Electrochemistry

Electrochemistry Remember from CHM151… A redox reaction in one in which electrons are transferred… Reduction For example: Oxidation 0 Lose Electrons O xidation G ain Electrons R eduction We can determine which element is oxidized or reduced by assigning oxidation numbers. 0 +1 -2 2H2(g) + O2(g) → 2H2O(g) ox red Since hydrogen went from a 0 to +1 oxidation each hydrogen, lost an electron Since oxygen went from a 0 to -2 reduction each oxygen, gained two electrons 1 Balancing Redox Reactions – Half-Reaction Method In CHM151, we encountered equations such as: Zn(s) + Al3+(aq) → Zn2+(aq) + Al(s) 3+ total charge on this side… 2+ total charge on this side… Notice that this is the net-ionic equation. That is, I removed any spectator ions (such as NO3- ). Without adding the spectator ion, the chemical equation might look like it is balanced. But is it? 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. 2Al3+(aq) + 2é → Al(s) x3 2Al3+(aq) + 6é → 2Al(s) Zn(s) → Zn2+(aq) + 3é x2 3Zn(s) → 3Zn2+(aq) + 6é 3Zn(s) + 2Al3+(aq) → 3Zn2+(aq) + 2Al(s) 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+ + 2F0 Well, it looks balanced…what’s wrong? -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. F2+ 2é → 2FAg → Ag+ + é F2+ 2é → 2Fx2 2Ag → 2Ag+ + 2é 2Ag + F2 → 2Ag+ + 2F- 2 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 Cr2O72-(aq) + HNO2(aq) → Cr3+(aq) + NO3-(aq) (acidic) Reduction Half-Reaction Cr2O72-(aq) → Cr3+(aq) Oxidation Half-Reaction HNO2(aq) → NO3-(aq) Cr2O72-(aq) → 2Cr3+(aq) Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l) HNO2(aq) + H2O(l) → NO3-(aq) Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l) HNO2(aq) + H2O(l) → NO3-(aq) + 3H+(aq) Cr2O72-(aq) + 14H+(aq) + 6é → 2Cr3+(aq) + 7H2O(l) HNO2(aq) + H2O(l) → NO3-(aq) + 3H+(aq) + 2é Cr2O72-(aq) + 514H+(aq) + 6é → 2Cr3+(aq) + 47H2O(l) 3HNO2(aq) + 3H2O(l) → 3NO3-(aq) + 9H+(aq) + 6é 3(HNO2(aq) + H2O(l) → NO3-(aq) + 3H+(aq) + 2é) 3HNO2(aq) + 3H2O(l) → 3NO3-(aq) + 9H+(aq) + 6é Cr2O72-(aq) + 5H+(aq) 3HNO2(aq) → 2Cr3+(aq) + 4H2O(l) + 3NO3-(aq) 3 Balancing Redox Reactions – Half-Reaction Method Cr(OH)3(s) + ClO3-( aq) → CrO42-(aq) + Cl-(aq) (basic) Reduction Half-Reaction ClO3-( aq) → Cl-(aq) Oxidation Half-Reaction Cr(OH)3(s) → CrO42-(aq) ClO3-( aq) → Cl-(aq) Cr(OH)3(s) → CrO42-(aq) ClO3-( aq) → Cl-(aq) + 3H2O Cr(OH)3(s) +H2O → CrO42-(aq) ClO3-( aq) + 6H+ → Cl-(aq) + 3H2O Cr(OH)3(s) +H2O → CrO42-(aq) + 5H+ ClO3-( aq) +6H+ + 6é → Cl-(aq) + 3H2O Cr(OH)3(s) +H2O → CrO42-(aq) + 5H+ + 3é ClO3-( aq) +6H+ + 6é → Cl-(aq) + 3H2O 2Cr(OH)3(s) + 2H2O → 2CrO42-(aq) + 410H+ + 6é 2(Cr(OH)3(s) +H2O → CrO42-(aq) + 5H+ + 3é) 2Cr(OH)3(s) + 2H2O → 2CrO42-(aq) + 10H+ + 6é 4OH- + ClO3-( aq) + 2Cr(OH)3(s) → Cl-(aq) + 2CrO42-(aq) + 4H2O + H2O 4OH- + ClO3-( aq) + 2Cr(OH)3(s) → Cl-(aq) + 2CrO42-(aq) + 5H2O Balancing Redox Reactions – Half-Reaction Method Cd(s) + Ni2O3(s) → Cd(OH)2(s)+ Ni(OH)2(s) Reduction Half-Reaction Ni2O3(s) → Ni(OH)2(s) Oxidation Half-Reaction Cd(s) → Cd(OH)2(s) (basic) 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 4 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- ee- (-) Anode e- Zn NO3- KNO3 K+ Cu Salt Bridge NO3- Zn2+ NO3- 2+ NO3Zn NO3- NO3- 1M (+) Cathode Cu2+ NO3- Cu2+ NO3- NO3- 1M It is traditional to draw the anode on the left hand side, cathode on the other 5 As the Voltaic/Galvanic Cell runs… e- ee- (-) e- Anode NO3- Zn (+) Cathode K+ KNO3 Cu Salt Bridge NO3- - NO3 Zn2+ Zn2+ NO NO33 NO3Zn2+ NO3- NO3- Cu2+ Cu2+ NO3K- + NO3- K+ NO3- ?M >1M ?M <1M 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- Anode e- Cathode e- e- Ag Ag+ NO3- Ag+ NO3- 1M NO3- KNO3 K+ F2 @ 1ATM Pt Salt Bridge Na+ FNa+ F- Platinum (Pt) and graphite (C) are used for gases or solutions. 