1. Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Ch17.1 – Galvanic Cells
Ch4: Redox involves the transfer of electrons (OIL RIG)
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Half rxns:
Ox: Red agent: _____
Red: Ox agent: _____

2. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
lose 2e- (oxidized)
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
gained 2e- (reduced)
Ox ½ Reaction: Zn(s)  Zn(aq)+2 +2e Red agent: Zn
Red ½ Reaction: Cu+2(aq) + 2e-  Cu(s) Ox agent: Cu+2

3. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
lose 2e- (oxidized)
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
gained 2e- (reduced)
Oxidized ½ Reaction: Zn(s)  Zn(aq)+2 +2e Red agent: Zn
Reduced ½ Reaction: Cu+2(aq) + 2e-  Cu(s) Ox agent: Cu+2
e- e e- e-
Cu Zn
CuSO4 soln ZnSO4 soln V

4. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
lose 2e- (oxidized)
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
gained 2e- (reduced)
Oxidized ½ Reaction: Zn(s)  Zn(aq)+2 +2e Red agent: Zn
Reduced ½ Reaction: Cu+2(aq) + 2e-  Cu(s) Ox agent: Cu+2
e- e e- e-
Cu Zn
CuSO4 soln ZnSO4 soln V

5. into electrical energy Electrodes - metals in voltaic cells
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-
Cu Na+ (necessary to maintain Na Zn
Cl- an ion charge balance) Cl-
Cu Na Na Zn
Cl Cl-
Cu Zn
CuSO4 soln ZnSO4 soln V

6. Cu+2 + 2e– → Cu Zn → Zn+2 + 2e– Zn → Zn+2 + 2e- E˚ = + .76V
electrons transfer
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na+ thru the wire
Cl (salt bridge) Cl-
Cu Na+ (necessary to maintain Na Zn
Cl- an ion charge balance) Cl-
Cu Na Na Zn
Cl Cl-
Cu Zn
Cu+2 SO Zn+2 SO4-2
SO Cu SO Zn+2
CuSO4 soln ZnSO4 soln
cathode cell anode cell V
Cell Potential,εcell
Reduction Half Cell Oxidation Half Cell
Cu+2 + 2e– → Cu Zn → Zn+2 + 2e–
Zn → Zn+2 + 2e- E˚ = V
Cu+2 + 2e- → Cu E˚ = V
Ecell=+1.10 V
E˚ → standard conditions
25˚C, I Molar

7. Cell Potentials

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

9. Al+3 + 3e-  Al Eo = –1.66 V Mg+2 + 2e-  Mg Eo = –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-  Al Eo = –1.66 V
Mg+2 + 2e-  Mg Eo = –2.37 V

10. 2Al+3(aq) + 3Mg(s)  2Al(s) + 3Mg+2(aq) Eo = +0.71 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-  Al Eo = –1.66 V
Mg+2 + 2e-  Mg Eo = –2.37 V
2Al+3 + 6e-  2Al Eo = –1.66 V
3Mg  3Mg+2 + 6e- Eo = V
2Al+3(aq) + 3Mg(s)  2Al(s) + 3Mg+2(aq) Eo = V
Mg metal Al+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 e’s head to is the cathode
Anode cell Cathode cell
Don’t multiply! V e- e- Mg Al
Al+3
Mg+2

11. ClO4- + 2e- + 2H+  ClO3- + H2O Eo = +1.19 V
Ex2) MnO4-(aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn+2(aq) + H2O(l)
Give the balanced cell rxn and EMF for the cell.
MnO4- + 5e- + 8H+  Mn+2 + 4H2O Eo = V
ClO4- + 2e- + 2H+  ClO3- + H2O Eo = V .

12. ClO4- + 2e- + 2H+  ClO3- + H2O Eo = +1.19 V
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
Ex2) MnO4-(aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn+2(aq) + H2O(l)
MnO4- + 5e- + 8H+  Mn+2 + 4H2O Eo = V
ClO4- + 2e- + 2H+  ClO3- + H2O Eo = V
2MnO e- + 16H+  2Mn+2 + 8H2O Eo = V
5ClO H2O  5ClO e- + 10H+ Eo = –1.19 V
2MnO4-(aq) + 6H+(aq) + 5ClO3-(aq)  5ClO4-(aq) + 2Mn+2(aq) + 3H2O(l)
Eo = V
| || |
Anode cell Cathode cell
Cl+5 loses 2e- to become Cl+7: oxidized → anode cell V e- e- ? ? ? ? .

