Chm152 Electrochemistry

Remember from CHM151… A redox reaction in one in which electrons are transferred… For example: Electrochemistry Reduction Oxidation 0 Lose Electrons O xidation 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 NO 3- ). Without adding the spectator ion, the chemical equation might look like it is balanced. But is it? red Since hydrogen went from a 0 to +1 oxidation each hydrogen, lost an electron 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) Since oxygen went from a 0 to -2 reduction each oxygen, gained two electrons 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 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”… -2 Balancing Redox Reactions – Half-Reaction Method In CHM151, we encountered equations such as: Zn(s) + Al3+(aq) → Zn2+(aq) + Al(s) +1 ox We can determine which element is oxidized or reduced by assigning oxidation numbers. Balancing Redox Reactions – Half-Reaction Method 0 2H2(g) + O2(g) → 2H2O(g) G ain Electrons R eduction 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 → 2F- F2+ 2é → 2F- Ag → Ag+ + é 2Ag → 2Ag+ + 2é x2 2Ag + F2 → 2Ag+ + 2F- Charges on both sides are now balanced Balancing Redox Reactions – Half-Reaction Method Balancing Redox Reactions – Half-Reaction Method Cr2O72-(aq) + HNO2(aq) → Cr3+(aq) + NO3-(aq) (acidic) 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 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) 1 Balancing Redox Reactions – Half-Reaction Method Cr(OH)3(s) + ClO3-( aq) → CrO42-(aq) + Cl-(aq) Balancing Redox Reactions – Half-Reaction Method (basic) Cd(s) + Ni2O3(s) → Cd(OH)2(s)+ Ni(OH)2(s) Reduction Half-Reaction ClO3-( aq) → Cl-(aq) Reduction Half-Reaction Ni2O3(s) → Ni(OH)2(s) Oxidation Half-Reaction Cr(OH)3(s) → CrO42-(aq) Oxidation Half-Reaction Cd(s) → Cd(OH)2(s) (basic) ClO3-( aq) → Cl-(aq) Cr(OH)3(s) → CrO42-(aq) Ni2O3(s) → 2Ni(OH)2(s) ClO3-( aq) → Cl-(aq) + 3H2O Cr(OH)3(s) +H2O → CrO42-(aq) Ni2O3(s) + H2O → 2Ni(OH)2(s) Cd(s) + 2H2O(l) → Cd(OH)2(s) ClO3-( aq) + 6H+ → Cl-(aq) + 3H2O Cr(OH)3(s) +H2O → CrO42-(aq) + 5H+ Ni2O3(s) + H2O + 2H+ → 2Ni(OH)2(s) Cd(s) + 2H2O → Cd(OH)2(s) + 2H+(aq) ClO3-( aq) +6H+ + 6é → Cl-(aq) + 3H2O Cr(OH)3(s) +H2O → CrO42-(aq) + 5H+ + 3é Ni2O3(s) + H2O + 2H+ + 2é → 2Ni(OH)2(s) Cd(s) + 2H2O → Cd(OH)2(s) + 2H+(aq) + 2é 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 Ni2O3(s) + H2O + 2H+ + 2é → 2Ni(OH)2(s) Cd(s) + 2H2O → Cd(OH)2(s) + 2H+(aq) + 2é 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 Anatomy of a Voltaic/Galvanic Cell Redox Reactions eZn(NO3)2 Cu(NO3)2 ee- (-) e- Anode NO3- Zn Cu Salt Bridge NO3- Zn2+ NO3- NO3- NO3- Zn + Cu2+ → Cu + Zn2+ Cu2+ Cu2+ Zn2+ NO3- NO3- NO3- NO3- 1M Although electrons are transferred, we are unable to use these electrons. Is there a way we can harness these electrons?? (+) Cathode K+ KNO3 1M It is traditional to draw the anode on the left hand side, cathode on the other As the Voltaic/Galvanic Cell runs… Anatomy of a Voltaic/Galvanic Cell What about a voltaic cell made from this REDOX reaction? e- e- 2Ag(s) + F2(g) → 2Ag+(aq) + 2F-(aq) e- (-) e- Anode NO3- Zn KNO3 (+) Cathode K+ Cu Salt Bridge How do we use a gas?? (-) (+) e- Anode Zn2+ NO 3 NO3NO3- NO3NO3Zn2+ NO3- ?M >1M NO3Cu2+ NO3NO3- Cathode e- eZn2+ e- Cu2+ Ag NO3K- + NO3- KNO3 K+ F2 @ 1ATM Pt Salt Bridge K+ ?M <1M Ag+ NO3- Ag+ NO3- It is traditional to draw the anode on the left hand side, cathode on the other 1M Na+ FNa+ F- Platinum (Pt) and graphite (C) are used for gases or solutions. 