2014 Electrolytic Cells and Electroplating

Chemistry Electrolytic Cells and Electroplating Advanced Placement

2014 Periodic Table of the Elements

1 2 H He 1.0079 4.0026 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 6.941 9.012 10.811 12.011 14.007 16.00 19.00 20.179 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar 22.99 24.30 26.98 28.09 30.974 32.06 35.453 39.948 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.90 50.94 52.00 54.938 55.85 58.93 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 93.94 (98) 101.1 102.91 106.42 107.87 112.41 114.82 118.71 121.75 127.60 126.91 131.29 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba * La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.2 192.2 195.08 196.97 200.59 204.38 207.2 208.98 (209) (210) (222) 87 88 89 104 105 106 107 108 109 110 111 112 † Fr Ra Ac Rf Db Sg Bh Hs Mt § § § §Not yet named (223) 226.02 227.03 (261) (262) (263) (262) (265) (266) (269) (272) (277)

58 59 60 61 62 63 64 65 66 67 68 69 70 71 *Lanthanide Series: Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.12 140.91 144.24 (145) 150.4 151.97 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97 90 91 92 93 94 95 96 97 98 99 100 101 102 103 †Actinide Series: Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.04 231.04 238.03 237.05 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) 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 = = millimeters millimeters 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 pH pH == −−log[Hlog[H++]] , , pOHpOH == −−log[OHlog[OH−−]] 1414 == pHpH ++ pOHpOH -- pH pH == 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 t t == 0.6930.693 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 KE per molecule = 1 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 Electrolytic Cells and Electroplating

What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with electrochemical processes:

E°cell; 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

ELECTROCHEMICAL CELLS

Electrochemical Cells: A Comparison Electrodes made Battery its cell Galvanic (voltaic) spontaneous − Is separated into 2 from metals (inert Pt potential drives the cells oxidation-reduction half-cells or C if ion to ion or reaction and thus the reaction gas) e− Battery charger − non-spontaneous requires an external Usually occurs in a Usually inert Electrolytic cells oxidation-reduction energy source to single container electrodes reaction drive the reaction and e−

Electrolysis and Non-spontaneous Cells

The Electrolytic Cell: What is what and what to know

An electrolytic cell is an electrochemical cell in which the reaction occurs only after adding electrical energy; thus it is not thermodynamically favorable – it requires a driving force

Electrolytic reactions often occur in a single container; not 2 separate ½ cells like Voltaic cells

The calculated E° for an electrolytic cell is negative (i.e. that is the amount of potential required to drive the reaction).

This driving force causes the electrons to travel from the positive electrode to the negative electrode, the exact opposite of what you would expect. Remember: EPA – Electrolytic Positive Anode

Anode − the electrode where oxidation occurs. Cathode − the electrode where reduction occurs.

Electron Flow − From Anode To CAThode

§ Electrolytic reactions typically occur aqueous solutions (or in molten liquids) § If there is no water present, you have a pure molten ionic compound thus § the cation will be reduced § the anion will be oxidized § If water is present, you have an aqueous solution of the ionic compound, thus § You must decide which species is being oxidized and reduced; the ions or the water! § No alkali or alkaline earth metal can be reduced in an aqueous solution - water is more easily reduced. § Polyatomic ions are typically NOT oxidized in an aqueous solution - water is more easily oxidized.

§ When it comes to water, be familiar with the following § REDUCTION OF WATER: − − 2 H2O() + 2 e → H2(g) + 2 OH E° = −0.83 V

§ OXIDATION OF WATER: + − 2 H2O()→ O2(g) + 4 H + 4 e E° = −1.23 V

Electrolysis: What to Do and How to Do it!

An electric current is applied to a 1.0 M KI solution. The possible reduction half reactions are listed in the table.

Reduction Half Reactions E° (V) − 2 H2O() + 2 e− → H2(g) + 2 OH −0.83 V

+ − O2(g) + 4 H + 4 e → 2 H2O() +1.23 V

− − I2(g) + 2 e → 2 I +0.53 V K+ + e− → K(s) −2.92 V

(A) Write the balanced half-reaction for the reaction that takes place at the anode.

(B) Write the balanced overall reaction for the reaction that takes place.

+ − First realize the solution contains 3 important species: K , I , and H2O. From those 3 species, you must decide which is being oxidized and which is being reduced.

Where to start – Water can be both oxidized and reduced in an aqueous solution. The I− cannot be further reduced (it has a 1– charge) The K+ cannot be further oxidixed (it has a 1+ charge)

− For OXIDATION: either I or H2O will be OXIDIZED… Write the oxidation half-reactions – Remember, the more positive oxidation potential will be oxidized. + − 2 H2O → O2 + 4 H + 4 e E° = −1.23 V − − 2 I → I2 + 2 e E° = −0.53 V (will be oxidized)

+ Must decide which is being reduced, K or H2O… The more positive reduction potential gets to be reduced; plus a great rule of thumb to remember is NO alkali metal will be reduced in aqueous solution − water will be. − − 2 H2O+ 2 e → H2 + 2 OH E° = −0.83 V (will be reduced) + − K + e → K E° = −2.92 V

Put BOTH half-reactions TOGETHER:

− − 2 I → I2 + 2 e E° = −0.53 V (oxidation) (occurs at the anode) − − 2 H2O+ 2 e → H2 + 2 OH E° = −0.83 V (reduction) (occurs at the cathode) Balanced Overall Reaction:

− − 2 I + 2 H2O → H2 + 2 OH + I2 E° = −1.36 V

(D) What would you observe occurring at the cathode when current is applied to this solution?

