ch17.1 – galvanic cells ch4: redox involves the transfer of electrons (oil rig) zn (s) + cuso...

Ch17.1 Galvanic Cells Ch4: Redox involves the transfer of electrons (OIL RIG) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s) Half rxns: Ox:Red agent: _____ Red: Ox agent: _____

lose 2e- (oxidized) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s) gained 2e- (reduced) Ox Reaction: Zn (s) Zn (aq) +2 +2e - Red agent: Zn Red Reaction: Cu +2 (aq) + 2e - Cu (s) Ox agent: Cu +2

lose 2e- (oxidized) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s) gained 2e- (reduced) Oxidized Reaction: Zn (s) Zn (aq) +2 +2e - Red agent: Zn Reduced Reaction: Cu +2 (aq) + 2e - Cu (s) Ox agent: Cu +2 e - e - e - e - Cu Zn CuSO 4 soln ZnSO 4 soln V

Galvanic cell (Voltaic cell, wet cell battery) - convert chemical potential energy into electrical energy Electrodes - metals in voltaic cells Anode - negative electrode, electrons produced here (Reducing agent - OIL) Cathode - positive electrode, electrons head here (Ox agent - RIG) e - e - e - e - electrons transfer Na + Cl - Na + Cl - Na + Cl - Na + Cl - Na + thru the wire Cl - (salt bridge) Cl - CuNa + (necessary to maintain Na + Zn Cl - an ion charge balance) Cl - CuNa + Na + Zn Cl - Cu Zn CuSO 4 soln ZnSO 4 soln V

Cell Potential, cell Reduction Half Cell Oxidation Half Cell Cu +2 + 2e Cu Zn Zn +2 + 2e Zn Zn +2 + 2e - E = +.76V Cu +2 + 2e - CuE = +.34 V E cell =+1.10 V electrons transfer Na + Cl - Na + Cl - Na + Cl - Na + Cl - Na + thru the wire Cl - (salt bridge) Cl - CuNa + (necessary to maintain Na + Zn Cl - an ion charge balance) Cl - CuNa + Na + Zn Cl - Cu Zn Cu +2 SO 4 -2 Zn +2 SO 4 -2 SO 4 -2 Cu +2 SO 4 -2 Zn +2 CuSO 4 soln ZnSO 4 soln cathode cell anode cell E standard conditions 25C, I Molar V

Cell Potentials

Ex1) Al +3 (aq) + Mg (s) Al (s) + Mg +2 (aq) Give the balanced cell rxn and EMF for the cell.

Ex1) Al +3 (aq) + Mg (s) Al (s) + Mg +2 (aq) Give the balanced cell rxn and EMF for the cell. Al +3 + 3e - AlE o = 1.66 V Mg +2 + 2e - Mg E o = 2.37 V

Ex1) Al +3 (aq) + Mg (s) Al (s) + Mg +2 (aq) Give the balanced cell rxn and EMF for the cell. Al +3 + 3e - AlE o = 1.66 V Mg +2 + 2e - Mg E o = 2.37 V 2Al +3 + 6e - 2AlE o = 1.66 V 3Mg 3Mg +2 + 6e - E o = +2.37 V 2Al +3 (aq) + 3Mg (s) 2Al (s) + 3Mg +2 (aq) E o = +0.71 V Mg metalAl +3 gained e - s so lost e - s so it was reduced it is oxidized. in charge. Formed Anode more Al on the electrode. The metal the Mg | Mg +2 || Al +3 | Al es head to is the cathode Anode cell Cathode cell V e-e- e-e- Mg Al Al +3 Mg +2 Dont multiply!

Ex2) MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) ClO 4 - (aq) + Mn +2 (aq) + H 2 O (l) Give the balanced cell rxn and EMF for the cell. MnO 4 - + 5e - + 8H + Mn +2 + 4H 2 O E o = +1.51 V ClO 4 - + 2e - + 2H + ClO 3 - + H 2 O E o = +1.19 V.