13. ClO4- + 2e- + 2H+  ClO3- + H2O Eo = +1.19 V
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
Ex2) MnO4-(aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn+2(aq) + H2O(l)
MnO4- + 5e- + 8H+  Mn+2 + 4H2O Eo = V
ClO4- + 2e- + 2H+  ClO3- + H2O Eo = V
2MnO e- + 16H+  2Mn+2 + 8H2O Eo = V
5ClO H2O  5ClO e- + 10H+ Eo = –1.19 V
2MnO4-(aq) + 6H+(aq) + 5ClO3-(aq)  5ClO4-(aq) + 2Mn+2(aq) + 3H2O(l)
Eo = V
Since REDOX occurs w
aqueous ions, we need
a non-reactive metal as
electrodes, so Pt makes
a great choice.
H2 gas bubbles off
Pt | ClO3- , ClO4- , H+ || Mn+2, MnO4- , H+ | Pt of cathode
Anode Cathode
Cl+5 loses 2e- to become Cl+7: oxidized → anode cell V e- e- Pt
ClO3-
ClO4- H+ Pt
Mn+2
MnO4- H+

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

15. Fe+3 + e-  Fe+2 Eo = +0.77V Ag+ + e-  Ag RIG Eo = +0.80V
Ex3) Ag+ + e-  Ag Eo = +0.80V
Fe+3 + e-  Fe+2 Eo = +0.77V
Setup a galvanic cell, give the balanced cell rxn and EMF for the cell.
Ag+ + e-  Ag RIG Eo = +0.80V
Fe+2  Fe+3 + e- OIL Eo = -0.77V
Ag+ + Fe+2  Ag + Fe+3 Eo = +0.03V
Since there’s 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- Pt Ag
Fe+3
Ag+
Fe+2

16. MnO4- + 5e-  Mn+2 + 4H2O Eo = +1.51V Ex4) Fe+2 + 2e-  Fe Eo = –0.44V
Setup a galvanic cell, give the balanced cell rxn and EMF for the cell.
Ch17 HW#1 p880 29,31,33 V e- e- ? ? ? ?

17. b. MnO4 + 8H+ + 5e- → Mn2 + 4H2O E˚= 1.51 V
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. Cl2 + 2e- → 2Cl- E˚= 1.36 V
Br2 + 2e- → 2Br- E˚= 1.09 V
b. MnO4 + 8H+ + 5e- → Mn2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O E˚= 1.60

18. a. Cl2 + 2e- → 2Cl- E˚= 1.36 V Br2 + 2e- → 2Br- E˚= 1.09 V
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. Cl2 + 2e- → 2Cl- E˚= 1.36 V
Br2 + 2e- → 2Br- E˚= 1.09 V
Cl2 + 2e- → 2Cl E˚= 1.36 V
2Br- → Br2 + 2e- E˚= –1.09 V
Cl2 + 2Br- → Br2 + 2Cl- E˚= V V e- e- Pt Pt
Br2
Cl-
Br-

19. b. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
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. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O E˚= 1.60 V

20. b. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O E˚= 1.60 V
5(IO4- + 2H+ +2e- → IO3- + H2O) E˚= V
2(Mn+2 + 4H2O → MnO4- + 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

21. b. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O E˚= 1.60 V
5IO H+ + 10e- → 5IO3- + 5H2O E˚= V)
2 Mn+2 + 8H2O → 2MnO H+ + 10e E˚= –1.51 V)
5IO Mn+2 + 3H2O → 5IO3- + 2MnO4- + 6H+ E˚= V)
H2 gas bubbles out
Anode Cathode
Pt | IO4-(aq), IO3-(aq) ,H+ || MnO4-(aq), Mn+2(aq) ,H+ | Pt
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
I+7 loses 2e- to become I+5: oxidized → anode cell V e- e- Pt Pt
IO4-
MnO4- H+ H+
IO3-
Mn+2

22. Br2 + 2e- → Br- E˚= 1.09 V Cl2 + 2e- → 2Cl- E˚= 1.36 V
31. Give the standard notation for each cell in Exercise 29.
a. Cl2 + 2e- → 2Cl- E˚= 1.36 V
Br2 + 2e- → Br- E˚= 1.09 V
Cl2 + 2e- → 2Cl E˚= 1.36 V
2Br- → Br2 + 2e- E˚= –1.09 V
Cl2 + 2Br- → Br2 + 2Cl- E˚= V
Pt | Br- , Br2 || Cl2 | Cl- | Pt
Cl2 V e- e- Pt Pt
Br2
Cl-
Br-

