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V Chem 210 Jasperse Ch. 19 Handouts^15 Ag Cu Zn2+2++^ (^) Ag (^) ZnCu +0.80+0.34-0. Al Mg3+2+^ (^) Al Mg - -1.662.
- Which species react with Cu2+?
- Which species react with Zn°?
- Which element loves e’s the most? Least?
- NiCl a. Product favored or not? 2 + H 2 Ni + 2 HCl E° = - 0.28 V
b. Is reduction potential for Ni2+^ positive? 19.6 E° Product Favoredcell and ΔG°, E° and K ΔnegG E°poscell Klarge Reactant Favored **Equilibrium pos 0 neg 0 small 1 ΔG° and E° • • both provide measurements for the favorability or unfavorability of a reactionobviously E°cell have opposite signs, but are relatedcell is more limited, to redox reactions
- • • K is also related, since it too relates to how favorable or unfavorable a reaction isΔE°G° = “free energy” available to do be released and do workcell also reflects the amount of energy that is released to do work when a favorable redox o transfer occursThe “free energy” in a cell is really the free energy to do the work of moving electrons and to the work that flowing electricity can do
Δ G° = - nFE° cell Δ G° = - 96.5nE° cell
n = number of electrons transferred in the balanced equat • crucial that you have a correctly balanced redox reaction, and can count how manyion (now coefficients matter!!) F = Faraday’s constant^ electrons transfer = 96.5 to get ΔG in kJ/mol
Chem 210 Jasperse Ch. 19 Handouts 16
Units Δ G°^ =^ - nFE° cell^ Δ G°^ =^ - 96.5nE° cell
F =
96, mole500C e" !
V = (^) CJ so C = !
(^) V^ J C = coulomb, unit of electricity, amount of charge Substituting (^) F =
!
(^) mole^ 96,500J e" (^) • V F = !
(^) mole^ 96.5 e^ "kJ (^) • V Thus when “n” is moles of electrons, and E° volts, the units cancel andcell is in Electrochemistry-Related Units/Terms: For interest, not for test^ only kJ are left. C = Coulomb = A = amp = • 1 mole of electrons = 96,500 C rate of charge flow quantity of electrical per time = Coulombs/secon charge = 6.24 • 10^18 electronsd V = volt = • • Force for moving electrons and chargeNot all Coulombs of charge have the same energy/power/force/ability to do work electrical power/force/strength; difference in electrical potential energy = J/C
- Just like dropping a brick from one cm has le high ss force than dropping it from two meters F = Faraday = charge per chemical amount (the mole) = !
96, mole500C e" = !
mole^ 96.5 e^ "kJ^ • V Watt = amount of energy 1. Balance the reaction, and find ΔG° given the reduction potentials shown Cu + (^) +0.77VFe3+^ (^) +0.34VCu2+^ + Fe2+
- Zn + Cr3+^ Zn2+^ + Cr ΔG° = - 11.6 kJ/mol a. Balance the reaction, and calculate E°cell. b. If the reduction potential for Zn2+^ is - 0.76V, what is the reduction potential for Cr3+?
Chem 210 Jasperse Ch. 19 Handouts 18 Nernst Equation: Eactual = E° -.^0592 n log Q
- Calculate actual voltage for Cu/Cu Ered° 2+(1.0M)//Ag+(0.032M)/Ag Ag Cu2++^ +0.34V+0.80V
- E If a cell with [Agred (^) ° 2 Ag+0.80+(aq) + Zn(s) (^) +] = 0.20 M has E → 2 Ag(s) + Znactual - = 1.63V, what is [Zn0.762+(aq) 2+]
Cell Potential and Equilibrium At equilibrium: a. Eactual = 0 V b. Q=K So, at equilibrium 0 = E°- At equilibrium: E°^. 0592 = n log K Rearranged: log K^. 0592 = n nE°logcell /^ K(0.0592)^ Finding K givenFinding^ E° given K E°cell
“Concentration cells concentrations • at equilibrium, the concentrations would be equal, so the voltage drive is to equalize”: anode and cathode use the same things, but with ions at different Example: So Eactual = ( H-0.592/n)logQ 2 + 2 H+^ H 2 + 2 H+^ E° = 0 This kind of voltage is key to pH meters, neurons (19.8) • • pH meter: dip meneurons: The H+ ter with known [Hconcentration differs inside and outside cell membranes. This creates a+] into a solution, measured voltage reflects solution [H+] voltage which is the key for nerve sensation
19.9 Common Batteries Chem 210 Jasperse Ch. 19 Handouts^19 A. Primary (“Dry Cell 1. Alkaline batteries • run till concentration achieves equilibrium = dead = toss”): Nonrechargeable H 2 O + Zn(s) + MnO oxidized anode cathodereduced 2 (s) ZnO(aq) + 2 MnO(OH)(s) E° = 1.54V
- • • reduction occurs at a graphite elethis is common when an electrode doesn’t involve a redoxflashlights, radio, toys, Jasperse insulin pump, Jasperse blood testers, tooth brush,ctrode
- Mercury battery^ etc.
