AP Chemistry Chapter 20 Notes Electrochemistry Applications of Redox

Review Oxidation reduction reactions involve a transfer of electrons. OIL- RIG Oxidation Involves Loss Reduction Involves Gain LEO-GER Lose Electrons Oxidation Gain Electrons Reduction

Applications Moving electrons is electric current. 8H++MnO4-+ 5Fe+2 +5e- ® Mn+2 + 5Fe+3 +4H2O Helps to break the reactions into half rxns. 8H++MnO4-+5e- ® Mn+2 +4H2O 5Fe+2 ® 5Fe+3 + 5e- ) In the same mixture it happens without doing useful work, but if separate

Connected this way the reaction starts Stops immediately because charge builds up. H+ MnO4- Fe+2

Galvanic Cell Salt Bridge allows current to flow H+ MnO4- Fe+2

Electricity travels in a complete circuit Instead of a salt bridge e- H+ MnO4- Fe+2

Porous Disk H+ MnO4- Fe+2

e- e- e- e- Anode Cathode e- e- Reducing Agent Oxidizing Agent

Cell Potential Oxidizing agent pushes the electron. Reducing agent pulls the electron. The push or pull (“driving force”) is called the cell potential Ecell Also called the electromotive force (emf) Unit is the volt(V) = 1 joule of work/coulomb of charge Measured with a voltmeter

0.76 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl

Standard Hydrogen Electrode This is the reference all other oxidations are compared to Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. H2 in H+ Cl- 1 M HCl

Cell Potential Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s) The total cell potential is the sum of the potential at each electrode. Eº cell = EºZn® Zn+2 + Eº Cu+2 ® Cu We can look up reduction potentials in a table. One of the reactions must be reversed, so change it sign.

Cell Potential Determine the cell potential for a galvanic cell based on the redox reaction. Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq) Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V 2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V

Line Notation solid½Aqueous½½Aqueous½solid Anode on the left½½Cathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are aqueous, a platinum electrode is indicated. For the last reaction Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

Galvanic Cell The reaction always runs spontaneously in the direction that produced a positive cell potential. Four things for a complete description. Cell Potential Direction of flow Designation of anode and cathode Nature of all the components- electrodes and ions

Practice Completely describe the galvanic cell based on the following half-reactions under standard conditions. MnO4- + 8 H+ +5e- ® Mn+2 + 4H2O Eº=1.51 Fe+3 +3e- ® Fe(s) Eº=0.036V

Potential, Work and DG emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q Charge is measured in coulombs. -w = qE Faraday = 96,485 C/mol e- q = nF = moles of e- x charge/mole e- w = -qE = -nFE = DG

Potential, Work and DG DGº = -nFE º if E º < 0, then DGº > 0 spontaneous if E º > 0, then DGº < 0 nonspontaneous In fact, reverse is spontaneous. Calculate DGº for the following reaction: Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq) Fe+2(aq) + e-® Fe(s) Eº = 0.44 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

Cell Potential and Concentration Qualitatively - Can predict direction of change in E from LeChâtelier. 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M if [Al+3] = 1.0 M and [Mn+2] = 1.5M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

The Nernst Equation E = Eº - RTln(Q) nF DG = DGº +RTln(Q) -nFE = -nFEº + RTln(Q) E = Eº - RTln(Q) nF 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Eº = 0.48 V Always have to figure out n by balancing. If concentration can gives voltage, then from voltage we can tell concentration.

The Nernst Equation 0 = Eº - RTln(K) nF Eº = RTln(K) nF As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and Ecell = 0 0 = Eº - RTln(K) nF Eº = RTln(K) nF nFEº = ln(K) RT

Batteries are Galvanic Cells Car batteries are lead storage batteries. Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O Dry Cell Zn + NH4+ +MnO2 ® Zn+2 + NH3 + H2O Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base) NiCad NiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

Corrosion Rusting - spontaneous oxidation. Most structural metals have reduction potentials that are less positive than O2 . Fe ® Fe+2 +2e- Eº= 0.44 V O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V Fe+2 + O2 + H2O ® Fe2 O3 + H+ Reaction happens in two places.