23. b. MnO4 + 8H+ + 5e- → Mn2 + 4H2O E˚= 1.51 V
31. Give the standard notation for each cell in Exercise 29.
a. Cl2 + 2e- → 2Cl- E˚= 1.36 V
Br2 + 2e- → Br- E˚= 1.09 V
Pt | Br- , Br2 || Cl2 | Cl- | Pt
b. MnO4 + 8H+ + 5e- → Mn2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O E˚= 1.60 V
Pt | IO4-(aq), IO3-(aq) ,H+ || MnO4-(aq), Mn+2(aq) ,H+ | Pt

24. 2VO2+ + 4H+ + Cd → Cd2+ + 2VO2+ + 2H2O E0 = +1.40
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. Cu2+ + e-→ Cu+ E0 = +0.16
Au3+ + 3e- → Au E0 = +1.50
Au3+ + 3e- → Au E0 = +1.50
3Cu+ → 3Cu2+ + 3e- E0 = –0.16
Au3+ + 3Cu+ → 3Cu2+ + Au E0 = +1.34
b Cd2+ + 2e- → Cd E0 = –0.40
VO2+ + 2H+ + e- → VO2+ +H2O E0 = +1.00
2VO2+ + 4H+ + 2e- → 2VO2+ + 2H2O E0 = +1.00
Cd → Cd2+ + 2e- E0 = +0.40
2VO2+ + 4H+ + Cd → Cd2+ + 2VO2+ + 2H2O E0 = +1.40

25. a. H2O2 + 2H+ + 2e- → 2H2O E˚= 1.78 V O2 + 2H+ + 2e- → H2O2 E˚= 0.68 V
Ch17.1 cont Ch17 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. H2O2 + 2H+ + 2e- → 2H2O E˚= 1.78 V
O2 + 2H+ + 2e- → H2O2 E˚= 0.68 V
b. Mn2+ + 2e- → Mn E˚= V
Fe3+ + 3e- → Fe E˚= V
32. Give the standard line notation for each cell in Exercise 30.

26. a. H2O2 + 2H+ + 2e- → 2H2O E˚= 1.78 V O2 + 2H+ + 2e- → H2O2 E˚= 0.68 V

27. a. H2O2 + 2H+ + 2e- → 2H2O E˚= 1.78 V O2 + 2H+ + 2e- → H2O2 E˚= 0.68 V
H2O2 → O2 + 2H+ + 2e- E˚= –0.68 V
2H2O2 → 2H2O + O2 E˚= V
Anode Cathode
Pt | H2O2 , H+ || H2O2 , H+ , H2O | Pt O2 V e- e- Pt
H2O Pt
H2O2
H2O2 H+ H+

28. b. Mn2+ + 2e- → Mn E˚= -1.18 V Fe3+ + 3e- → Fe E˚= -0.036 V
Anode Cathode V e- e-

29. b. Mn2+ + 2e- → Mn E˚= -1.18 V Fe3+ + 3e- → Fe E˚= -0.036 V
3Mn → 3Mn2+ + 6e- E˚= V
2Fe3+ + 3Mn → 3Mn2+ + 2Fe E˚= V
Anode Cathode
Fe | Fe+3 || Mn+2 | Mn V e- e- Fe Mn
Mn+2
Fe+3

30. a. 2Ag+(aq) + Cu(s) ↔ Cu2+(aq) + 2Ag(s)
35. Calculate εo 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) ↔ Cu2+(aq) + 2Ag(s)
b. Zn2+ (aq) + Ni(s) ↔ Ni2+ (aq) + Zn(s)

31. a. 2Ag+(aq) + Cu(s) ↔ Cu2+(aq) + 2Ag(s)
35. Calculate εo 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) ↔ Cu2+(aq) + 2Ag(s)
2Ag+(aq) + 2e- ↔ 2Ag(s) E˚= V
Cu(s) ↔ Cu2+(aq) + 2e- E˚= –0.16 V
E˚= V
spontaneous
b. Zn2+ (aq) + Ni(s) ↔ Ni2+ (aq) + Zn(s)

