AP Chemistry Chapter 20 Notes Electrochemistry Applications of

AP Chemistry Chapter 20 Notes Electrochemistry Applications of Redox

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

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

Ø Connected this way the reaction starts Ø Stops immediately because charge builds up. H+ Mn. O 4 - Fe+2

Galvanic Cell Salt Bridge allows current to flow H+ Mn. O 4 - Fe+2

Ø Electricity travels in a complete circuit e- Ø Instead of a salt bridge H+ Mn. O 4 - Fe+2

Porous Disk H+ Mn. O 4 - Fe+2

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

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

0. 76 H 2 in Anode Zn+2 SO 4 -2 1 M Zn. SO 4 Cathode H+ Cl - 1 M HCl

Standard Hydrogen Electrode H 2 in Ø This is the reference all other oxidations are compared to ØE º = 0 Ø º indicates standard states of 25ºC, 1 atm, 1 M solutions. H+ Cl - 1 M HCl

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

Cell Potential Ø Determine the cell potential for a galvanic cell based on the redox reaction. Ø Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq) Ø Fe+3(aq) + e-® Fe+2(aq) Eº = 0. 77 V Ø Cu+2(aq)+2 e- ® Cu(s) Eº = 0. 34 V Ø Cu(s) ® Cu+2(aq)+2 e. Eº = -0. 34 V Ø 2 Fe+3(aq) + 2 e-® 2 Fe+2(aq) Eº = 0. 77 V

Line Notation solid½Aqueous½solid Ø Anode on the left½½Cathode on the right Ø Single line different phases. Ø Double line porous disk or salt bridge. Ø If all the substances on one side are aqueous, a platinum electrode is indicated. Ø For the last reaction Ø Cu(s)½Cu+2(aq)½½Fe+2(aq), Fe+3(aq)½Pt(s) Ø

Galvanic Cell Ø Ø 1) 2) 3) 4) 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 Ø Completely describe the galvanic cell based on the following half-reactions under standard conditions. Ø Mn. O 4 - + 8 H+ +5 e- ® Mn+2 + 4 H 2 O Eº=1. 51 Ø Fe+3 +3 e- ® Fe(s) Eº=0. 036 V

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

Potential, Work and DG DGº = -n. FE º FIXED Ø if E º > 0, then DGº < 0 spontaneous Ø if E º < 0, then DGº > 0 nonspontaneous Ø Ø In fact, reverse is spontaneous. Ø Calculate DGº for the following reaction: Ø Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq) Ø Fe+2(aq) + e-® Fe(s) Ø Cu+2(aq)+2 e- ® Cu(s) Eº = 0. 44 V Eº = 0. 34 V

Cell Potential and Concentration Ø Qualitatively - Can predict direction of change in E from Le. Châtelier. Ø 2 Al(s) + 3 Mn+2(aq) ® 2 Al+3(aq) + 3 Mn(s) Ø Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1. 5 M and [Mn+2] = 1. 0 M Ø if [Al+3] = 1. 0 M and [Mn+2] = 1. 5 M Ø if [Al+3] = 1. 5 M and [Mn+2] = 1. 5 M

The Nernst Equation Ø DG = DGº +RTln(Q) Ø -n. FE = -n. FEº + RTln(Q) Ø E = Eº - RTln(Q) n. F Ø 2 Al(s) + 3 Mn+2(aq) ® 2 Al+3(aq) + 3 Mn(s) Eº = 0. 48 V Ø Always have to figure out n by balancing. Ø If concentration can gives voltage, then from voltage we can tell concentration.

