Chapter

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#1 → ← #2 #3 ↑ #3 ↑ #4 ↑ half-reaction #5 ↑ half-reaction Identify the 5 items shown below:

Galvanic Cell

What happens at the anode? cathode? AN OX: Anode Oxidation RED CAT: Reduction at Cathode

Can you fill in the blanks below? ____ ∣ ____ ∥ ____ ∣ ____ Mg (s) Mg 2+ (aq) Cu 2+ (aq) Cu (s)

What is E° cell for Mg ∣ Mg 2+ ∥ Cu 2+ ∣ Cu ? = 2.71 vSpontaneous?Yes, since +

E° cell standard conditions: [ion] = 1 M P = 1 atm T = 25°C Standard Cell Potential: E° cell E° cell = E° ox + E° red

NO! Standard reduction potentials are intensive properties. They do NOT depend on quantities. Should you balance the exchange of electrons for cell potentials?... for redox equations? Yes! This is an example of an extensive property that does depend on the quantities.

Problem: Calculate the voltage produced by a voltaic cell with an Al electrode in Al 3+ solution and an iron electrode in a solution of Fe 2+ ions. − =1.25 volts

… will be less than what you calculate. Some will be converted to unusable sources. Work Electrochemical work: …moving electrons

EMF ( ℰ ) Electromotive force (emf ), ℰ, is the difference in electrical potential (between anode and cathode) per unit charge. It is measured in volts. ℰ = −w/q Beware! Electro “q” = charge (coulombs) Thermo “q” = heat (J) w = − ℰ ∙q

∆G° Reaction proceeds?KpKp E°E° = 0 < 0 > 0 at equilibrium → spontaneous ( towards products ) ← nonspontaneous ( is spontaneous towards reactants ) =1 >1 <1 =0 + −

ΔG = –n∙F∙E or ∆G° = standard free energy (J) n = moles of electrons (exchanged in balanced redox equation) F = Faraday constant = 96,500 C/mol e – E° = standard cell potential (J/C) ΔG° = –n∙F∙E°

Problem: Could Zn ions plate on a strip of Cu? Calculate ∆G° for a half cell of Zn and Zn ions connected to another half cell of Cu and Cu 2+ ions. ∆G° = −nFE° = −2 moles (96,500 C/mol) ( ) = − 212,300 J = −212 kJ Cu 2+ (aq) + Zn (s) ⇌ Zn 2+ (aq) + Cu (s) No; −E° is nonspontaneous

Cu 2+ (aq) + Zn (s) ⇌ Zn 2+ (aq) + Cu (s) E° = = 1.10 volts What if [Zn 2+ ] > 1 M? Shifts left; opposes cell reaction E° < 1.10 volts What if [Cu 2+ ] > 1 M? Shifts right; E° > 1.10 volts

Concentration Cell Same half cells, but different solution concentrations Why are the electrons flowing to the more concentrated solution? To neutralize the higher concentration of + charge.

Nernst Equation – looks at relationship between cell potential and concentration E = E° − ln(Q) At 25° we can combine R, T, F, and ln to log conversion to get: E = E° − log(Q)

Problem: Calculate E cell for a galvanic cell based on: FeO 4 2− + 8H + + 3e − → Fe H 2 O E°=+2.20 v O 2 + 4H + + 4e − → 2H 2 O E° = v if [FeO 4 2− ] = M, [O 2 ] = 1.0x10 −5 M, [Fe 3+ ] = M; pH = 5.2 E cell = 2.20 − 1.23 = 0.97v. Q = 6.25x10 87 n = 12e − 4FeO 4 2− +32H e − +6H 2 O→4Fe H 2 O+3O 2 +12H + +12e − gives us: 4FeO 4 2− + 20 H + → 4Fe H 2 O + 3O 2 E = E°− ( / n) log(Q) = v

Batteries – galvanic cells Leclanché cell Dry cell Zn (s) Zn 2+ (aq) + 2e - Anode: Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) + Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) → Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s)

Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Anode: Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) → ZnO (s) + Hg (l) Mercury Battery

Batteries Anode: Cathode: Lead storage battery PbO 2 (s) + 4H + (aq) + SO 4 2- (aq) + 2e - → PbSO 4 (s) + 2H 2 O (l) Pb (s) + SO 4 2- (aq) → PbSO 4 (s) + 2e - Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 4 2- (aq) → PbSO 4 (s) + 2H 2 O (l)

Batteries Solid State Lithium Battery

Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - 2H 2 (g) + O 2 (g) → 2H 2 O (l)

Corrosion

Cathodic Protection of an Iron Storage Tank

Electrolysis: Use of electrical energy to bring about a chemical reaction that would otherwise be nonspontaneous Electrolysis of Water: I = q/t I = current (amperes) q= charge (Coulombs) t=time (sec.)

Electrolytic Cell: an electrochemical cell where electrical energy drives nonspontaneous reactions. Electroplating: depositing one metal on another by use of electric current.

Problem: How many grams of Cu can be reduced by applying a 3.0-amp current for 16.2 min. to a solution containing Cu 2+ ions? (3.0 C/s )x (972 s) x (1 moles e-/96500 C) x (1 mol Cu/(2 moles e-))x(63.54 g Cu/mol) = 0.96 g Cu