32. a. 2Ag+(aq) + Cu(s) ↔ Cu2+(aq) + 2Ag(s)
35. Calculate εo 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) ↔ Cu2+(aq) + 2Ag(s)
2Ag+(aq) + 2e- ↔ 2Ag(s) E˚= V
Cu(s) ↔ Cu2+(aq) + 2e- E˚= –0.16 V
E˚= V
spontaneous
b. Zn2+ (aq) + Ni(s) ↔ Ni2+ (aq) + Zn(s)
Zn2+ (aq) + 2e- ↔ Zn(s) E˚= –0.76 V
Ni(s) ↔ Ni2+ (aq) + 2e- E˚= V
E˚= –0.53 V
not spontaneous

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

34. Equations you might remember: ΔG = ΔH – TΔS ΔE = q + w
internal
energy
work
potential
(voltage)
heat
charge
enthalpy
entropy

35. Equations you might remember: ΔG = ΔH – TΔS ΔE = q + w
w = ΔE – q
ΔG = w
ΔG = -qεmax
internal
energy
work
potential
(voltage)
heat
charge
enthalpy
entropy
These equations are ideally related

36. Equations you might remember: ΔG = ΔH – TΔS ΔE = q + w
w = ΔE – q
ΔG = w
ΔG = -qεmax = -nFεmax
internal
energy
work
potential
(voltage)
heat
charge
enthalpy
entropy
These equations are ideally related
If you feel so inclined
u may wanna look at ex 17.3,
pg850 to c this in action.
Farady: the charge of
one mole of electrons
96,485 Coulombs / 1 mole
moles

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

38. 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.

39. Cd2+, IO3-, K+, H2O, AuCl4-, I2 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)
Cd2+, IO3-, K+, H2O, AuCl4-, I2
Reminder: ox agents get reduced in charge (They add electrons).

40. IO3- + 6H+ + 5e- → ½I2 + 3H2O E˚= +1.20 V strongest
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)
Cd2+, IO3-, K+, H2O, AuCl4-, I2
Reminder: ox agents get reduced in charge (They add electrons).
Cd2+ + 2e- → Cd E˚= –0.40 V
IO3- + 6H+ + 5e- → ½I2 + 3H2O E˚= V strongest
K+ + e- → K E˚= –2.92 V weakest
2H2O + 2e- → 2H2 + OH- E˚= –0.83 V
AuCl4- + 3e- → Au + 4Cl- E˚= V
I2 + 2e- → 2I- E˚= V

41. 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 Cu2+ ? The lower will oxidize.
b. Is H+ (aq) capable of oxidizing Mg(s) ?
c. Is Fe3+ (aq) capable of oxidizing I-?
d. Is Fe3+ (aq) capable of oxidizing Br -?

42. 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 Cu2+ ? The lower will oxidize. No
b. Is H+ (aq) capable of oxidizing Mg(s) ?
Yes
c. Is Fe3+ (aq) capable of oxidizing I-?
d. Is Fe3+ (aq) capable of oxidizing Br -?

43. Na+: Cl- : Ag+: Zn2+: Zn : Pb :
51. Consider only the species (at standard conditions)
Na+, Cl-, Ag+, Zn2+, 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 SO42- (aq)
in acid?
d. Which species can be reduced by Al(s)?
Al itself gets oxidized (flipped), everything above it will reduce.
Na+:
Cl- :
Ag+:
Zn2+:
Zn :
Pb :

44. Na+:–2.70 Cl- :–1.36 Ag+:+0.80 Zn2+:–0.76 Zn :+0.76 Pb :+0.13
51. Consider only the species (at standard conditions)
Na+, Cl-, Ag+, Zn2+, 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 SO42- (aq)
in acid?
Zn and Pb
their reduction potentials are less than SO42-
d. Which species can be reduced by Al(s)?
Al itself gets oxidized (flipped), everything above it will reduce.
Ag+, Zn2+ are willing to reduce, not Na+
Na+:–2.70
Cl- :–1.36
Ag+:+0.80
Zn2+:–0.76
Zn :+0.76
Pb :+0.13

45. 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 Br2 but not oxidize Cl- to Cl2

46. 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 Mn2+ but not oxidize Ni to Ni2+

47. H2O2 + 2H+ + 2Ag → 2Ag+ + 2H2O Eo = +0.98 V
55. A galvanic cell is based on the following half-reactions at 25˚C
Ag+ + e- → Ag Eo = V
H2O2 + 2H+ + 2e- → 2H2O Eo = V
Predict whether Gcell is larger or smaller than Gocell for the following cases
Ag → Ag+ + e- Eo = –0.80 V
H2O2 + 2H+ + 2Ag → 2Ag+ + 2H2O Eo = V
a. [Ag+] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M
b. [Ag+] = 2.0 M, [H2O2] = 1.0 M, [H+ ] = 1 x M