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

Batteries are Galvanic Cells Ø Car batteries are lead storage batteries. Ø Pb +Pb. O 2 +H 2 SO 4 ®Pb. SO 4(s) +H 2 O Ø Dry Cell Zn + NH 4+ +Mn. O 2 ® Zn+2 + NH 3 + H 2 O Ø Alkaline Zn +Mn. O 2 ® Zn. O+ Mn 2 O 3 (in base) Ø Ni. Cad Ø Ni. O 2 + Cd + 2 H 2 O ® Cd(OH)2 +Ni(OH)2

Corrosion Ø Rusting - spontaneous oxidation. Ø Most structural metals have reduction potentials that are less positive than O 2. Ø Fe ® Fe+2 +2 e. Eº= 0. 44 V Ø O 2 + 2 H 2 O + 4 e- ® 4 OHEº= 0. 40 V Ø Fe+2 + O 2 + H 2 O ® Fe 2 O 3 + H+ Ø Reaction happens in two places.

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

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

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

1. 10 e- e- Zn 1. 0 M Zn+2 Anode 1. 0 M Cu+2 Cathode Cu

e- A battery >1. 10 V Zn 1. 0 M Zn+2 Cathode 1. 0 M Cu+2 e- Cu Anode

Calculating plating Ø Have to count charge. Ø Measure current I (in amperes) Ø 1 amp = 1 coulomb of charge per second Øq = I x t Ø q/n. F = moles of metal Ø Mass of plated metal Ø How long must 5. 00 amp current be applied to produce 15. 5 g of Ag from Ag+

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

Balancing Redox Equations

2. in base 3+ 2 Am (aq) + S 2 O 8 (aq) ----> Am. O 2+(aq) + SO 42 -(aq)

3. Mn. O 4 -(aq) + H 2 C 2 O 4(aq) Mn 2+(aq) + CO 2(g)

4. Bi(OH)3 + Sn. O 2 2 Bi(s) + Sn. O 3 2 -

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

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

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

Coulomb a unit of electric charge equal to the quantity of charge in 19 about 6 x 10 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. 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 Na. Cl. 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. 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 = n. F thus if: wmax = - q Emax. then G = - n. FE

G = 0 G + RT ln Q G = - n. FE = - 0 n. FE + RT ln Q

NERNST EQUATION

if: a. A + b. B c. C + d. D

0 25 C IF T = = 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 0 40 C for the cell: 2+ Cd(s) Cd (1 M) 2+ Pb (1 M) Pb(s)

7 a: Calculate the standard free energy for the cell: Cr(s) Cr 3+ (1 M) Fe 2+ (1 M) Fe(s)

7 a: Calculate the standard free energy for the cell: Cr(s) Cr 3+ (1 M) Fe 2+ (1 M) Fe(s) 7 b: What will be the 2+ voltage if [Fe ] = 0. 50 M and [Cr 3+] = 0. 30 M at 200 C?

8: Through electrochemical calculations, determine the Ksp for silver bromide. Ag. Br + e Ag + Br 0 E = 0. 10 V

Review of Redox & Electrochemical Cells

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

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

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

Anode and Cathode Ø OXIDATION occurs at the ANODE. Ø 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 ? ? ?

M 1 +(aq) X-(aq) M 2

M 1 +(aq) X-(aq) Anode M 1 + + e - M 2+(aq) X-(aq) M 2 Cathode M 2 + + e - M 2

K+(aq) NO 3 -(aq) M 1 +(aq) X-(aq) Anode M 1 + + e - M 2+(aq) X-(aq) M 2 Cathode M 2 + + e - M 2

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

ee- K+(aq) NO 3 -(aq) M 1 +(aq) X-(aq) Anode M 1 + + e - M 2+(aq) X-(aq) M 2 Cathode M 2 + + e - M 2

shorthand notation oxidation reduction + + M 1 || M 2 anode cathode e flow

e this flow can accomplish work

Electrochemical Standard State Conditions [ions] = 1 M 0 T = 25 C Pgas = 1 atm

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

2 b. Which of the oxidizing agents listed is (are) capable of oxidizing Br to Br. O 3 ?

Line Notation: ANODE CATHODE Ni(s)|Ni 2+ (aq, 1 M)||Au 3+(aq, 1 M)|Au(s) oxidation reduction

Line Notation: ANODE CATHODE Al(s) | Al 3+(aq, 1 M) || Ni 2+(aq, 1 M) | Ni (s) oxidation reduction