Ex2) MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) ClO 4 - (aq) + Mn +2 (aq) + H 2 O (l) MnO 4 - + 5e - + 8H + Mn +2 + 4H 2 O E o = +1.51 V ClO 4 - + 2e - + 2H + ClO 3 - + H 2 O E o = +1.19 V 2MnO 4 - + 10e - + 16H + 2Mn +2 + 8H 2 O E o = +1.51 V 5ClO 3 - + 5H 2 O 5ClO 4 - + 10e - + 10H + E o = 1.19 V 2MnO 4 - (aq) + 6H + (aq) + 5ClO 3 - (aq) 5ClO 4 - (aq) + 2Mn +2 (aq) + 3H 2 O (l) E o = +0.32 V | || | Anode cell Cathode cell. V e-e- e-e- ? ? ? ? Mn +7 gains 5e - to become Mn +2 : reduced in charge cathode cell Cl +5 loses 2e - to become Cl +7 : oxidized anode cell

Ex2) MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) ClO 4 - (aq) + Mn +2 (aq) + H 2 O (l) MnO 4 - + 5e - + 8H + Mn +2 + 4H 2 O E o = +1.51 V ClO 4 - + 2e - + 2H + ClO 3 - + H 2 O E o = +1.19 V 2MnO 4 - + 10e - + 16H + 2Mn +2 + 8H 2 O E o = +1.51 V 5ClO 3 - + 5H 2 O 5ClO 4 - + 10e - + 10H + E o = 1.19 V 2MnO 4 - (aq) + 6H + (aq) + 5ClO 3 - (aq) 5ClO 4 - (aq) + 2Mn +2 (aq) + 3H 2 O (l) E o = +0.32 V Since REDOX occurs w aqueous ions, we need a non-reactive metal as electrodes, so Pt makes a great choice. H 2 gas bubbles off Pt | ClO 3 -, ClO 4 -, H + || Mn +2, MnO 4 -, H + | Pt of cathode Anode Cathode V e-e- e-e- Pt Mn +2 MnO 4 - H + ClO 3 - ClO 4 - H + Mn +7 gains 5e - to become Mn +2 : reduced in charge cathode cell Cl +5 loses 2e - to become Cl +7 : oxidized anode cell

Ex3) Ag + + e - AgE o = +0.80V Fe +3 + e - Fe +2 E o = +0.77V Setup a galvanic cell, give the balanced cell rxn and EMF for the cell.

Ex3) Ag + + e - AgE o = +0.80V Fe +3 + e - Fe +2 E o = +0.77V Setup a galvanic cell, give the balanced cell rxn and EMF for the cell. Ag + + e - Ag RIGE o = +0.80V Fe +2 Fe +3 + e - OILE o = -0.77V Ag + + Fe +2 Ag + Fe +3 E o = +0.03V Since theres no Fe (s), we need another metal to be the electrode. Pt is non-reactive, so makes a great choice. Pt | Fe +2 (aq),Fe +3 (aq) || Ag + (aq) | Ag Anode Cathode V e-e- e-e- Pt Ag Ag + Fe +2 Fe +3

Ex4) Fe +2 + 2e - FeE o = 0.44V MnO 4 - + 5e - Mn +2 + 4H 2 OE o = +1.51V Setup a galvanic cell, give the balanced cell rxn and EMF for the cell. Ch17 HW#1 p880 29,31,33 V e-e- e-e- ? ? ? ?

29. Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine E for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. a. Cl 2 + 2e - 2Cl - E= 1.36 V Br 2 + 2e - 2Br - E= 1.09 V b. MnO 4 + 8H + + 5e - Mn 2 + 4H 2 OE= 1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60

Ch17 HW#1 p880 29,31,33 29. Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine E for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. a. Cl 2 + 2e - 2Cl - E= 1.36 V Br 2 + 2e - 2Br - E= 1.09 V Cl 2 + 2e - 2Cl - E= 1.36 V 2Br - Br 2 + 2e - E= 1.09 V Cl 2 + 2Br - Br 2 + 2Cl - E= +0.27 V V e-e- e-e- Pt Cl - Br - Br 2

