Chemistry Oxidation-Reduction and Galvanic Cells Advanced Placement 2014 2 Ar Kr 10 18 36 54 86 Xe He Ne Rn (222) 83.80 4.0026 20.179 39.948 131.29 ) 3 I 9 F Cl At Br 17 35 53 85 Lr 71 Lu 10 (210 19.00 79.90 (260) 35.453 126.91 174.97 ) 8 S O Te 16 34 52 84 Se Po 70 Yb No 102 (209 16.00 32.06 78.96 (259) 127.60 173.04 7 P N Bi 15 33 51 83 As Sb 69 Md Tm 101 74.92 (258) 14.007 30.974 121.75 208.98 168.93 1 2 0 6 C Si 8.71 14 32 50 82 Sn Pb Er Ge 68 Fm 10 28.09 72.59 207. (257) 12.01 11 167.26 11 §Not yet named 5 B Tl In Al 4.82 13 31 49 81 67 99 Es Ga Ho (252) 26.98 69.72 10.8 11 204.38 164.93 2 § 30 48 80 Zn Cd Hg Cf 66 98 Dy 11 12.41 (277) 65.39 (251) 1 200.59 162.50 § 29 47 79 Ag Au Cu 65 97 Tb Bk 111 (272) 63.55 (247) 107.87 196.97 158.93 ) 0 § Ni Pt 28 46 78 Pd 64 96 Gd 11 Cm (269 (247) 58.69 106.42 195.08 157.25 2 Ir 27 45 77 Mt Co Rh 63 95 Eu 109 Am ble of the Elements (266) 58.93 192. (243) 102.91 151.97 Ta 1 2 4 26 44 76 Fe Hs Ru Os 62 94 Pu 108 Sm (265) 55.85 101. 190. (244) 150. Tc 25 43 75 61 93 Bh Re Np Mn Pm 107 (98) Periodic (262) (145) 54.938 186.21 237.05 U W Cr 24 42 74 Sg 60 92 Mo Nd 106 (263) 52.00 93.94 183.85 144.24 238.03 ) 4 1 5 V Pr Ta 23 41 73 59 91 Pa Nb Db 10 (262 50.9 92.9 180.95 140.91 231.04 ) 0 2 4 Ti Zr Hf Rf 22 40 72 58 90 Th Ce 10 (261 47.9 91.2 178.49 140.12 232.04 : : Y 21 39 57 89 La Sc Ac * † 44.96 88.91 138.91 227.03 0 8 2 4 Sr 12 20 38 56 88 Be Ba Ca Ra Mg 9.012 24.3 40.0 87.6 137.33 226.02 ) 1 †Actinide Series 1 3 K H Li Fr 11 19 37 55 87 *Lanthanide Series Cs Na Rb (223 6.94 22.99 39.10 85.47 1.0079 132.91 ADVANCEDADVANCED PLACEMENTPLACEMENT CHEMISTRYCHEMISTRY EQUATIONSEQUATIONS ANDAND CONSTANTSCONSTANTS ThroughoutThroughout thethe testtest thethe followingfollowing symbolssymbols havehave ththee definitionsdefinitions specifiedspecified unlessunless otherwiseotherwise noted.noted. L,L, mLmL == liter(s),liter(s), milliliter(s)milliliter(s) mmmm HgHg == millimetersmillimeters ofof mercurymercury gg == gram(s)gram(s) J,J, kJkJ == joule(s),joule(s), kilojoule(s)kilojoule(s) nmnm == nanometer(s)nanometer(s) VV == volt(s)volt(s) atmatm == atmosphere(s)atmosphere(s) molmol == mole(s)mole(s) ATOMICATOMIC STRUCTURESTRUCTURE EE == energyenergy EE == hhνν νν == frequencyfrequency cc == λνλν λλ == wavelengthwavelength Planck’sPlanck’s constant,constant, hh == 6.6266.626 × × 1010−−3434 JJ s s SpeedSpeed ofof light,light, cc == 2.9982.998 ×× 101088 msms−−11 Avogadro’sAvogadro’s numbernumber == 6.0226.022 ×× 10102323 molmol−−11 ElectronElectron charge,charge, e e == −−1.6021.602 ××1010−−1919 coulombcoulomb EQUILIBRIUMEQUILIBRIUM [C][C]cdcd [D] [D] KKcc == ,, wherewhere aa AA ++ bb BB RR cc CC ++ dd DD EquilibriumEquilibrium Constants Constants [A][A]abab [B] [B] cdcd KKcc (molar(molar concentrations)concentrations) ((PPPPCC )( )(DD ) ) KKpp == abab KKpp (gas(gas pressures)pressures) ((PPPPABAB )( )( ) ) KKaa (weak(weak acid)acid) [H[H+-+- ][A ][A ] ] KK == K (weak base) aa [HA][HA] Kbb (weak base) KK (water)(water) [OH-+-+ ][HB ] ww KK == [OH ][HB ] bb [B][B] ++ −− −−1414 KKww == [H[H ][OH][OH ]] == 1.01.0 ××1010 atat 2525°°CC == KKaa ×× KKbb pHpH == −−log[Hlog[H++]] , , pOHpOH == −−log[OHlog[OH−−]] 1414 == pHpH ++ pOHpOH -- pHpH == ppKK ++ loglog[A[A ] ] aa [HA][HA] ppKKaa == −−loglogKKaa,, ppKKbb == −−loglogKKbb KINETICSKINETICS ln[A]ln[A] −− ln[A]ln[A] == −−ktkt tt 00 kk == raterate constantconstant 1111 tt == timetime -- == ktkt AAAA [][]tt [][]00 tt½½ == half-lifehalf-life 0.6930.693 tt == ½½ kk GASES, LIQUIDS, AND SOLUTIONS P = pressure V = volume PV = nRT T = temperature moles A n = number of moles PA = Ptotal × XA, where XA = total moles m = mass M = molar mass Ptotal = PA + PB + PC + . D = density n = m KE = kinetic energy M Ã = velocity K = °C + 273 A = absorbance a = molarabsorptivity m D = b = path length V c = concentration 1 KE per molecule = mv2 2 Gas constant, R = 8.314 J mol--11 K Molarity, M = moles of solute per liter of solution --11 = 0.08206 L atm mol K --11 A = abc = 62.36 L torr mol K 1atm= 760 mm Hg = 760 torr STP= 0.00D C and 1.000 atm THERMOCHEMISTRY/ ELECTROCHEMISTRY q heat = m = mass q= mcD T c = specific heat capacity T temperature DSSDD= ÂÂ products- S D reactants = SD = standard entropy DDHHDD= ÂÂproducts- D HfD reactants f HD = standard enthalpy GD standard free energy DDGGDD= ÂÂf products- D GfD