Electrochemistry Remember from CHM151 A redox reaction in one in which electrons are transferred Reduction Oxidation For example: L E O ose lectrons xidation G E R ain lectrons eduction We can determine which element is oxidized or reduced by assigning oxidation numbers. 0 0 +1 2 2H 2 (g) + O 2 (g) 2H 2 O(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 HalfReaction Method In CHM151, we encountered equations such as: Zn(s) + Al + (aq) Zn 2+ (aq) + Al(s) + total charge on this side 2 + total charge on this side Notice that this is the netionic equation. That is, I removed any spectator ions (such as ). 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 halfreactions Reduction HalfReaction Al + (aq) Al(s) gain electrons Al + (aq) + é Al(s) Oxidation HalfReaction Zn(s) Zn 2+ (aq) lose electrons Zn(s) Zn 2+ (aq) + 2é Total electrons transferred must be the same. 2Al + (aq) + 2é Al(s) x 2Al + (aq) + 6é 2Al(s) Zn(s) + 2Al + (aq) Zn 2+ (aq) + 2Al(s) Zn(s) Zn 2+ (aq) + é x2 Zn(s) Zn 2+ (aq) + 6é Charges on both sides are now balanced Balancing Redox Reactions HalfReaction Method Let s do a little more complicated one between Ag and F 2 gas Ag + F 2 Ag + + F Ag + F 2 Ag + + 2F Well, it looks balanced what s wrong? 0 2 Charge imbalance Reduction HalfReaction F 2 2F gain electrons F 2 + 2é 2F Oxidation HalfReaction Ag Ag + lose electrons Ag Ag + + é Total electrons transferred must be the same. F 2 + 2é 2F Ag Ag + + é x2 F 2 + 2é 2F 2Ag 2Ag + + 2é 2Ag + F 2 2Ag + + 2F 2
Balancing Redox Reactions HalfReaction Method Step 1: Balance all elements that are not H or O by using coefficients Step 2: Balance O by using H 2 O Step : Balance H by using H + Step 4: If there is a charge imbalance, use electrons to fix the imbalance Step 5: Step 6: Step 7: Make sure both halfreactions have the same number of electrons Recombine both halfreactions and mathematically cancel and species present on both sides If the reactions occurs in basic media, change all H + to H 2 O by adding OH. You must add the same amount of OH to both sides of the equation Balancing Redox Reactions HalfReaction Method 2 (aq) + H 2 (aq) Cr + (aq) + (aq) (acidic) Reduction HalfReaction Oxidation HalfReaction 2 (aq) Cr + (aq) H 2 (aq) (aq) 2 (aq) 2Cr + (aq) 2 (aq) 2Cr + (aq) + 7H 2 O(l) 2 (aq) + 14H + (aq) 2Cr + (aq) + 7H 2 O(l) H 2 (aq) + H 2 O(l) (aq) H 2 (aq) + H 2 O(l) (aq) + H + (aq) 2 (aq) + 14H + (aq) + 6é 2Cr + (aq) + 7H 2 O(l) H 2 (aq) + H 2 O(l) (aq) + H + (aq) + 2é 2 (aq) + 514H + (aq) + 6é 2Cr + (aq) + 47H 2 O(l) H 2 (aq) + H 2 O(l) (aq) + 9H + (aq) + 6é (H 2 (aq) + H 2 O(l) (aq) + H + (aq) + 2é) H 2 (aq) + H 2 O(l) (aq) + 9H + (aq) + 6é 2 (aq) + 5H + (aq) H 2 (aq) 2Cr + (aq) + 4H 2 O(l) + (aq)
Balancing Redox Reactions HalfReaction Method Cr(OH) (s) + ClO ( aq) CrO 4 2 (aq) + Cl (aq) (basic) Reduction HalfReaction Oxidation HalfReaction ClO ( aq) Cl (aq) Cr(OH) (s) CrO 4 2 (aq) ClO ( aq) Cl (aq) ClO ( aq) Cl (aq) + H 2 O Cr(OH) (s) CrO 4 2 (aq) Cr(OH) (s) +H 2 O CrO 4 2 (aq) ClO ( aq) + 6H + Cl (aq) + H 2 O Cr(OH) (s) +H 2 O CrO 4 2 (aq) + 5H + ClO ( aq) +6H + + 6é Cl (aq) + H 2 O ClO ( aq) +6H + + 6é Cl (aq) + H 2 O 2Cr(OH) (s) + 2H 2 O 2CrO 4 2 (aq) + 410H + + 6é Cr(OH) (s) +H 2 O CrO 4 2 (aq) + 5H + + é 2(Cr(OH) (s) +H 2 O CrO 4 2 (aq) + 5H + + é) 2Cr(OH) (s) + 2H 2 O 2CrO 4 2 (aq) + 10H + + 6é 4OH + ClO ( aq) + 2Cr(OH) (s) Cl (aq) + 2CrO 4 2 (aq) + 4H 2 O + H 2 O 4OH + ClO ( aq) + 2Cr(OH) (s) Cl (aq) + 2CrO 4 2 (aq) + 5H 2 O Balancing Redox Reactions HalfReaction Method Cd(s) + Ni 2 O (s) Cd(OH) 2 (s)+ Ni(OH) 2 (s) (basic) Reduction HalfReaction Oxidation HalfReaction Ni 2 O (s) Ni(OH) 2 (s) Cd(s) Cd(OH) 2 (s) Ni 2 O (s) 2Ni(OH) 2 (s) Ni 2 O (s) + H 2 O 2Ni(OH) 2 (s) Ni 2 O (s) + H 2 O + 2H + 2Ni(OH) 2 (s) Ni 2 O (s) + H 2 O + 2H + + 2é 2Ni(OH) 2 (s) Cd(s) + 2H 2 O(l) Cd(OH) 2 (s) Cd(s) + 2H 2 O Cd(OH) 2 (s) + 2H + (aq) Cd(s) + 2H 2 O Cd(OH) 2 (s) + 2H + (aq) + 2é Ni 2 O (s) + H 2 O + 2H + + 2é 2Ni(OH) 2 (s) Cd(s) + 2H 2 O Cd(OH) 2 (s) + 2H + (aq) + 2é 4