Ch17 HW#1 p880 29,31,33 29. Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine E for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. b. MnO 4 - + 8H + + 5e - Mn +2 + 4H 2 OE= 1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60 V

b. MnO 4 - + 8H + + 5e - Mn +2 + 4H 2 OE= 1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60 V 5(IO 4 - + 2H + +2e - IO 3 - + H 2 O) E= +1.60 V 2(Mn +2 + 4H 2 O MnO 4 - + 8H + + 5e - ) E= 1.51 V Mn +7 gains 5e - to become Mn +2 : reduced in charge cathode cell I +7 loses 2e - to become I +5 : oxidized anode cell

b. MnO 4 - + 8H + + 5e - Mn +2 + 4H 2 OE= 1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60 V 5IO 4 - + 10H + + 10e - 5IO 3 - + 5H 2 O E= +1.60 V) 2 Mn +2 + 8H 2 O 2MnO 4 - + 16H + + 10e - E= 1.51 V) 5IO 4 - + 2 Mn +2 + 3H 2 O 5IO 3 - + 2MnO 4 - + 6H + E= +0.09 V) H 2 gas bubbles out Anode Cathode Pt | IO 4 - (aq), IO 3 - (aq),H + || MnO 4 - (aq), Mn +2 (aq),H + | Pt V e-e- e-e- Pt IO 4 - Mn +7 gains 5e - to become Mn +2 : reduced in charge cathode cell I +7 loses 2e - to become I +5 : oxidized anode cell IO 3 - MnO 4 - Mn +2 H+H+ H+H+

31. Give the standard notation for each cell in Exercise 29. a. Cl 2 + 2e - 2Cl - E= 1.36 V Br 2 + 2e - Br - E= 1.09 V Cl 2 + 2e - 2Cl - E= 1.36 V 2Br - Br 2 + 2e - E= 1.09 V Cl 2 + 2Br - Br 2 + 2Cl - E= +0.27 V Pt | Br -, Br 2 || Cl 2 | Cl - | Pt V e-e- e-e- Pt Cl - Br - Br 2 Cl 2

31. Give the standard notation for each cell in Exercise 29. a. Cl 2 + 2e - 2Cl - E= 1.36 V Br 2 + 2e - Br - E= 1.09 V Pt | Br -, Br 2 || Cl 2 | Cl - | Pt b. MnO 4 + 8H + + 5e - Mn 2 + 4H 2 OE= 1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60 V Pt | IO 4 - (aq), IO 3 - (aq),H + || MnO 4 - (aq), Mn +2 (aq),H + | Pt

33. Give the balanced cell reaction and determine E for the galvanic cells based on the following half-reactions. Standard reaction potentials are found in Table 17.1. a. Cu 2+ + e - Cu + E 0 = +0.16 Au 3+ + 3e - AuE 0 = +1.50 3Cu + 3Cu 2+ + 3e - E 0 = 0.16 Au 3+ + 3Cu + 3Cu 2+ + AuE 0 = +1.34 b. Cd 2+ + 2e - CdE 0 = 0.40 VO 2 + + 2H + + e - VO 2+ +H 2 OE 0 = +1.00 2VO 2 + + 4H + + 2e - 2VO 2+ + 2H 2 OE 0 = +1.00 Cd Cd 2+ + 2e - E 0 = +0.40 2VO 2 + + 4H + + Cd Cd 2+ + 2VO 2+ + 2H 2 OE 0 = +1.40

Ch17.1 contCh17 HW#2 p880 30,32,35 30. Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine E for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. a. H 2 O 2 + 2H + + 2e - 2H 2 O E= 1.78 V O 2 + 2H + + 2e - H 2 O 2 E= 0.68 V b. Mn 2+ + 2e - MnE= -1.18 V Fe 3+ + 3e - FeE= -0.036 V 32. Give the standard line notation for each cell in Exercise 30.