48. H2O2 + 2H+ + 2Ag → 2Ag+ + 2H2O Eo = +0.98 V
55. A galvanic cell is based on the following half-reactions at 25˚C
Ag+ + e- → Ag Eo = V
H2O2 + 2H+ + 2e- → 2H2O Eo = V
Predict whether Gcell is larger or smaller than Gocell for the following cases
Ag → Ag+ + e- Eo = –0.80 V
H2O2 + 2H+ + 2Ag → 2Ag+ + 2H2O Eo = V
a. [Ag+] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M
[H2O2] and [H+] increased favors products, Gcell is larger
b. [Ag+] = 2.0 M, [H2O2] = 1.0 M, [H+ ] = 1 x M
[Ag+] increased, favors reactants AND
[H2O2] and [H+] decreased favors reactants,
Gcell is much smaller

49. 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 + HSO4-  PbSO4 + H+ + 2e-
Cathode rxn: PbO2 + HSO4- + 3H+ + 2e-  PbSO4 + 2H2O
Cell reaction:

50. Lead storage battery cell reaction:
Pb(s) + PbO2(s) + 2HSO4-(aq) + 2H+(aq)  2PbSO4(s) + 2H2O(l)

51. Dry cell battery (alkaline version):
Anode reaction: Zn + 2OH-  ZnO + H2O + 2e-
Cathode rxn: 2MnO2 + H2O + 2e-  Mn2O3 + 2OH-
Cell reaction:

52. Corrosion - the oxidation of metals.

53. O2 + 2H2O + 4e-  4OH- Oxidation of Iron Fe  Fe+2 + 2e- Prevention
Galvanization: Fe  Fe+2 + 2e- –EMF = 0.44V
Zn  Zn+2 + 2e- –EMF = 0.76V
Ch17 HW#4 p883 48,50,52,73a

54. 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
Cr3+, H2, Zn, Li, F-, Fe2+

55. 3. H2 → 2H+ + 2e- 0.00 2. Zn → Zn+2 + 2e- +0.76 1. Li → Li+ + e- +3.05
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
Cr3+, H2, Zn, Li, F-, Fe2+
5. 2Cr3+ + 7H2O → Cr2O H+ + 6e –1.33
3. H2 → 2H+ + 2e
2. Zn → Zn+2 + 2e
1. Li → Li+ + e
6. 2F- → F2 + 2e –2.87
4. Fe2+ → Fe3+ + e- –0.77

56. 50. Answer the following questions using data from table 17.1
(all under standard conditions)
a. Is H2(g) capable of reducing Ag+ (aq)?
H2 → 2H+ + 2e
b. Is H+(g) capable of reducing Ni2+ (aq)?
c. Is Fe2+(aq) capable of reducing VO2+ (aq)?
Fe2+ → Fe3+ + e- –0.77
d. Is Fe2+(aq) capable of reducing Cr3+ (aq) to Cr2+(aq)?

57. 50. Answer the following questions using data from table 17.1
(all under standard conditions)
a. Is H2(g) capable of reducing Ag+ (aq)?
H2 → 2H+ + 2e
No.
b. Is H+(g) capable of reducing Ni2+ (aq)?
Yes.
c. Is Fe2+(aq) capable of reducing VO2+ (aq)?
Fe2+ → Fe3+ + e- –0.77
d. Is Fe2+(aq) capable of reducing Cr3+ (aq) to Cr2+(aq)?

58. 52. Consider only the species (at standard condition)
Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, 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 Ce4+ ?
d. Which of the species can be oxidized by H+(aq)?
e. Which of the species can be reduced by H2(g)?

59. Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, Sn
52. Consider only the species (at standard condition)
Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, 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 Ce4+ ?
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 H2(g)?
Any rxn with a positive voltage (E > 0.00) in the forward direction
will be reduced:

60. Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, Sn
52. Consider only the species (at standard condition)
Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, 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 → Ce4+ (+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 Ce4+ ?
Yes. Ce4+ 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: Ce3+(-1.70V), Fe2+(-0.77)
e. Which of the species can be reduced by H2(g)?
Any rxn with a positive voltage (E > 0.00) in the forward direction
will be reduced: Ce4+ (+1.70V), Fe3+ (+0.77V)

