e- e- Anode Cathode e- e- - + NO3 KNO3 K F2 @ 1ATM
Ag Salt Bridge Pt
Platinum (Pt) and graphite (C) are used Ag+ Na+ - + for gases or NO3 Ag F- - + - solutions. NO3 Na F
1M 1M
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Voltaic/Galvanic Cell Shorthand notation Another way we can specify the species in our voltaic cell if to use “shorthand notation.” Let’s take our Zn/Cu cell, for example. With the species used it would become:
Zn (s)│Zn2+ (1M)││ Cu2+ (1M)│ Cu (s) How do we know which directions the elections will travel in a voltaic cell?
+ - Ag (s)│Ag (1M)││ F (1M)│ F2 (1atm)│Pt(s)
Standard Reduction Potentials
All values are based on whether Positive they will donate or accept electrons values to hydrogen. Because of this, Stronger indicate hydrogen has a reduction potential oxidizing chemical of zero. agents species will accept Notice that all values are for electrons reaction in the reduction direction. from We can also write them in the hydrogen oxidation direction (reverse) if we need to. Hydrogen “zero”
Species lower on Negative the list will donate values electrons to species Stronger indicate reducing chemical higher on the list agents species will donate electrons to hydrogen
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Standard Cell Potential
E°cell = E° reduction +E° oxidation
or E°cell = E°red cathode – E°red anode
Remember that oxidation happens at the anode.
Let’s determine E°cell between Cu and Zn. Which will most 2+ - Cu + 2e → Cu E°red = 0.34 V likely give up an 2+ - Zn + 2e → Zn E°red = -0.76V electron and become oxidized? 2+ - Zn → Zn + 2e E°ox = 0.76V
Since Zn will more easily give up its E°cell = 0.34V+ 0.76V = 1.10V electron, it will become oxidized. The reaction in the table for Zn will Remember that this voltage will actually take place in the opposite only take place under standard direction than written, thus E°ox = - thermodynamic conditions (25.0°C, E°red. 1atm, aqueous solutions 1.0M.
Standard Cell Potential
Determine the E°cell for the following reaction
+ - Ag + F2 → Ag + F
- F2 + 2é → 2F E°red = 2.87 V + Ag + é → Ag E°red = 0.80 V
Al(s) + Mn2+ (aq) → Zn(s) + Al3+(aq)
3+ Al + 3é → Al E°red = -1.66 V 2+ Mn + 2é → Mn E°red = -1.18V
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Cell Potential and Free Energy
ΔG° = -nFE° cell Standard Cell Potential (V)
Faraday’s Constant mols of electrons (96,500 C/mol) transferred C = J/V
Need to balance reactions to determine correct Half -reactions number of electrons 2+ - - Cu + 2e → Cu Cu2+ + Zn→ Cu + Zn2+ ΔG° = -(2mol e )(96500 J/V mol)(1.10 V)= Zn → Zn2+ + 2e- -2.12x105 J or -212 kJ Reaction is already balanced, 2 electrons transferred SPONTANEOUS
3+ 2+ 3+ Al + 3é → Al Al(s) + Zn (aq) → Zn(s) + Al (aq) ΔG° = -(6mol e-)(96500 J/V mol)(2.00 V)=
2+ Cu + 2é → Cu -1.56x106 J or -1560 kJ
SPONTANEOUS
Cell Potential under real conditions – The Nernst Equation
Since cells are not always under standard conditions, we need an equation that we can calculate cell potential under different conditions (higher/lower molarity, higher/lower temperature) RT Ecell E cell lnQ Assume we have the Zn/Cu cell with the anode at a nF concentration of 1.50M and the cathode at 0.50M.
(8.314 J/mol K)(298.15K) (1.50M) Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) E 1.10V ln (2 mol e)(96,500 J/V mol) (0.50M) [Zn2 ] Q ? Q E 1.09V [Cu 2 ]
Assume we have the Al/Zn cell with the anode at a Al(s) + Zn2+ (aq) → Zn(s) + Al3+(aq) concentration of 0.50M and the cathode at 3.0M at 70.0°C.
[Al3 ]2 (8.314 J/mol K)(343.15K) (0.25M)2 Q ? Q E 2.00V ln [Zn2 ]3 (6 mol e)(96,500 J/V mol) (3.0M)3 E 2.03V
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Concentration Cells Will this cell work? Let’s say you want to make a cell, and you only have one metal (Cu) and only one ionic
solution (Cu(NO)3), cold you make a cell that would generate any voltage?
Well, not in it’s current condition of 1.00M and 1.00M Let’s change the concentrations to 0.10M and 1.00M
(8.314 J/molK)(298.15K) 0.10M RT Ecell 0.00V ln Ecell E cell lnQ (2)(96500 J/V mol) 1.00M nF
Ecell 0.030V
G (2mol e)(96,500 J/Vmol)(0.030V)
G 5.8x103 J or - 5.8kJ
What would the value of Q need to be if you wanted 1.00V?
Cell Potential and K
Now that we can calculate ΔG for a cell, we should also be able to calculate the equilibrium constant like we did in the last chapter using:
ΔG° = -RTlnK or…also knowing… ΔG° = -nFE°cell -nFE°= -RTlnK
RT Standard Cell Potential (V) E cell lnK nF
Faraday’s Constant mols of electrons (96,500 C/mol) transferred C = J/V
(8.314 J/molK)(298.15K) 1.10V lnK 84.6 lnK 36 (2 mol e)(96500 J/V mol) e e K 5.67x10
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ΔG°
ΔG° = -nFE°cell ΔG° = -RTlnK
RT E°cell E cell lnK K nF
Some Shortcuts From the Book
(8.314J/V mol)(298.15K) E cell lnK n (96,500 J/V mol)
0.0257V E cell lnK n
0.0592V E cell logK n
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Electrolytic Cell
A voltaic/galvanic cell with eventually run down to 0V. This is because eventually, one of the reactants will run out.
We could “recharge” our cell by running the REDOX reactions in reverse Dead Cell
An electrolytic cell is one where electrons flow in the non- spontaneous direction.
Electrolytic Cell
Direct comparison – Electrolytic vs. Voltaic/Galvanic
Voltaic/Galvanic Cell Electrolytic Cell
Reaction at Anode Reaction at Cathode Reaction at Cathode Reaction at Anode Zn → Zn2+ + 2e- Cu2+ + 2e-→ Cu Zn2+ + 2e- → Zn Cu → Cu2+ + 2e- Oxidation Reduction Reduction Oxidation
2+ 2+ 2+ 2+ Zn + Cu → Zn + Cu E°cell = 1.10V Zn + Cu → Zn + Cu E°cell = -1.10V
Notice: oxidation always happens at the anode!
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Other Applications of Electrolytic Cells
Faraday’s Law (Electrolysis)
The amount of a substance produced at an electrode is proportional to the quantity of electricity (in Coulombs) transferred to that electrode. 1 ampere = 1 C/s
1 ampere · s What mass of Zn will be produced if a Zn/Cu = electrolytic cell is run at 3.0amps for 5.0hrs? 1 Coulomb (C)
60 min 60 sec 1 C 96,500 C 3.00 amp 5.00 hr 54,000 C 1 hr 1 min 1 amps = - 1mol e 1 mol e 54,000 C 5.60x101 mol e 96,500 C
1 mol 1 mol Cu 65.39 g Zn 5.60x10-1 mol e 18.3g Zn = 2 mol e 1 mol Zn molar mass
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