30/32a) a. H 2 O 2 + 2H + + 2e - 2H 2 O E= 1.78 V O 2 + 2H + + 2e - H 2 O 2 E= 0.68 V V e-e- e-e-

30/32a) a. H 2 O 2 + 2H + + 2e - 2H 2 O E= 1.78 V O 2 + 2H + + 2e - H 2 O 2 E= 0.68 V H 2 O 2 + 2H + + 2e - 2H 2 O E= +1.78 V H 2 O 2 O 2 + 2H + + 2e - E= 0.68 V 2H 2 O 2 2H 2 O + O 2 E= +1.10 V AnodeCathode Pt | H 2 O 2, H + || H 2 O 2, H +, H 2 O | Pt V e-e- e-e- Pt H2O2H2O2 H+H+ H2O2H2O2 O2O2 H2OH2O H+H+

30/32b) b. Mn 2+ + 2e - MnE= -1.18 V Fe 3+ + 3e - FeE= -0.036 V Anode Cathode V e-e- e-e-

30/32b) b. Mn 2+ + 2e - MnE= -1.18 V Fe 3+ + 3e - FeE= -0.036 V 2Fe 3+ + 6e - 2FeE= -0.036 V 3Mn 3Mn 2+ + 6e - E= +1.18 V 2Fe 3+ + 3Mn 3Mn 2+ + 2FeE= +1.14 V Anode Cathode Fe | Fe +3 || Mn +2 | Mn V e-e- e-e- Fe Mn Mn +2 Fe +3

35. Calculate o values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the reactions that are not already balanced. Standard reduction potentials are found in Table 17.1. a. 2Ag + (aq) + Cu(s) Cu 2+ (aq) + 2Ag(s) b. Zn 2+ (aq) + Ni(s) Ni 2+ (aq) + Zn(s)

35. Calculate o values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the reactions that are not already balanced. Standard reduction potentials are found in Table 17.1. a. 2Ag + (aq) + Cu(s) Cu 2+ (aq) + 2Ag(s) 2Ag + (aq) + 2e - 2Ag(s) E= +0.80 V Cu(s) Cu 2+ (aq) + 2e - E= 0.16 V E= +0.64 V spontaneous b. Zn 2+ (aq) + Ni(s) Ni 2+ (aq) + Zn(s)

35. Calculate o values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the reactions that are not already balanced. Standard reduction potentials are found in Table 17.1. a. 2Ag + (aq) + Cu(s) Cu 2+ (aq) + 2Ag(s) 2Ag + (aq) + 2e - 2Ag(s) E= +0.80 V Cu(s) Cu 2+ (aq) + 2e - E= 0.16 V E= +0.64 V spontaneous b. Zn 2+ (aq) + Ni(s) Ni 2+ (aq) + Zn(s) Zn 2+ (aq) + 2e - Zn(s) E= 0.76 V Ni(s) Ni 2+ (aq) + 2e - E= +0.23 V E= 0.53 V not spontaneous

Ch17.2 Electric Work and Free Energy EMF In your text, work is defined as flowing out of a system.

Equations you might remember: G = H TS E = q + w enthalpyentropy internal energy heat workpotential (voltage) charge

Equations you might remember: G = H TS E = q + w G = H TS w = E q G = w G = -q max enthalpy entropy internal energy heat workpotential (voltage) charge These equations are ideally related

Equations you might remember: G = H TS E = q + w G = H TS w = E q G = w G = -q max = -nF max enthalpy entropy internal energy heat workpotential (voltage) charge These equations are ideally related moles Farady: the charge of one mole of electrons 96,485 Coulombs / 1 mole If you feel so inclined u may wanna look at ex 17.3, pg850 to c this in action.

Dependence of Cell Potential on Concentration Ex1) For the cell rxn: 2Al (s) + 3Mn +2 (aq) 2Al +3 (aq) + 3Mn (s) E o cell = 0.48V predict whether E cell is larger or smaller than E o cell if: a) [Al +3 ] = 2.0M, [Mn +2 ] = 1.0M b) [Al +3 ] = 1.0M, [Mn +2 ] = 3.0M

Concentration Cells - nature will try to equalize the concentration in the 2 cells, but only a small voltage will be produced. Ex2) Determine the direction of electron flow and designate the anode and cathode.

Ch17 HW#3 p881 47,49,51,53,55 (Do 47 in class) 47. Using data from Table 17.1, place the following in order of increasing strength as oxidizing agents. (all under standard conditions) Cd 2+, IO 3 -, K +, H 2 O, AuCl 4 -, I 2 Reminder: ox agents get reduced in charge (They add electrons).