reactants = n = number of moles D DDGD = HDD - TS D E = standard reduction potential I = current (amperes) = -RTln K q = charge (coulombs) t = time (seconds) = -nFED q Faraday’s constant , F = 96,485 coulombs per mole I ϭ t of electrons 1 joule 1volt = 1coulomb Electrochemistry Redox and Galvanic Cells What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with electrochemical processes: E°cell; reduction and oxidizing agent; cell potential; reduction or oxidation; anode; cathode; salt bridge; electron flow; voltage; electromotive force; galvanic/voltaic; electrode; battery; current; amps; time; grams (mass); plate/deposit; electroplating; identity of metal; coulombs of charge ELECTROCHEMICAL TERMS Electrochemistry the study of the interchange of chemical and electrical energy OIL RIG Oxidation Is Loss, Reduction Is Gain (of electrons) LEO the lion says GER Lose Electrons in Oxidation; Gain Electrons in Reduction Oxidation the loss of electrons, increase in charge Reduction the gain of electrons, reduction of charge Oxidation number the assigned charge on an atom Balancing Oxidation−Reduction Reactions: The half-reaction method 1 Split the overall reaction into its oxidation and reduction half-reactions. 2 One half-reaction at a time, balance the atoms, except hydrogen and oxygen. 3 Balance oxygen atoms by adding H2O molecules to the side that is short atoms of oxygen. 4 Balance hydrogen atoms by adding H+ ions to the side missing atoms of hydrogen. Add up the total charge on each side of the half reactions. Then add electrons, to the side that is most positive, 5 until the charge on both sides is equal. Make sure both the oxidation and reduction half reactions have the same number of electrons. If not, multiply 6 each half-reaction by the coefficient required to produce an equal number of electrons being transferred in the two half-reactions. 7 Add the two half-reactions; simplify by cancelation (often there are common species on both sides). If the solution is neutral or acidic the reaction is balanced. If the solution is basic continue with step 8 Add OH− ions (equal to the number of H+ ions) to both sides of the equation. One side of the equation will + 8 also have H ; combine the OH− ion with those H+ and form H2O. Water, which appears on both sides of the equation, can be canceled out. NOTE: No electrons should appear in the final balanced equation! Copyright © 2014 National Math + Science Initiative, Dallas, Texas. 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Visit us online at www.nms.org Page 1 Electrochemistry – Redox and Galvanic Cells Balancing Examples Balance the following redox reactions 2− − 3+ In acidic solution: Cr2O7 (aq) + Cl (aq) → Cr (aq) + Cl2(g) 2− 3+ Cr2O7 → Cr Reduction half-reaction 2− 3+ Cr2O7 → 2 Cr The coefficient of 2 balances the Cr atoms 2− 3+ Cr2O7 → 2 Cr + 7 H2O The 7 H2O molecules balance the O atoms + 2− 3+ + 14 H + Cr2O7 → 2 Cr + 7 H2O The 14 H ions balance the H atoms − − + 2− 3+ The 6 e are added because the charge on the left side of 6 e + 14 H + Cr2O7 → 2 Cr + 7 H2O the reaction is +12 and the charge on the right is +6 − Cl → Cl2 Oxidation half-reaction − 2 Cl → Cl2 The coefficient of 2 balances the Cl atoms − − − The 2 e are added because the charge on the right side of 2 Cl → Cl2 + 2 e the reaction is 0 and the charge on the left side is −2 This half-reaction was multiplied by 3 so there are equal 6 Cl− 3 Cl + 6 e− → 2 numbers of electrons being lost and gained + 2− − 3+ 14 H + Cr2O7 + 6 Cl → 2 Cr + 3 Cl2 + 7 H2O Balanced for atoms and charge − − In basic solution: Cl2(g) → Cl (aq) + OCl (aq) − Cl2 → Cl Reduction half-reaction − Cl2 → 2 Cl The coefficient of 2 balances the Cl atoms The 2 e− are added because the charge on the right − − Cl2 + 2 e → 2 Cl side of the reaction is 0 and the charge on the left side is −2 − Cl2 → OCl Oxidation half-reaction − Cl2 → 2 OCl The coefficient of 2 balances the Cl atoms − 2 H2O + Cl2 → 2 OCl The 2 H2O molecules balance the O atoms − + + 2 H2O + Cl2 → 2 OCl + 4 H The 4 H ions balance the H atoms The 2 e− are added because the charge on the left side − + − 2 H2O + Cl2 → 2 OCl + 4 H + 2 e of the reaction is 0 and the charge on the right side is +2 − + − − − + − The 4 OH ions are added to neutralize the H ions in 4 OH + 2 H2O + 2 Cl2 → 2 Cl + 2 OCl + 4 H + 4 OH the basic solution.