Redox Reactions Cu() 2 Zn() 2 Zn + Cu 2+ Cu + Zn 2+ 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 (+) Anode e Zn K K + Salt Bridge Cu Cathode Zn 2+ Zn 2+ Cu 2+ Cu 2+ 1M 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 e e () Anode e Zn K K + Salt Bridge Cu (+) Cathode Cu 2+ Zn 2+ Zn 2+ Cu 2+ Zn 2+ K + >1M?M <1M?M K + 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) + F 2 (g) 2Ag + (aq) + 2F (aq) How do we use a gas?? e e () (+) Anode Cathode e e Ag K K + Salt Bridge Pt F 2 @ 1ATM Ag + Ag + Na + F Na + F Platinum (Pt) and graphite (C) are used for gases or solutions. 1M 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: Zn (s) Zn 2+ (1M) Cu 2+ (1M) Cu (s) How do we know which directions the elections will travel in a voltaic cell? Ag (s) Ag + (1M) F (1M) F 2 (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. 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. Species lower on the list will donate electrons to species higher on the list Stronger oxidizing agents Hydrogen zero Stronger reducing agents Positive values indicate chemical species will accept electrons from hydrogen 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. Cu 2+ + 2e Cu E red = 0.4 V Zn 2+ + 2e Zn E red = 0.76V Zn Zn 2+ + 2e E ox = 0.76V Which will most likely give up an electron and become oxidized? E cell = 0.4V+ 0.76V = 1.10V Remember that this voltage will only take place under standard thermodynamic conditions (25.0 C, 1atm, aqueous solutions 1.0M. 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 + F 2 Ag + + F F 2 + 2é 2F Ag + + é Ag E red = 2.87 V E red = 0.80 V Al(s) + Mn 2+ (aq) Zn(s) + Al + (aq) Al + + é Al Mn 2+ + 2é Mn E red = 1.66 V 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 Half reactions Cu 2+ + 2e Cu Zn Zn 2+ + 2e Need to balance reactions to determine correct number of electrons Cu 2+ + Zn Cu + Zn 2+ Reaction is already balanced, 2 electrons transferred ΔG = (2mol e )(96500 J/V mol)(1.10 V)= 2.12x10 5 J or 212 kj SPONTANEOUS Al + + é Al Cu 2+ + 2é Cu Al(s) + Zn 2+ (aq) Zn(s) + Al + (aq) ΔG = (6mol e )(96500 J/V mol)(2.00 V)= 1.56x10 6 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) Assume we have the Zn/Cu cell with the anode at a concentration of 1.50M and the cathode at 0.50M. E cell E cell RT nf lnq (8.14 J/molK)(298.15K) (1.50M) E 1.10V ln (2mole)(96,500 J/Vmol) (0.50M) E 1.09V Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) Q? 2 [Zn ] [Cu ] Q 2 Assume we have the Al/Zn cell with the anode at a concentration of 0.50M and the cathode at.0m at 70.0 C. Al(s) + Zn 2+ (aq) Zn(s) + Al + (aq) 2 (8.14 J/mol K)(4.15K) (0.25M) E 2.00V ln (6 mol e)(96,500 J/V mol) (.0M) E 2.0V Q? 2 [Al ] Q 2 [Zn ] 9
Concentration Cells Let s say you want to make a cell, and you only have one metal (Cu) and only one ionic solution (Cu() ), cold you make a cell that would generate any voltage? Will this cell work? 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 (8.14 J/molK)(298.15K) 0.10M 0.00V ln (2)(96500 J/V mol) 1.00M E cell E cell RT lnq nf Ecell 0. 00V G ( 2mol e)(96,500 J/Vmol)(0.00V) G 5. 8x10 Jor 5.8kJ What would the value of Q need to be if you wanted 1.00V? ΔG = RTlnK 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: or also knowing nfe = RTlnK ΔG = nfe cell Standard Cell Potential (V) RT E cell nf lnk mols of electrons transferred Faraday s Constant (96,500 C/mol) C = J/V (8.14 J/molK)(298.15K) 1.10V lnk 84.6 lnk (2 mol e)(96500 J/V mol) e e K 5.67x10 6 10
ΔG ΔG = nfe cell ΔG = RTlnK E cell RT E cell nf lnk K Some Shortcuts From the Book E cell (8.14J/V mol)(298.15k) lnk n (96,500 J/V mol) 0.0257V E cell n lnk 0.0592V E cell n logk 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 Electrolytic Cell Reaction at Anode Zn Zn 2+ + 2e Reaction at Cathode Cu 2+ + 2e Cu Reaction at Cathode Zn 2+ + 2e Zn Reaction at Anode Cu Cu 2+ + 2e Oxidation Reduction Reduction Oxidation Zn + Cu 2+ Zn 2+ + Cu E cell = 1.10V Zn 2+ + Cu Zn + Cu 2+ 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) What mass of Zn will be produced if a Zn/Cu electrolytic cell is run at.0amps for 5.0hrs? 96,500 C = 1mol e 1 mol = molar mass 60 min 60 sec 1C.00 amp 5.00 hr 54,000 C 1hr 1min 1amps 1mol e 54,000 C 5.60x10 96,500 C molcu 65.9g Zn 5.60x10 1 mole 1 2mole 1mol Zn 1 mol e 18.g Zn 1