Ch17 HW#3 p881 47,49,51,53,55 (Do 47 in class) 47. Using data from Table 17.1, place the following in order of increasing strength as oxidizing agents. (all under standard conditions) Cd 2+, IO 3 -, K +, H 2 O, AuCl 4 -, I 2 Reminder: ox agents get reduced in charge (They add electrons). Cd 2+ + 2e - CdE= 0.40 V IO 3 - + 6H + + 5e - I 2 + 3H 2 O E= +1.20 V strongest K + + e - K E= 2.92 V weakest 2H 2 O + 2e - 2H 2 + OH - E= 0.83 V AuCl 4 - + 3e - Au + 4Cl - E= +0.99 V I 2 + 2e - 2I - E= +0.54 V

49. Answer the following questions using the data in table 17.1 (all under standard conditions) The higher potential will reduce. a. Is H + (aq) capable of oxidizing Cu(s) to Cu 2+ ? The lower will oxidize. b. Is H + (aq) capable of oxidizing Mg(s) ? c. Is Fe 3+ (aq) capable of oxidizing I - ? d. Is Fe 3+ (aq) capable of oxidizing Br - ?

49. Answer the following questions using the data in table 17.1 (all under standard conditions) The higher potential will reduce. a. Is H + (aq) capable of oxidizing Cu(s) to Cu 2+ ? The lower will oxidize. No b. Is H + (aq) capable of oxidizing Mg(s) ? Yes c. Is Fe 3+ (aq) capable of oxidizing I - ? Yes d. Is Fe 3+ (aq) capable of oxidizing Br - ? No

Na + : Cl - : Ag + : Zn 2+ : Zn : Pb : 51. Consider only the species (at standard conditions) Na +, Cl -, Ag +, Zn 2+, Zn, Pb In answering the following questions, give reasons for you answers (use data from table 17.1) a. Which is the strongest oxidizing agent? Itself gets reduced: b. Which is the strongest reducing agent? Itself gets oxidized: c. Which species can be oxidized by SO 4 2- (aq) in acid? d. Which species can be reduced by Al(s)? Al itself gets oxidized (flipped), everything above it will reduce.

Na + :2.70 Cl - :1.36 Ag + :+0.80 Zn 2+ :0.76 Zn :+0.76 Pb :+0.13 51. Consider only the species (at standard conditions) Na +, Cl -, Ag +, Zn 2+, Zn, Pb In answering the following questions, give reasons for you answers (use data from table 17.1) a. Which is the strongest oxidizing agent? Itself gets reduced: Ag + b. Which is the strongest reducing agent? Itself gets oxidized: Zn c. Which species can be oxidized by SO 4 2- (aq) in acid? Zn and Pb their reduction potentials are less than SO 4 2- d. Which species can be reduced by Al(s)? Al itself gets oxidized (flipped), everything above it will reduce. Ag +, Zn 2+ are willing to reduce, not Na +

53. Use the table of standard reduction potential (table 17.1) to pick a reagent that is capable of each of the following oxidations (under standard conditions in acidic solutions). a. Oxidize Br to Br 2 but not oxidize Cl - to Cl 2

53. Use the table of standard reduction potential (table 17.1) to pick a reagent that is capable of each of the following oxidations (under standard conditions in acidic solutions). b. Oxidize Mn to Mn 2+ but not oxidize Ni to Ni 2+

55. A galvanic cell is based on the following half-reactions at 25C Ag + + e - AgE o = +0.80 V H 2 O 2 + 2H + + 2e - 2H 2 OE o = +1.78 V Predict whether G cell is larger or smaller than G o cell for the following cases H 2 O 2 + 2H + + 2e - 2H 2 OE o = +1.78 V Ag Ag + + e - E o = 0.80 V H 2 O 2 + 2H + + 2Ag 2Ag + + 2H 2 O E o = +0.98 V a. [Ag + ] = 1.0 M, [H 2 O 2 ] = 2.0 M, [H + ] = 2.0 M b. [Ag + ] = 2.0 M, [H 2 O 2 ] = 1.0 M, [H + ] = 1 x 10 -7 M