- Zn + HgOLess power than alkaline batteries, but^ ^ ZnO + Hg^ E° = 1.35 V mercury batteries are physically smaller
- • used inmercury is poisonous, so battery disposal an environmental issue small things (calculators, watches, cameras,…) B Secondary Batteries (“nicad” and “car”) = 1. Lead-acid (car battery) +4 Rechargeable Pb°(s) + PbO Anode cathode 2 (s) + 2 H ↑ CAREFUL! 2 SO 4 2 PbSO 4 (s) + 2 H 20 + energy E° = 2.0V
- The PbSO 4 Leakage caused corrosion!(s) coats electrodes, so reaction can be reversed when “recharged”
- • • Each cell is 2.0V: six alternating cathode/anoEnergy during a recharge drives it in the reverse direction, to the leftSide products during recharge des in series sums to 12V Ox: 6 H Red: 4 H 2 O 2 O + 4e O (^2) - + 4 H 2 H 3 O 2 + + 4 OH+ 4e- -
- Side Both H an explosion. Why no sparks or cigarette lighting is allowed around a car battery 2 and O 2 are produced during the recharge. These are a perfect recipe for
- NiCad (^) • E°=1.3electric shavers, dustbusters, video camcorders, rechargeable power toothbrush, any rechargeable cordless appliances Cd(s) + 2 NiO(OH)(s) + 2 H Anode^0 cathode+3^20 +2Cd(OH)^2 + 2 Ni(OH)+2 2 (s) + energy E°=1.
- again, solid products stay on electrodes, so the reaction can reverse upon treatment with electrical energy
Chem 210 Jasperse Ch. 19 Handouts 21 Problem electrolysis in water? For w to reduce them?: Which of the following metal cations could be converted into elemental metal byhich metal cations would you need to use molten salt if you wanted Hg Cu2+2+ HgCu 0.9V0.2V Co 2 H2+ 2 O Co H (^2) + 2 OH- - -0.3V0.4V Mn Mg2+2+^ MnMg - -1.2V2.4V
- • Easily reduced cations (Zn water.Cations of Active metals can’t (K2+, Ni2++, Cr, Mg3+2+, Sn, Na2+, etc.) c+,…). If they are to be reduced to elementalan be reduced to elemental form in
Oxidation/Anode^ form, they must be reduced as molten salts. o o If a reduced species is harder to oxidize than water itself, insteadIf you want to oxidize something that is harder to oxidize than water, you need to do water will just get oxidized o it as a molten salt rather than in waterOnly reduced species with oxidation potentials more positive than water) or - 0.82 (in neutral water) can be oxidized in water. - 1.23 V (in acidic Water Oxidation: 2 H 2 H 22 OO OO 22 + 4 H+ 4 H++^ + 4e+ 4e-- (^) E =E° = - (^) 0.82 ([H-1.23 ([H+] = 1 x+] = 1.0 M) 10 - (^7) M) Problem which processes would require molten salts??: Which of the following oxidations could be conducted by electrolysis in water? And 2F 2Cl-^ - (^) Cl F 22 + 2e+ 2e - -2.871. 2H^ 2Br 2 O-^ ^ Br O^2 2 + 2e + 4 H^ +^ (10-^7 M) + 4e - -1.080. 2I Cr-2+^ (^) I 2 Cr + 2e3+ (^) + e - +0.410. Al Al3+^ + 3e +1.
