1. Electrochemistry Chapter 11 E-mail: benzene4president@gmail.com
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2. Electrochemistry – ch 11 1. Consider the following reaction:
a. What substance is getting reduced?
b. How many moles of electrons are transferred?
2 VO H+ + Cd → 2 VO H2O + Cd2+

3. Electrochemistry – ch 11 2. Which is the strongest oxidizing agent?
a. Mn2+
b. Br –
c. Br2
d. Ag+
e. Li
Half-Reaction
E°red (V)
Br2 + 2 e – → 2 Br –
1.09
Ag+ + e – → Ag
0.80
Mn e – → Mn
–1.18
Li+ + e – → Li
–3.05

4. Electrochemistry – ch 11 3. Which is the strongest reducing agent?
a. Na+
b. Al
c. Zn2+
d. F-
e. Mn
Half-Reaction
E°red (V)
F2 + 2 e – → 2 F –
2.87
Zn e – → Zn
–0.76
Mn e – → Mn
–1.18
Al e – → Al
–1.66
Na+ + e – → Na
–2.71

5. Electrochemistry – ch 11 4. True or false:
a. Galvanic cells spontaneously produce a current under standard conditions
b. Electrons flow from the anode to cathode in a voltaic cell
c. Oxidation occurs at the cathode
d. Corrosion of a metal occurs at the anode

6. Electrochemistry – ch 11
5. You designed a galvanic cell with silver and gold electrodes. Assign the electrodes, write the overall reaction, and calculate the standard cell potential. As the cell operates what happens to the masses of the silver and gold electrodes?
1.0 M Ag+
1.0 M Au3+ Ag Au
Half-Reaction
E°red (V)
Au e – → Au
1.50
Ag+ + e – → Ag
0.80

7. Electrochemistry – ch 11
6. Using reduction potentials answer the following:
a. Is Cl2 able to reduce Cr3+?
b. Is Pb2+ able to oxidize Ni?
c. Will Au dissolve in an acidic solution?
d. Will Zn dissolve in an acidic solution?
e. What can oxidize Al but not Zn?
f. What can reduce Zn2+ but not Na+?
Half-Reaction
E°red (V)
F2 + 2 e – → 2 F –
2.87
Au e – → Au
1.50
Cl2 + 2 e– → 2 Cl–
1.36
2 H+ + 2 e – → H2
0.00
Pb2+ + 2e– → Pb
–0.13
Ni e– → Ni
–0.23
Cr3+ +3 e– → Cr
–0.73
Zn e – → Zn
–0.76
Mn e – → Mn
–1.18
Al e – → Al
–1.66
Na+ + e – → Na
–2.71

8. Electrochemistry – ch 11
7. Consider the following reaction for a voltaic cell at 75 °C. Pb2+(aq) + 2 Cr2+(aq) → Pb(s) + 2 Cr3+(aq)
a. Calculate the initial cell voltage for the above reaction if the initial concentrations are [Pb2+] = 0.25 M, [Cr2+] = 0.20M and [Cr3+] = M.
b. Will the initial potential increase, decrease or remain the same if the volume of electrodes is doubled by adding water?
c. Will the initial potential increase, decrease or remain the same if the solid lead is cut in half?
d. Will the initial potential increase, decrease or remain the same if you add NaCl causing PbCl2 to precipitate out of solution?
Half-Reaction
E°red (V)
Pb2+ + 2e– → Pb
–0.13
Cr3+ + e– → Cr2+
–0.50

9. Electrochemistry – ch 11
8. Consider the following cell: Cu (s) | Cu2+(aq)(0.001 M) || Fe 3+(aq)(0.02M) , Fe2+(aq) (? M) | Pt(s)
Determine the concentration of the Fe2+ if the measured cell voltage at 27°C is 0.5 V.
Half-Reaction
E°red (V)
Fe3+ + e– → Fe2+
0.77
Cu2+ + 2e– → Cu
0.34

10. Electrochemistry – ch 11 9. Consider the following cell:
Al(s) | Al3+ (1.0 M) | | Pb2+(1.0 M) | Pb (s)
Calculate the cell potential after the reaction has operated long enough for the [Al3+] to have changed by 0.66 M at 25 °C.
Half-Reaction
E°red (V)
Pb2+ + 2e– → Pb
–0.13
Al e – → Al
–1.66

11. Electrochemistry – ch 11
10. Consider the Galvanic cell at 25 °C described as follows: X | X2+ || Y3+ | Y Where X and Y are unknown metals. Given the standard reduction potential for Y3+ is 1.5 V and that K for the overall reaction in this Galvanic cell is 1.2 x 1020 what is the standard reduction potential of X2+?

12. Electrochemistry – ch 11 Fe2+ + 2 e– → Fe E°red = –0.44 V
11. Consider two electrodes connected by a wire. One side has M Fe2+/Fe (s) and the other side has 10 M Fe2+/Fe(s).
a. Assign the electrodes
b. Calculate the cell voltage at 25 °C.
Fe e– → Fe E°red = –0.44 V

13. Electrochemistry – ch 11
12. What mass of Co (58.93 g/mol) will plate out from a solution of Co2+ when a current of 15 amps is applied for 1.15 hours?

14. Electrochemistry – ch 11
13. How long will it take (in min) to plate out 10.0 g of Bi (209 g/mol) from a solution of BiO+ using a current of 25.0 A?

15. Electrochemistry – ch 11
14. It takes 24 min to plate out 9.8 g of an unknown metal (M) from a solution of MCl3 when a current of 10 amps is applied. Identify the metal.