55. A galvanic cell is based on the following half-reactions at 25C Ag + + e - AgE o = +0.80 V H 2 O 2 + 2H + + 2e - 2H 2 OE o = +1.78 V Predict whether G cell is larger or smaller than G o cell for the following cases H 2 O 2 + 2H + + 2e - 2H 2 OE o = +1.78 V Ag Ag + + e - E o = 0.80 V H 2 O 2 + 2H + + 2Ag 2Ag + + 2H 2 O E o = +0.98 V a. [Ag + ] = 1.0 M, [H 2 O 2 ] = 2.0 M, [H + ] = 2.0 M [H 2 O 2 ] and [H + ] increased favors products, G cell is larger b. [Ag + ] = 2.0 M, [H 2 O 2 ] = 1.0 M, [H + ] = 1 x 10 -7 M [Ag + ] increased, favors reactants AND [H 2 O 2 ] and [H + ] decreased favors reactants, G cell is much smaller

Ch17.3 Batteries A connection of galvanic cells in series, such that the potentials of the individual cells add together to give a total potential. Lead storage battery Anode reaction: Pb + HSO 4 - PbSO 4 + H + + 2e - Cathode rxn: PbO 2 + HSO 4 - + 3H + + 2e - PbSO 4 + 2H 2 O Cell reaction:

Lead storage battery cell reaction: Pb (s) + PbO 2(s) + 2HSO 4 - (aq) + 2H + (aq) 2PbSO 4(s) + 2H 2 O (l)

Dry cell battery (alkaline version): Anode reaction: Zn + 2OH - ZnO + H 2 O + 2e - Cathode rxn: 2MnO 2 + H 2 O + 2e - Mn 2 O 3 + 2OH - Cell reaction:

Corrosion - the oxidation of metals.

Oxidation of Iron Fe Fe +2 + 2e - O 2 + 2H 2 O + 4e - 4OH - Prevention Galvanization: Fe Fe +2 + 2e - EMF = 0.44V Zn Zn +2 + 2e - EMF = 0.76V Ch17 HW#4 p883 48,50,52,73a

48. Using the data from table 17.1, place the following in order of increasing strength as reducing agents (all under standard conditions). A reducing agent itself will get oxidized Cr 3+, H 2, Zn, Li, F -, Fe 2+

Ch17 HW#4 p883 48,50,52,73a 48. Using the data from table 17.1, place the following in order of increasing strength as reducing agents (all under standard conditions). A reducing agent itself will get oxidized Cr 3+, H 2, Zn, Li, F -, Fe 2+ 5. 2Cr 3+ + 7H 2 O Cr 2 O 7 2- + 14H + + 6e - 1.33 3. H 2 2H + + 2e - 0.00 2. Zn Zn +2 + 2e - +0.76 1. Li Li + + e - +3.05 6. 2F - F 2 + 2e - 2.87 4. Fe 2+ Fe 3+ + e - 0.77

50. Answer the following questions using data from table 17.1 (all under standard conditions) a. Is H 2 (g) capable of reducing Ag + (aq)? H 2 2H + + 2e - 0.00 b. Is H + (g) capable of reducing Ni 2+ (aq)? H 2 2H + + 2e - 0.00 c. Is Fe 2+ (aq) capable of reducing VO 2 + (aq)? Fe 2+ Fe 3+ + e - 0.77 d. Is Fe 2+ (aq) capable of reducing Cr 3+ (aq) to Cr 2 +(aq)? Fe 2+ Fe 3+ + e - 0.77

50. Answer the following questions using data from table 17.1 (all under standard conditions) a. Is H 2 (g) capable of reducing Ag + (aq)? H 2 2H + + 2e - 0.00 No. b. Is H + (g) capable of reducing Ni 2+ (aq)? H 2 2H + + 2e - 0.00 Yes. c. Is Fe 2+ (aq) capable of reducing VO 2 + (aq)? Fe 2+ Fe 3+ + e - 0.77 No. d. Is Fe 2+ (aq) capable of reducing Cr 3+ (aq) to Cr 2 +(aq)? Fe 2+ Fe 3+ + e - 0.77 No.

52. Consider only the species (at standard condition) Ce 4+, Ce 3+, Fe 2+, Fe 3+, Fe, Mg 2+, Mg, Ni 2+, Sn in answering the following questions, give reasons for your answers (use data from table 17.1) a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Will iron dissolve in a 1.0 M solution of Ce 4+ ? d. Which of the species can be oxidized by H + (aq)? e. Which of the species can be reduced by H 2 (g)?