Chem 210 Jasperse Ch. 19 Handouts 22 Cl O I 2 22 + 2e + 4H+ 2e– (^) –+^ + 4e 2I 2Cl– – (^) –^ 2H 2 O E˚ = 1.36E˚ = 1.23 VE˚ = 0.54 V Sn 2H Mn2+ (^2) 2+O + 2e^ + 2e (^) + 2e–^ – – (^) Sn MnH (^2) (g) + 2OH – E˚ =E˚ =E˚ = – ––0.14 V0.83 V1.18 V
- Given the eduction potentials, what is the product at the when a current is passed through an aqueous solution of SnCl which chemicals and ions are really in the solution and subject to the electrolysis.) anode 2? (Hint: remember and at the cathode Anode a. Sn b. Cl 2 Cathode c. O d. H e. none of the above (^22) e. none of the above^ a. Sn^ b. Cl c. O d. H 222
- Given the eduction potentials, what is the product at the when a current is passed through an aqueous solution of MnI chemicals and ions are really in the solution and subject to the electrolysis.) anode 2? (Hint: remember which and at the cathode Anode a. Mn b. I 2 Cathode c. O d. H e. none of the above (^22) e. none of the above^ a. Mn^ b. I c. O d. H^222 Electroplating • • metal forms on surface of cathodemany metals are “plated” on outside of things in their way: metal cation elemental metal (reduction at cathode)
- • • “Silverware” for a long time involved plating a coating of silver over something elseArt objects, etc.Materials that are otherwise subject to rust, corrosions are often electroplated with a
Some Famous Electrolyses^ coating that is resistant to air, rain, and acid. (trivia):
- NaCl in H production 2 O NaOH (anode) and HCl and O 2 (cathode) NaOH, HCl
- NaCl (molten) Na metal (cathode) + Cl 2 (anode) Cl 2 production
19.13 Corrosion Chem 210 Jasperse Ch. 19 Handouts^24 Corrosion involves a product H The metal being oxidized always functions as the oxidation half 2 O,…) -favored oxidation of a metal exposed to environment (O 2 , H+, Molecular oxygen is reduced to water in the presence of acid as the reductio As for any favorable redox reaction, the sum of the two half reactions must give positive E Thus, the favorability of the oxygen reduction half is critical in determining which metals cann half and cannot be oxidized The reduction of oxygen in the presenc O 2 + 4H+ (^) 2 H 2 O E°e of acid is a rather favorable reduction half reactionred= +1.23V very good under standard conditions!! Obviously acid that is “standard conditions” 1.0 M is rare The more acidic the water environment, the more favorable oxygen redu metals can be corroded ction is and the more Under 1.0 M acid conditions (pH = 1), any metal that has an oxidation potential better than V can be oxidized in air Under neutral pH = 7 conditions, any metal that has an oxidation potential better than - 0.82V can - 1. be oxidized in air Most metals are included, especially under acidic conditions!! Why most metals are not found in their elemental form in nature, but rather as ions Exception: gold!! Metals usually end as metal oxides or sometimes metal hydrox Ag tarnish ides Cu “greening” Fe rusting Rust: 2 Fe + O 2 + 2 H 2 O 2 Fe(OH) 2 Fe 2 O 3 red-brown rust Practical notes: 1. 2. Corrosion often speeded by HGold has always been valued because unlike other oxidizable m+ (^) and/or ionic salts that acidity wateretals, it retains it’s elemental
- form and it’s lustrous golden elemental surface appearance.Most metals get coated with a film of hard metal oxide, which ends up protecting the interior or the metal.
- • The interior stays elemental metal, but is protecte exposure to air.Sometimes it takes chemical activation to clear the oxide film and enable the elementald by sheath of hard metal oxide from
- Why does iron have such a special rusting problem? • metal inside to be exposed for chemical reactions.Iron is bad because iron oxide (rust) forms flakes that break off.
- As a result, the interior iron is corrosion. not protected and is continuously exposed for further Prevention 1. Coat iron surface with something that resists corrosion and protects. Development of improved and more resistant sealants has been a major priority of auto-industry
- “Galvanized iron” iron but oxidizes to give a hard, protecti[e Zn(OH)-Iron materials are electroplated with Zn, which is more easily oxidized than 2 coating.
Chem 210 Jasperse Ch. 19 Handouts 25
Relating Standard Cell Potential to Standard Half Cell Potentials^ Chapter^ 19 Electrochemistry Math Summary
Eºcell=Eºoxidation + Eºreduction (standard conditions assume 1.0 M concentrations) Relating Half Cell Potentials when Written in Opposite Directions Eºox = - Eºred for half reactions written in opposite directions Relating Standard Cell Potentials to ∆Gº = - nFE˚cell (to give answer in kJ, use F = 96.485) ∆G F = 96,500 C/mol n=number of electrons transferred Relating Actual Cell Potential to Standard Ecell = Eºcell - [0.0592/n] log Q Cell Potential when Concentrations aren't 1.0(Q = ratio of actual concentrations) -M Relating Standard Cell Potential to Equilibrium Constant log K = nEº/0. Relating Actual Cell Potential to Actual Concentration Ecell = - [0.0592/n] log Q for concentration cells, where anode and cathode differ only ins in Concentration Cells concentration, but otherwise have same ions Relating # of Moles of Electrons Transferred as a Function of Time and Current in Electro 1 mol e- (^) = 96,500 C lysis moles of electrons = [current (A)•time (sec)]/96,500 rearranged: time (sec)=(moles of electrons)(96500)/current (in A) for electrolysis, moles, current, and time are related. so time (hours)=(moles of electrons)(26.8)/current (in A)^ Note: 3600 sec/hour Electrochemistry C = Coulomb = quantity of electrical charge = 6.24 • 10 • 1 mole of electrons = 96,500 C-Related Units (^18) electrons A = amp = rate of charge flow per time = C/sec V = volt = electrical power/force/strength = J/C F = Faraday = !