Salt speeds up process by increasing conductivity Water Rust e- Iron Dissolves- Fe ® Fe+2

Preventing Corrosion Coating to keep out air and water. Galvanizing - Putting on a zinc coat Has a lower reduction potential, so it is more. easily oxidized. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

Electrolysis Running a galvanic cell backwards. Put a voltage bigger than the potential and reverse the direction of the redox reaction. Used for electroplating.

1.10 e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Anode Cathode

A battery >1.10V e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Cathode Anode

Calculating plating Have to count charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second q = I x t q/nF = moles of metal Mass of plated metal How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+

Other uses Electroysis of water. Seperating mixtures of ions. More positive reduction potential means the reaction proceeds forward. We want the reverse. Most negative reduction potential is easiest to plate out of solution.

Balancing Redox Equations

Redox Reactions Environment [acidic or basic] is very important Particles available but sometimes are not given in the reaction

hydrogen cation & water acid hydrogen cation & water base hydroxide ion & water

Identify species undergoing 1. oxidation 2. reduction

Split overall reaction into 1. Oxidation half-reaction 2. Reduction half-reaction

Half-reaction in ACID Sol’n 1. Balance species changing oxidation # 2. Balance oxygen by adding H2O

Half-reaction in ACID Sol’n 3. Balance hydrogen by adding H+ 4. Balance charge by adding e-

Half-reaction in ACID Sol’n 5. Add the two half- reactions eliminating all electrons (least common multiple concept)

CHECK final equation for BALANCE of ATOMS and CHARGE

The Redox Blues

1. in acid AuCl4-(aq) + AsH3(g) ---> H3AsO3(aq) + Au(s) + Cl-(aq)

Half-reaction in BASIC Sol’n 1. Balance species changing oxidation # 2. Balance oxygen by adding twice as many OH-

Half-reaction in BASIC Sol’n 3. Balance hydrogen by adding H2O 4. Balance charge by adding e-

Half-reaction in BASIC Sol’n 5. Add the two half- reactions eliminating all electrons (least common multiple concept)

OR Balance by acidic method & then “neutralize” the H+ by adding OH- and adjusting the H2O

2. in base Am3+(aq) + S2O82-(aq) ----> AmO2+(aq) + SO42-(aq)

3. MnO4-(aq) + H2C2O4(aq)  Mn2+(aq) + CO2(g)

4. Bi(OH)3 + SnO22-  Bi(s) + SnO32-

ELECTROLYTIC CELLS

Electrolytic Cell a cell that uses electrical energy to produce a chemical change that would otherwise NOT occur spontaneously

Process referred to as electrolysis

(+) (-) M+(aq) M M X-(aq)

e- e- M+(aq) M M X-(aq) Anode M M+ + e- oxidation Cathode M+ + e- M (+) (-) M+(aq) M M X-(aq) Anode M M+ + e- oxidation Cathode M+ + e- M reduction

a unit of electrical current equal to one coulomb of charge per second Ampere a unit of electrical current equal to one coulomb of charge per second coul sec 1 amp = 1

Coulomb a unit of electric charge equal to the quantity of charge in about 6 x 1019 electrons

a constant representing the charge on one mole of electrons Faraday a constant representing the charge on one mole of electrons 1 F = 96,485 C 96,500 C

3: It is necessary to replate a silver teapot with 15. 0 g of silver 3: It is necessary to replate a silver teapot with 15.0 g of silver. If the electrolytic cell runs at 2.00 amps, how long will it take to plate the teapot?

4: Sodium metal and chlorine gas are prepared industrially in a Down’s Cell from the electrolysis of molten NaCl. What mass of metal and volume of gas can be made per day if the cell operates at 7.0 volts and 4.0 x 104 amps if the cell is 75% efficient?