52. Consider only the species (at standard condition) Ce 4+, Ce 3+, Fe 2+, Fe 3+, Fe, Mg 2+, Mg, Ni 2+, Sn In answering the following questions, give reasons for your answers (use data from table 17.1) a. Which is the strongest oxidizing agent? Itself gets reduced, look for largest V b. Which is the strongest reducing agent? Itself gets oxidized, look for largest V when rxn flips c. Will iron dissolve in a 1.0 M solution of Ce 4+ ? d. Which of the species can be oxidized by H + (aq)? Any rxn when flipped over, that has a negative voltage (E < 0.00), will be oxidized: e. Which of the species can be reduced by H 2 (g)? Any rxn with a positive voltage (E > 0.00) in the forward direction will be reduced:

52. Consider only the species (at standard condition) Ce 4+, Ce 3+, Fe 2+, Fe 3+, Fe, Mg 2+, Mg, Ni 2+, Sn In answering the following questions, give reasons for your answers (use data from table 17.1) a. Which is the strongest oxidizing agent? Itself gets reduced, look for largest V Ce 4+ (+1.70V) b. Which is the strongest reducing agent? Itself gets oxidized, look for largest V when rxn flips Mg (+2.37V) c. Will iron dissolve in a 1.0 M solution of Ce 4+ ? Yes. Ce 4+ wants to reduce (+1.70V), Fe will oxidize (+0.44V) d. Which of the species can be oxidized by H + (aq)? Any rxn when flipped over, that has a negative voltage (E < 0.00), will be oxidized: Ce 3+ (-1.70V), Fe 2+ (-0.77) e. Which of the species can be reduced by H 2 (g)? Any rxn with a positive voltage (E > 0.00) in the forward direction will be reduced: Ce 4+ (+1.70V), Fe 3+ (+0.77V)

73. Consider a galvanic cell based on the following half reactions Zn 2+ + 2e - Zn E = -0.76 V Fe 2+ + 2e - Fe E = -0.44 V a. Determine the overall cell reaction and calculate E cell. Zn Zn 2+ + 2e - E = +0.76 V Fe 2+ + 2e - Fe E = -0.44 V

73. Consider a galvanic cell based on the following half reactions Zn 2+ + 2e - Zn E = -0.76 V Fe 2+ + 2e - Fe E = -0.44 V a. Determine the overall cell reaction and calculate E cell. Zn Zn 2+ + 2e - E = +0.76 V Fe 2+ + 2e - Fe E = -0.44 V Zn (aq) + Fe 2+ (aq) Fe (aq) + Zn 2+ (aq) E = +0.32 V

Ch17.4 Electrolysis Electrolytic Cell uses electrical energy to produce a chemical change. Electrolysis forces a current thru a cell to produce chemical change for which the cell potential is negative. (Use what we know to develop a new twist): Ex of galvanic cell: Anode: Zn Zn +2 + 2e - +0.76 V Cathode: Cu +2 + 2e - Cu +0.34 V +1.10 V

Ex of galvanic cell: Anode: Zn Zn +2 + 2e - +0.76 V Cathode: Cu +2 + 2e - Cu +0.34 V +1.10 V charge current time Units: 1 mol of electrons = 1 farady of charge = 96,485 coulombs

Ex1) How long must a current of 5.00A be applied to a soln of Ag + to produce 10.5g of silver?

HW#79a) How long will it take to plate out each of the following with a current of 100.0A? a. 1.0kg Al from aqueous Al 3+

Electrolysis of water 2H 2 O O 2 + 4H + + 4e - EMF = 1.23V 4H 2 O + 4e - 2H 2 + 4OH - EMF = 0.83V

Electroplating metals In an electrolytic cell, a soln contains the following metal ions: Ag +, Cu +2, Zn +2. The voltage is increased gradually. In which order will the metals be plated onto the cathode?