1M 2 Standard Reduction Potentials 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) All values are based on whether they will donate or accept electrons to hydrogen. Because of this, hydrogen has a reduction potential of zero. Positive values indicate chemical species will accept electrons from hydrogen 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. Hydrogen “zero” Species lower on the list will donate electrons to species higher on the list Negative values indicate chemical species will donate electrons to hydrogen Stronger reducing agents Ag (s)│Ag+ (1M)││ F- (1M)│ F2 (1atm)│Pt(s) Standard Cell Potential Standard Cell Potential Determine the E°cell for the following reaction E°cell = E° reduction +E° oxidation or Ag + F2 → Ag+ + F- E°cell = E°red cathode – E°red anode F2 + 2é → 2F- E°red = 2.87 V E°red = 0.80 V Ag+ + é → Ag 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. Al(s) + Mn2+ (aq) → Zn(s) + Al3+(aq) Which will most likely give up an electron and become oxidized? Al3+ + ΔG° = -nFE°cell mols of electrons transferred Half -reactions Al3+ + 3é → Al E°red = -1.18V Cell Potential under real conditions – The Nernst Equation Standard Cell Potential (V) Faraday’s Constant (96,500 C/mol) C = J/V Cu2+ + Zn→ Cu + Zn2+ Al(s) + Zn2+ (aq) → Zn(s) + Al3+(aq) 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. Need to balance reactions to determine correct number of electrons Reaction is already balanced, 2 electrons transferred Cu2+ + 2é → Cu E°red = -1.66 V 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. Cell Potential and Free Energy Cu2+ + 2e-→ Cu Zn → Zn2+ + 2e- 3é → Al Mn2+ + 2é → Mn E  1.10V  ΔG° = -(2mol e-)(96500 J/V mol)(1.10 V)= (8.314 J/mol K)(298.15K) (1.50M) ln (2 mol e)(96,500 J/V mol) (0.50M) -2.12x105 J or -212 kJ E  1.09V Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) Q ? Q SPONTANEOUS ΔG° = -(6mol e-)(96500 J/V mol)(2.00 V)= 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. -1.56x106 J or -1560 kJ E  2.00V  SPONTANEOUS (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 [Zn2 ] [Cu 2 ] Al(s) + Zn2+ (aq) → Zn(s) + Al3+(aq) Q ? Q [Al3 ]2 [ Zn 2 ]3 E  2.03V 3 Concentration Cells Cell Potential and K 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? 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 ΔG° = -nFE°cell or…also knowing… -nFE°= -RTlnK 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  Standard Cell Potential (V) 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 RT lnK nF Faraday’s Constant (96,500 C/mol) C = J/V mols of electrons transferred G  ( 2mol e)(96,500 J/V  mol)(0.030V) G  5.8 x103 J or - 5.8kJ 1.10V  (8.314 J/mol  K)(298.15K) lnK (2 mol e)(96500 J/V  mol) e 84.6   e lnK  K  5.67 x1036 What would the value of Q need to be if you wanted 1.00V? Some Shortcuts From the Book ΔG° ΔG° = -nFE°cell E°cell E  cell  E  cell  (8.314J/V  mol)(298.15K) lnK n (96,500 J/V  mol) ΔG° = -RTlnK RT lnK nF E  cell  0.0257V lnK n E  cell  0.0592V logK n K Direct comparison – Electrolytic vs. Voltaic/Galvanic 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 Voltaic/Galvanic Cell An electrolytic cell is one where electrons flow in the nonspontaneous direction. Reaction at Cathode Reaction at Cathode Reaction at Anode Zn → Zn2+ + 2e- Cu2+ + 2e-→ Cu Zn2+ + 2e- → Zn Cu → Cu2+ + 2e- Oxidation Reduction Zn + Cu2+ → Zn2+ + Cu Electrolytic Cell Electrolytic Cell Reaction at Anode E°cell = 1.10V Reduction Zn2+ + Cu → Zn + Cu2+ Oxidation E°cell = -1.10V Notice: oxidation always happens at the anode! 4 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 1C  54,000 C 1 min 1 amp  s 1 hr  54,000 C 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 5