2014 Oxidation-Reduction and Galvanic Cells

Chemistry Oxidation-Reduction and Galvanic Cells 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 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!

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. − − − − + 4 OH + 2 H2O + 2 Cl2 → 2 Cl + 2 OCl + 4 H2O The 4 OH ions and the 4 H ions combine

− − − 4 OH + 2 Cl2 → 2 Cl + 2 OCl + 2 H2O

− − − 2 OH + Cl2 → Cl + OCl + H2O Balanced for atoms and charge

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−

The Galvanic Cell: What is what and what to know? Anode the electrode where oxidation occurs. Over time the mass of the anode ANODE: AN OX − may decrease as the metal is oxidized into ions. Cathode − the electrode where reduction occurs. Over time the mass of the CATHODE: RED CAT cathode may increase as the metal ions in the solution are reduced and plated onto it. FAT CAT Electron Flow − From Anode To CAThode Ca + hode Cathode is + galvanic cells Salt Bridge provides ions to balance the charge in each cell; contains a neutral Salt Bridge − salt that is very soluble (avoids precipitation issues). The salt cations flow into the

cathode and the salt anions flow into the anode.

The Galvanic Cell: How it Works!

The above picture shows the oxidation-reduction reaction between Zn and Cu. The diagram of the cell clearly shows: the anode and cathode; the half-reaction that occurs at each electrode, the direction of electron flow, the direction of ion movement from the salt bridge, the overall reaction, as well as the E°cell for the reaction.

Calculating Standard Cell Potential (E°cell) The difference in electrical potential between the two half-reactions is measured with a voltmeter. The difference between the cell potentials of the two half-reactions determines the overall cell potential for the reaction. Look at a table of Standard Reduction Potentials (or the reduction potentials that are provided). 1 Write both reduction reactions from the table with their voltages. THE MORE POSITIVE REDUCTION POTENTIAL IS REDUCED. 2 THE LEAST POSITIVE IS OXIDIZED.

3 Reverse the equation that will be oxidized; be sure to change the sign of the voltage [this is now E°oxidation] 4 Balance and add the two half-reactions together.

5 Now add the two cell potentials together. E°cell = E°oxidation + E°reduction ° indicates standard conditions (1 atm, 25°C; 1 M)

2+ − Zn (aq) + 2e → Zn(s) E°reduction = − 0.76 V 2+ − Cu (aq) + 2e → Cu(s) E°reduction = + 0.34 V

More positive is reduced (+ 0.34 V > − 0.76 V) – Cu2+ (+0.34) is reduced and Zn oxidized

2+ − Zn(s) → Zn (aq) + 2e E°oxidation = + 0.76 V 2+ − Cu (aq) + 2e → Cu(s) E°reduction = + 0.34 V

2+ 2+ Zn(s) + Cu (aq) → Zn (aq) + Cu(s) E°cell = + 1.10 V

Non Standard Conditions

§ The pressure, the temperature or the concentration changes from 1 atm, 25°C, or 1M…

§ Typically a concentration change from 1 M will result in a change in the cell potential § THINK about the changes in concentration – they only affect the ions in solution (aq) [PRODUCTion ] § ion [REACTANT ] § If the ratio of ions increases compared to standard conditions, the reverse reaction will become thermodynamically favored § If the ratio of ions decreases compared to standard conditions, the forward reaction will become thermodynamically favored

Non Standard Conditions Example For the cell

2 Al(s) + 3 Mn2+(aq) → 2 Al3+(aq) + 3 Mn(s)

The reaction above was carried out at 25°C. The concentration of the aluminum ion (Al3+ ) is 2.0 M, and the concentration 2+ of the manganese ion (Mn ) is 1.00 M. Indicate whether Ecell will increase, decrease, or remain the same (compared to the standard cell potential, E°cell). Justify your answer.

For the cell X(s) + Y2+(aq) → X2+(aq) + Y(s)

The concentration of the X ion is 0.20 M, and the concentration of the Y ion is 0.30 M. Indicate whether Ecell will increase, decrease, or remain the same (compared to E°cell). Justify your answer.

CELL POTENTIAL, ELECTRICAL WORK, EQUILIBRIUM & FREE ENERGY

Work The electromotive force (emf) or cell potential (E ) is the driving force responsible for the movement of electrons from cell the anode to the cathode in a voltaic cell. The unit for this potential is the Volt (V) − by definition, it requires 1 joule of energy (work) to transport 1 coulomb of electrical charge across a potential of 1 volt.

work(J ) emf() V = charge(C )

Summary OF Gibb’s Free Energy and Cell Potential

E°cell = − Implies non-spontaneous reaction (ΔG° = + and K < 1)

E°cell = + Implies spontaneous reaction (ΔG° = − and K > 1)

E°cell = 0 Reaction is at equilibrium (dead battery)