Electroplating metals In an electrolytic cell, a soln contains the following metal ions: Ag +, Cu +2, Zn +2. The voltage is increased gradually. In which order will the metals be plated onto the cathode? The higher the (+), the greater the tendency to occur. Ag + + e - Ag E = +0.80V Cu +2 + 2e - Cu E = +0.34V Zn +2 + 2e - Zn E = 0.76V The order of oxidizing ability (the order they are reduced): Ag + > Cu +2 > Zn +2

Ex2) An acidic soln contains the ions: Ce +4, VO 2 +, Fe +3. Using EMFs from Table 17.1, give the order of oxidizing ability. (The order they arte reduced.)

Ex2) An acidic soln contains the ions: Ce +4, VO 2 +, Fe +3. Using EMFs from Table 17.1, give the order of oxidizing ability. (The order they arte reduced.) Ce +4 + e - Ce +3 E = +1.70V VO 2 + + 2H + + e - VO 2 +2 + H 2 O E = +1.00V Fe +3 + e - Fe +2 E = +0.77V Oxidizing ability: Ce +4 > VO 2 + > Fe +3 Ce +4 is reduced at the lowest voltage. Ch17 HW#5 p883 79,91,95 + Ch17 Rev

79. How long will it take to plate out each of the following with a current of 100.0A? a. 1.0kg Al from aqueous Al 3+ (In class?) b. 1.0g Ni from aqueous Ni 2+ c. 5.0 mol Ag from aqueous Ag +

91. A solution at 25C contains 1.0 M Cd 2+, 1.0 M Ag +, 1.0 M Au 3+, and 1.0 M Ni 2+ in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

95. In the electrolysis of an aqueous solution of Na 2 SO 4, what reactions occur at the anode and the cathode? (Assume standard conditions.) S 2 O 8 2- + 2e - 2SO 4 2- E = 2.01 V O 2 + 4H + = 4e - 2H 2 O E = 1.23 V 2H 2 O + 2e - H 2 + 2OH - E = -0.83 V Na + + e - Na E = -2.71 V

Ch17 Rev p880+ 34a,36a,54,59 + Bonus FRQ!!! 34. a. Give the balanced cell reaction and determine the E for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in table 17.1. a. Cr 2 O 7 2- + 14H + + 6e - 2Cr 3+ + 7H 2 O H 2 O 2 + 2H + + 2e - 2H 2 O

36. a. Calculate E values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the reactions. Standard reduction potentials are found in Table 17.1. a. MnO 4 - (aq) + I - (aq) I 2 (aq) + Mn 2+ (aq)

54. Use the table of standard reduction potentials (Table 17.1) to pick a reagent that is capable of each of the following reductions (under standard conditions in acidic solution). a. Reduce Cu 2+ to Cu but not reduce Cu 2+ to Cu + b. Reduce Br 2 to Br - but not to reduce I 2 to I -

59. Consider the concentration cell shown below. Calculate the cell potential at 25C when the concentration of Ag + in the compartment on the right is the following. a. 1.0 Mb. 2.0 Mc. 0.10 Md. 4.0 x 10 -5 M e. Calculate the potential when both solutions are 0.10 M in Ag +. For each case, also identify the cathode, the anode, and the direction in which electrons flow. V Ag [Ag + ] = 1.0M

AP Chemistry - Ch17 FRQ Review 1. An external direct-current power supply is connected to two platinum electrodes immersed in a beaker containing 1.0 M CuSO (ag) at 25C, as shown in the diagram. As the cell operates, copper metal is deposited onto one electrode and O (g) is produced as the other electrode. The two reduction half-reactions for the overall reaction that occurs in the cell are show in the table below. (a) On the diagram, indicate the direction of electron flow in the wire. (b) Write a balanced net ionic equation for the electrolysis reaction that occurs in the cell. (c) Predict the algebraic sign of G for the reaction. Justify your answer. An electric current of 1.50 Amps passes through the cell for 40.0 minutes. (d) Calculate the mass, in grams, of the Cu(s) that is deposited on the electrode. (e) Calculate the dry volume, in liters measured at 25C and 1.16 atm, of the O (g) that is produced. V Cu Half-reactionE (V) O (g) + 4 H (ag) + 4 e 2 H O(l) +1.23 Cu (ag) + 2 e Cu(s) +0.34

ch17.1 – galvanic cells ch4: redox involves the transfer of electrons (oil rig) zn (s) + cuso...

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