61. Fe2+ + 2e- → Fe E˚ = -0.44 V Zn → Zn2+ + 2e- E˚ = +0.76 V
73. Consider a galvanic cell based on the following half reactions
Zn2+ + 2e- → Zn E˚ = V
Fe2+ + 2e- → Fe E˚ = V
a. Determine the overall cell reaction and calculate E˚cell.
Zn → Zn2+ + 2e E˚ = V
Fe2+ + 2e- → Fe E˚ = V

62. Zn(aq) + Fe2+(aq) → Fe(aq) + Zn2+(aq) E˚ = +0.32 V
73. Consider a galvanic cell based on the following half reactions
Zn2+ + 2e- → Zn E˚ = V
Fe2+ + 2e- → Fe E˚ = V
a. Determine the overall cell reaction and calculate E˚cell.
Zn → Zn2+ + 2e E˚ = V
Fe2+ + 2e- → Fe E˚ = V
Zn(aq) + Fe2+(aq) → Fe(aq) + Zn2+(aq) E˚ = V

63. 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 V
Cathode: Cu+2 + 2e-  Cu V
+1.10 V

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

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

66. 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 Al3+

67. 4H2O + 4e-  2H2 + 4OH- –EMF = –0.83V Electrolysis of water
2H2O  O2 + 4H+ + 4e- –EMF = –1.23V
4H2O + 4e-  2H2 + 4OH- –EMF = –0.83V

68. 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?

69. 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

70. Ex2) An acidic soln contains the ions: Ce+4, VO2+, Fe+3.
Using EMF’s from Table 17.1, give the order of oxidizing ability.
(The order they arte reduced.)

71. VO2+ + 2H+ + e-  VO2+2 + H2O E = +1.00V
Ex2) An acidic soln contains the ions: Ce+4, VO2+, Fe+3.
Using EMF’s from Table 17.1, give the order of oxidizing ability.
(The order they arte reduced.)
Ce+4 + e-  Ce E = +1.70V
VO2+ + 2H+ + e-  VO2+2 + H2O E = +1.00V
Fe+3 + e-  Fe+2 E = +0.77V
Oxidizing ability: Ce+4 > VO2+ > Fe+3
Ce+4 is reduced at the lowest voltage.
Ch17 HW#5 p883 79,91,95 + Ch17 Rev

72. 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 Al3+ (In class?)
b. 1.0g Ni from aqueous Ni2+
c. 5.0 mol Ag from aqueous Ag+

73. 91. A solution at 25°C contains 1.0 M Cd2+, 1.0 M Ag+, 1.0 M Au3+,
and 1.0 M Ni2+ in the cathode compartment of an electrolytic cell.
Predict the order in which the metals will plate out as the voltage
is gradually increased.

74. S2O82- + 2e- → 2SO42- E° = 2.01 V O2 + 4H+ = 4e- → 2H2O E° = 1.23 V
95. In the electrolysis of an aqueous solution of Na2SO4, what reactions
occur at the anode and the cathode? (Assume standard conditions.)
S2O e- → 2SO42- E° = 2.01 V
O2 + 4H+ = 4e- → 2H2O E° = 1.23 V
2H2O + 2e- → H2 + 2OH- E° = V
Na+ + e- → Na E° = V

75. H2O2 + 2H+ + 2e- → 2H2O 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. Cr2O H+ + 6e- → 2Cr3+ + 7H2O
H2O2 + 2H+ + 2e- → 2H2O

76. 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. MnO4-(aq) + I-(aq) ↔ I2(aq) + Mn2+(aq)

77. 54. Use the table of standard reduction potentials (Table 17
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 Cu2+ to Cu but not reduce Cu2+ to Cu+
b. Reduce Br2 to Br- but not to reduce I2 to I-

78. at 25°C when the concentration of Ag+ in the compartment on the right
59. Consider the concentration cell shown below. Calculate the cell potential
at 25°C when the concentration of Ag+ in the compartment on the right
is the following.
a. 1.0 M b. 2.0 M c M d. 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
[Ag+]
= 1.0M

79. Cu 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 CuSO₄(ag) at 25°C,
as shown in the diagram. As the cell operates, copper metal is deposited
onto one electrode and O₂(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 25°C and 1.16 atm, of the O₂(g)
that is produced. V
Half-reaction
E° (V)
O₂(g) + 4 H⁺(ag) + 4 e⁻ → 2 H₂O(l)
+1.23
Cu²⁺(ag) + 2 e⁻ → Cu(s)
+0.34 Cu