96, mole500C e" = !
mole^ 96.5 e^ "kJ^ • V
Chem 210 Jasperse Ch. 19 Handouts 27 Balancing Redox: Simple Cases 1. 2. (^) Identify oxidation numbers for redox actorsSet coefficients for them so that the where all Reactants and Products are Provides #e’s released = #e’s accepted
- Then balance any redox spectatorsCheck at the end to make sure:^ •^ focus completely on the atoms whose oxidation numbers change Note: Test problems will give you all of the species involved. Some OWL problems will be^ •^ •^ Charges balanceAtoms balance harder and will not include all of the chemicals Some Harder OWL-Level Redox-Balancing Problems: When some necessary chemicals are ommitted a. Sometimes H 2 O, OH , H are omitted, and need to be added in order to balance b. oxygens and hydrogensIn knowing how to do this, it is helpful to distinguish acid versus base conditions c. d. Under acid conditions, it’s appropriUnder base conditions, it’s appropriate to have OHate to have H but not OHbut not H Acid Conditions 1. 2. Identify oxidation numbers for redox actorsSet coefficients for them so that the (^) #e’s Base Conditions -^ released = #e’s accepted focus completely on oxidation numbers change Add H the atoms whose 2 O’s to
- balance oxygenAdd HAdd H (^2) +O’s as needed to balance’s as needed to balance hydrogens oxygens
- and chargeCheck at the end to make sure: a. Charges balance b. Atoms balance 1. 2. Identify oxidation numbers for redoSet coefficients for them so that the released = #e’s accepted x actors #e’s - focus completely on the atoms whose oxidation numbers change Add OH ’s to 3. balance chargeAdd OH ’s as needed to balance charge 4. 5. Add HCheck at the end to 2 O’s to balance hydrogens a. Charges balance make sure: b. Atoms balance
Standard Reduction (Electrode) Potentials at 25˚ C (OWL)
- Chem 210 Jasperse Ch. 19 Handouts
- MnO Cl Cr 22 O+ 2e 47 - 2 - + 8 H + 14 H 2Cl+ + 5e+- + 6e Mn 2Cr2+ 3++ 4H + 7H 2 O 2 O 1.511.361. Half F Ce 2 4++ 2e + e - cell reaction 2F Ce- 3+ E 2.871.61 o (volts)
- O Br NO 22 + 4H 3 + 2e- + 4H+ + 4e+ 2Br + 3e - 2H NO + 2H 2 O 2 O 1.2291.080.
- 2Hg Hg O 2 2++ 4 H2+ + 2e+ 2e+ (10 Hg-Hg 7 M) + 4e 2 2+ 2H 2 O 0.9200.8550.
- Ag Hg Fe3++ 2 2+ + e + e+ 2e Ag Fe 2Hg2+ 0.7990.7890.
- I Fe(CN) Cu 2 + 2e2+ + 2e 63 - 2I+ e- Cu Fe(CN) 44 - 0.5350.480.
- Cu S + 2H 2H2++ + 2e+ e+ + 2e HCu 2 + H 2 S 0.1530.140.
- Pb Sn Ni2+2+2+ + 2e+ 2e+ 2e PbSnNi - --0.1260.140.
- Co Cd Cr3+2+2+ + e+ 2e+ 2e Cr CdCo2+ - --0.280.4030.
- 2H Fe Cr2+3+ 2 O + 2e + 2e+ 3e FeCrH 2 + 2OH- (10-^7 M) - --0.410.440.
- Zn 2H Mn2+ 2 2+O + 2e + 2e+ 2e Zn HMn 2 + 2OH - - --0.7630.831.
- Al Mg Na3++2+ + e + 3e+ 2e Na Al Mg - --1.662.372.
- K Li++ + e+ e KLi - -2.9253.