1M 6 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: How do we know which directions the elections will travel in a voltaic cell? Zn (s)│Zn2+ (1M)││ Cu2+ (1M)│ Cu (s) Ag (s)│Ag+ (1M)││ F- (1M)│ F2 (1atm)│Pt(s) Standard Reduction Potentials All values are based on whether they will donate or accept electrons to hydrogen. Because of this, hydrogen has a reduction potential of zero. Stronger oxidizing agents Notice that all values are for reaction in the reduction direction. We can also write them in the oxidation direction (reverse) if we need to. Positive values indicate chemical species will accept electrons from hydrogen Hydrogen “zero” Species lower on the list will donate electrons to species higher on the list Stronger reducing agents Negative values indicate chemical species will donate electrons to hydrogen 7 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. Cu2+ + 2e-→ Cu Zn2+ + 2e-→ Zn E°red = 0.34 V E°red = -0.76V Zn → Zn2+ + 2e- E°ox = 0.76V E°cell = 0.34V+ 0.76V = 1.10V Remember that this voltage will only take place under standard thermodynamic conditions (25.0°C, 1atm, aqueous solutions 1.0M. Which will most likely give up an electron and become oxidized? Since Zn will more easily give up its electron, it will become oxidized. The reaction in the table for Zn will actually take place in the opposite direction than written, thus E°ox = E°red. Standard Cell Potential Determine the E°cell for the following reaction Ag + F2 → Ag+ + FF2 + 2é → 2FAg+ + é → Ag E°red = 2.87 V E°red = 0.80 V Al(s) + Mn2+ (aq) → Zn(s) + Al3+(aq) Al3+ + 3é → Al E°red = -1.66 V Mn2+ + 2é → Mn E°red = -1.18V 8 Cell Potential and Free Energy ΔG° = -nFE°cell mols of electrons transferred Standard Cell Potential (V) Faraday’s Constant (96,500 C/mol) C = J/V Need to balance reactions to determine correct number of electrons Half -reactions Cu2+ + 2e-→ Cu Zn → Zn2+ + 2e- Cu2+ + Zn→ Cu + Zn2+ ΔG° = -(2mol e-)(96500 J/V mol)(1.10 V)= -2.12x105 J or -212 kJ Reaction is already balanced, 2 electrons transferred Al3+ + 3é → Al Cu2+ Al(s) + Zn2+ (aq) → Zn(s) + Al3+(aq) SPONTANEOUS ΔG° = -(6mol e-)(96500 J/V mol)(2.00 V)= + 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) E cell  E  cell  Assume we have the Zn/Cu cell with the anode at a concentration of 1.50M and the cathode at 0.50M. E  1.10V  (8.314 J/mol K)(298.15K) (1.50M) ln (2 mol e)(96,500 J/V mol) (0.50M) E  1.09V Assume we have the Al/Zn cell with the anode at a concentration of 0.50M and the cathode at 3.0M at 70.0°C. E  2.00V  (8.314 J/mol K)(343.15K) (0.25M) 2 ln (6 mol e)(96,500 J/V mol) (3.0M) 3 RT lnQ nF Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) Q Q ? [Zn2 ] [Cu 2 ] Al(s) + Zn2+ (aq) → Zn(s) + Al3+(aq) Q ? Q [Al3 ]2 [ Zn 2 ]3 E  2.03V 9 6/24/2014 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 E cell  0.00V  (8.314 J/mol  K)(298.15K)  0.10M  ln  (2)(96500 J/V  mol)  1.00M  E cell  E  cell  RT lnQ nF Ecell  0.030V G  ( 2mol e)(96,500 J/V  mol)(0.030V) G  5.8 x103 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 Standard Cell Potential (V) E  cell  RT lnK nF mols of electrons transferred 1.10V  (8.314 J/mol  K)(298.15K) lnK (2 mol e)(96500 J/V  mol) e 84.6   e lnK  Faraday’s Constant (96,500 C/mol) C = J/V K  5.67 x1036 10 ΔG° ΔG° = -nFE°cell E°cell ΔG° = -RTlnK E  cell  RT lnK nF K Some Shortcuts From the Book E  cell  (8.314J/V  mol)(298.15K) lnK n (96,500 J/V  mol) E  cell  0.0257 V lnK n E  cell  0.0592V logK n 11 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 nonspontaneous direction. Electrolytic Cell Direct comparison – Electrolytic vs. Voltaic/Galvanic Voltaic/Galvanic Cell Reaction at Anode Zn → Zn2+ + 2e- Oxidation Electrolytic Cell Reaction at Cathode Reaction at Cathode Cu2+ Zn2+ + + 2e-→ Cu Reduction Zn + Cu2+ → Zn2+ + Cu E°cell = 1.10V 2e- → Zn Reduction Zn2+ + Cu → Zn + Cu2+ Reaction at Anode Cu → Cu2+ + 2eOxidation E°cell = -1.10V Notice: oxidation always happens at the anode! 12 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 = 1 Coulomb (C) 96,500 C = 1mol e1 mol = molar mass What mass of Zn will be produced if a Zn/Cu electrolytic cell is run at 3.0amps for 5.0hrs? 3.00 amp 5.00 hr  60 min 60 sec 1 hr 54,000 C 1C  54,000 C 1 min 1 amp  s 1 mol e   5.60 x10 1 mol e  96,500 C 5.60x10-1 mol e  1mol Cu 65.39 g Zn  18.3g Zn 2 mol e  1 mol Zn 13