The larger the E°cell The more spontaneous the reaction

Gibb’s Free Energy and Cell Potential: Relevant Equations ΔG = Gibb’s free energy change ΔG° = −n ℑ Ε° n = number of moles of electrons. 96500C ℑ = Faraday’s constant mole− ΔG = Gibb’s free energy change ΔG° = −RT lnK R = Gas constant 8.315× 10−3 kJ mol−1 K−1 K = equilibrium constant T = Temperature (K)

Free Energy, Equilibrium, and Cell Potential

ΔG° K E° 0 at equilibrium 0 >1, products negative + favored <1, reactants positive favored −

Electrochemistry Cheat Sheet

Cathode is + galvanic cells OIL RIG – oxidation is loss of e−; reduction is gain “the more positive reduction potential gets to be FATCAT – e− from the anode to the cathode reduced”

ANOX – oxidation at the anode REDCAT – reduction at the cathode

Be able to… label the parts of a galvanic cell; such as The cathode will gain mass because it is the site of the anode; cathode; salt bridge; electron flow; half reduction; the anode will lose mass because it is the site reactions; overall reaction; direction of ions from salt of oxidation. (only true when electrodes are metals − not bridge true of inert electrodes)

E°cell = +; thermodynamically favored E°cell = −; non thermodynamically favored ΔG = (−); K>1 ΔG = (+); K<1

Be able to explain whether Ecell increases or decreases or Salt Bridge − provides ions to balance the charge in each remains the same when the concentrations of ions ½ cell; contains a neutral salt that is very soluble (avoids change from 1M solutions. Justify by describing which precipitation issues). The salt cations flow into the direction is more thermodynamically favorable (OR cathode and the salt anions flow into the anode. USE the Nernst equation.

For the Nernst Equation you MUST be able to relate The Nernst Equation – NOT required for the EXAM but how the sign of log Q affects the overall cell potential. it is very helpful in solving for cell potential under non- If Q is greater than 1 then the log of Q is standard conditions positive; thus E (+) decreases the E °cell − cell RT If Q is greater less 1 then the log of Q is E = E° − lnQ cell cell nℑ negative; thus E°cell − (−) increases the Ecell

Connections Thermo and Equilibrium Stoichiometry Connects EQUILIBRIUM to ELECTRO and THERMO Connects THERMO to ELECTRO E°cell = 0 at equilibrium thus ΔG = 0 ΔG° = −n ℑ Ε° Use this to connect EQUILIBRIUM to THERMO (and then THERMO to ELECTRO ΔG° = −RT lnK Potential Pitfalls

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

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

NMSI SUPER PROBLEM

Answer the questions below, which relate to reactions involving copper, Cu and copper(II) ion, Cu2+.

A standard voltaic cell is constructed using copper and metal X. The standard reduction potential for Cu is given below. o Cu2+(aq) + 2e− → Cu(s) E = 0.34 V

Immediately after closing the switch, the Voltmeter shows a reading of 0.47 V. Several minutes later it was noted that small flakes were adhering to the Cu electrode.

(a) Which metal, Cu or X, is the anode? Justify your answer.

(b) In the diagram of the cell shown above, label the

i. cathode

ii. direction of electron flow

(d) Determine the standard reduction potential for the X2+/X half-cell.

(e) Using the information provided, select the metal that was used for the X electrode. Explain your choice.

Ag+(aq) + e− → Ag(s) Eo = 0.80 V

Pb2+(aq) + 2e− → Pb(s) Eo = −0.13 V

Sn2+(aq) + 2e− → Sn(s) Eo = −0.14 V

(f) Write a balanced net ionic equation for this electrochemical cell.

(g) This galvanic cell has a salt bridge that is filled with a saturated solution of KNO3. i. As the cell operates, describe what happens in the salt bridge.

ii. Describe what you would observe in the anode half-cell if the salt bridge contained a saturated solution of KCl instead of KNO3.

(h) In the original galvanic cell, if the [Cu2+] is changed from 1.0 M to 0.1 M, would the new cell

potential, Ecell, at 25°C, increase, decrease, or remain the same. Justify your answer.

(i) For the original reaction in the galvanic cell above, indicate whether i. ΔG is positive or negative. Justify your choice.

ii. the equilibrium constant, K, is greater than one or less than one. Justify your choice.

In another experiment, a 1.019 gram piece of Cu was cut from the electrode used above and added to 250. mL of 0.25 M nitric acid, HNO3. An oxidation-reduction reaction between the copper and the nitrate ion occurs as indicated below. − 2+ Cu + NO3 → Cu + NO

(j) Write a complete and balanced net ionic equation for this redox reaction. Show work to support your answer.

(k) Identify the limiting reactant. Show work to support your answer.

(l) On the basis of the limiting reactant identified above, calculate the value of the concentration of Cu2+ ions after the reaction is complete.