Gas bubbles form from the production of hydrogen gas.

(E) Calculate the ΔG° for this reaction.

ΔGnE°=−ℑ° J ΔG°=−(2)(96500)( − 1.36)= 262000 mol kJ ΔG°=262 mol

(F) Is this reaction thermodynamically favorable or non-thermodynamically favorable? Justify your answer.

Since ΔG° is positive (and E°cell is negative) the reaction is non-thermodynamically favorable.

The Electrical Energy of Electrolysis

When running an electrical current through a solution to cause electrolysis, you can measure that current and determine how much of what substance is going to be produced – think stoichiometry too! They will ask you “how many grams of a metal can be plated” or “how long it will take to plate a given mass”

§ The amount of electrical charge flowing through an electrochemical cell is measured in coulombs (c). The rate at which the charge flows (per second) is called the current and is measured in amperes, or amps symbolized by I for inductance. By definition, Coulomb(c) amp(I) = sec(t) § A Faraday is the amount of charge associated with 1 mol of electrons. A Faraday has the value of 96,500 C.

Joule (J ) Volt 1 Coulomb(c) Coulomb(c) Amp 1 sec(t) Coulomb(c) Faraday 96,500 mol of e− mol of e− Balanced REDOX Equation mol of substance # of Coulombs = IT or c = IT

time (t) MUST BE IN SECONDS!!!!

§ Using these units you can measure how much substance is going to be produced, how many coulombs or how many amps were required, how long it takes, etc.… It § A shortcut is the formula… (Molar Mass) = grams plated nℑ Calculate the mass of copper metal produced during the passage of 2.50 amps of current through a solution of copper(II) sulfate for 50.0 minutes.

It (Molar Mass) = grams plated n ℑ (2.50)(50× 60) (63.55) = grams plated = 2.47g Cu (2)(96500)

Electrochemistry Cheat Sheet

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

Electrolytic Cell Relationships

The Mnemonics apply to Electrolytic cells too: Electrolytic Positive Anode ANOX; REDCAT; FATCAT… “the more positive reduction potential gets to be 96,500 Coulombs = 1 mole of electrons reduced” “the more positive oxidation potential gets to be # electrons in balanced equations = # moles of oxidized” electrons transferred

Reduction of H2O All time measurements must be in sec for − − 2 H2O() + 2 e → H2(g) + 2 OH E° = −0.83 V electroplating/electrolysis problems

Oxidation of H2O It + − (Molar Mass) = grams plated 2 H2O → O2 + 4 H + 4 e E° = −1.23 V nℑ No alkali or alkaline earth metal can be reduced in an Polyatomic ions are typically NOT oxidized in an aqueous solution - water is more easily reduced. aqueous solution - water is more easily oxidized. Coulomb() q amp() I = sec(t )

E = ; not thermodynamically favored; ΔG = (+); Joule (J ) °cell − Volt = 1 K<1 Coulomb(c) Coulomb(c) ΔG° = −n ℑ Ε° Amp = sec(t) Coulomb(c) ΔG° = −RT lnK Faraday (ℑ) = 96,500 mol of e− Inert electrodes are typically used in gas and ion to ion galvanic cells and in electrolytic cells Connections Thermo and Equilibrium Stoichiometry Potential Pitfalls

Watch signs on voltages!! BE SURE units cancel out in your calculations.

Balancing overall reactions − make sure # of Units on E are volts electrons is the same in both half reactions. °

NMSI SUPER PROBLEM

An electric current is applied to two separate solutions for 30 minutes, under the same conditions using inert electrodes. Observations are noted in the table below.

Solution A – 1.0 M K2SO4 Solution B – 1.0 M CuSO4 Anode: gas bubbles Anode: gas bubbles Cathode: gas bubbles Cathode: dark flakes formed on the electrode

In both reactions, water is oxidized according to the following oxidation half-reaction. + − 2 H2O ()→ O2(g) + 4 H + 4 e E° = −1.23 V

(a) Write the balanced equation for the half-reaction that occurs at the cathode in

(i) Solution A

(ii) Solution B

(b) For Solution A, is the reaction thermodynamically favorable or not thermodynamically favorable? Justify your answer.

(c) In the electrolysis of the K2SO4 solution, identify the gas produced and describe a test that can be used to identify the gas at the

(i) anode

(ii) cathode

(d) Describe in the box below, what observations, if any, would be noted if a couple of drops of phenolphthalein indicator were added around the cathode of both solutions. Phenolphthalein indicator is colorless in acidic solutions and turns pink in basic solutions.

Justify your observations.

Solution A – 1.0 M K2SO4 Solution B – 1.0 M CuSO4

(e) The dark flakes formed on the electrode in the electrolysis of Solution B were collected and dried. The mass of these flakes was determined to be1.019 grams.

(i) Identify the flakes.

(ii) Calculate the amount of current that was passed through Solution B.