5: At what current must a cell be run in order to produce 5 5: At what current must a cell be run in order to produce 5.0 kg of aluminum in 8.0 hours if the cell produces solid aluminum from molten aluminum chloride?

ELECTROCHEMISTRY, FREE ENERGY, & EQUILIBRIUM

thus: wmax = - q . Emax

but: wmax = G and q = nF thus if: wmax = - q . Emax then G = - nFE

G = G0 + RT ln Q G = - nFE - nFE = - nFE0 + RT ln Q

NERNST EQUATION

if: aA + bB  cC + dD

IF T = 250C = 298.15 K ln Q = 2.303 log Q R = 8.314 J/mol.K F = 96,485 C/mol

what if : Q = Keq ? then: E = 0.0 V

6: Calculate the equilibrium constant at 400C for the cell: Cd(s) Cd2+ (1M) Pb2+ (1M) Pb(s)

7a: Calculate the standard free energy for the cell: Cr(s) Cr3+ (1M) Fe2+ (1M) Fe(s)

7a: Calculate the standard free energy for the cell: Cr(s) Cr3+ (1M) Fe2+ (1M) Fe(s) 7b: What will be the voltage if [Fe2+] = 0.50M and [Cr3+] = 0.30M at 200C?

8: Through electrochemical calculations, determine the Ksp for silver bromide. AgBr + e-  Ag + Br- E0 = 0.10 V

Review of Redox & Electrochemical Cells

Review Oxidation: loss of e- [increase in ox #] [reducing agent] Reduction: gain of e- [decrease in ox #] [oxidizing agent]

The ease with which a chemical species can be reduced Reduction Potential The ease with which a chemical species can be reduced

Standard Reduction Potential Appendix M Table 20.1 in text

1. Which of the following elements listed is the best reducing agent? Cu Zn Fe Ag Cr

2a. Choosing from among the reactants in the given half reactions, identify the strongest and weakest oxidizing agents.

Anode and Cathode OXIDATION occurs at the ANODE. REDuction occurs at the CAThode.

Electrochemical Cell device in which chemical energy is spontaneously changed to electrical energy

battery voltaic cell galvanic cell

An electrochemical cell consists of ???

M2+(aq) M1+(aq) M1 X-(aq) X-(aq) M2

Anode M1  M1+ + e- Cathode M2+ + e-  M2 M2+(aq) M1+(aq) M1 X-(aq)

Cathode Anode M2+ + e-  M2 M1  M1+ + e- M2+(aq) M1+(aq) M1 X-(aq) K+(aq) NO3-(aq) M2+(aq) M1+(aq) M1 X-(aq) X-(aq) M2 Cathode M2+ + e-  M2 Anode M1  M1+ + e-

e- flow is from source of high “concentration” to source of low “concentration”

e- e- Cathode Anode M2+ + e-  M2 M1  M1+ + e- M2+(aq) M1+(aq) M1 K+(aq) NO3-(aq) M2+(aq) M1+(aq) M1 X-(aq) X-(aq) M2 Cathode M2+ + e-  M2 Anode M1  M1+ + e-

shorthand notation oxidation reduction M1 | M1+ || M2+ | M2 anode  cathode  e- flow 

this e- flow can accomplish work

Electrochemical Standard State Conditions [ions] = 1 M T = 250C Pgas = 1 atm

An electrochemical cell is spontaneous if: Oxidation-reduction occurs Ered + Eox > 0

2b. Which of the oxidizing agents listed is (are) capable of oxidizing Br- to BrO3- ?

Line Notation: ANODE CATHODE Ni(s)|Ni2+ (aq, 1 M)||Au3+(aq, 1 M)|Au(s) oxidation reduction

Line Notation: ANODE CATHODE Al(s) | Al3+(aq, 1 M) || Ni2+(aq, 1 M) | Ni (s) oxidation reduction