16. Electrochemistry – ch 11
15. What volume of gas at STP is produced from the electrolysis of water by a current of 3.5 amps in 15 minutes?
2 H2O (l) → 2 H2(g) + O2(g)

17. Electrochemistry – ch 11
You have completed ch 11

18. Electrochemistry – ch 11 – Answers
1. Consider the following reaction:
a. What substance is getting reduced? VO2+
b. How many moles of electrons are transferred? 2
2 VO H+ + Cd → 2 VO H2O + Cd2+ +5 +4
Oxidation number for V is decreasing ⇒ Getting reduced

19. Electrochemistry – ch 11 – Answers
2. Which is the strongest oxidizing agent? a. Mn2+ b. Br- √ c. Br2 d. Ag+ 3. Which is the strongest reducing agent? a. Na+ √ b. Al c. Zn2+ d. F- e. Mn

20. Electrochemistry – ch 11 – Answers
4. Which of the following are true about galvanic cells (aka. voltaic cells): √ a. Spontaneously produce a current b. A current must be provided in order to run c. Oxidation occurs at the cathode √ d. Have possible plating out of metals at the cathode √ e. The current flows from anode to cathode f. The concentrations at the electrodes are 1M

21. Electrochemistry – ch 11 – Answers
5. You want to design a galvanic cell with silver and gold electrodes. Show or describe how you would set it up? Assign the electrodes, write the overall reaction, and calculate the standard cell potential. As the cell operates what happens to the masses of the silver and gold electrodes?
Cathode ½: Au3+ + 3e– → Au E° = 1.5 V
Anode ½: Ag → Ag+ + e– E° = 0.8 V
Overall Rxn: Au3+(aq) + 3Ag(s) → Au(s) + 3Ag+(aq) Ecell° = 0.7 V
The cathode is getting plated with Au so the mass is getting heavier and at the anode Ag is corroding so the mass is going down.

22. Electrochemistry – ch 11 – Answers
6. Using reduction potentials answer the following: a. Is Cl2 able to reduce Cr3+? No Cl2 is an oxidizing reagent b. Is Pb2+ able to oxidize Ni? Yes c. Will Au dissolve in an HCl solution? No d. Will Zn dissolve in an HCl solution? Yes e. What can oxidize Al but not Zn? Mn2+ f. What can reduce Zn2+ but not Na+? Mn or Al

23. Electrochemistry – ch 11 – Answers
7. Consider the following reaction at 75 °C. Pb2+(aq) + 2 Cr2+(aq) → Pb(s) + 2 Cr3+(aq) a. Calculate the standard potential Eredº = -0.13V and Eoxº = +0.5 V ⇒ Erxnº = 0.37V b. Will the initial potential increase, decrease or remain the same if the volume of electrodes is doubled by adding water? adding water will cause the concentrations to decrease ⇒ Q = Cr3+ 2 Pb2+ Cr2+ 2 ⇒ the denominator will decrease more than the numerator causing Q to increase which causes the potential to decrease ⇒ E = E° - 𝑅𝑇 𝑛𝐹 lnQ c. Will the initial potential increase, decrease or remain the same if the solid lead is cut in half? solids are not in the expression for Q and will therefore not affect the potential d. Will the initial potential increase, decrease or remain the same if you add NaCl causing PbCl2 to precipitate out of solution? Q increases when precipitating out the lead (II) ion and therefore the potential decreases

24. Electrochemistry – ch 11 – Answers
8. Consider the following cell: Cu (s) | Cu2+(aq)(0.001 M) || Fe 3+(aq)(0.02M) , Fe2+(aq) (? M) | Pt(s)
Determine the concentration of the Fe2+ if the measured cell voltage at 27°C is 0.5 V.
anode: ⇒ Cu → Cu2+ + 2e– –0.34 V
cathode: ⇒ Fe 3+ → Fe2+ + e– V
overall: ⇒ Cu + 2 Fe 3+ → Cu Fe V
E = E° - 𝑅𝑇 𝑛𝐹 lnQ ⇒ 0.5 V = 0.43 V - ( 𝐽 𝑚𝑜𝑙𝐾 )(300𝐾) (2)(96,485 𝐶 𝑚𝑜𝑙𝑒 ) ln Fe
⇒ Fe2+ = M
Continue to next slide…

25. Electrochemistry – ch 11 – Answers
9. Consider the following cell: Al(s) | Al3+ (1.0 M) | | Pb2+(1.0 M) | Pb (s) Calculate the cell potential after the reaction has operated long enough for the [Al3+] to have changed by 0.66 M at 25 °C. Anode: Al → Al3+ + 3e- E°ox = +1.66V Cathode: Pb2+ + 2e- → Pb E°red = -0.13V Cell: 2Al + 3Pb2+ → 2Al3+ + 3Pb E°cell = 1.53V E = E° - 𝑅𝑇 𝑛𝐹 lnQ ⇒ E = 1.53V - ( 8.314𝐽 𝑚𝑜𝑙𝐾 )(298𝐾) (6 𝑚𝑜𝑙 𝑒−)(96,485 𝐶 𝑚𝑜𝑙𝑒− ) ln = 1.5 V
3Pb2+
2Al3+ 1M
-3/2(0.66)
+0.66
0.01
1.66

26. Electrochemistry – ch 11 – Answers
10. Consider the Galvanic cell at 25 °C described as follows: X | X2+ || Y3+ | Y Where X and Y are unknown metals. Given the standard reduction potential for Y3+ is 1.5 V and that K for the overall reaction in this Galvanic cell is 1.2 x 1020 what is the standard reduction potential of X2+? Anode: X → X2+ + 2e- E°ox = - E°red Cathode: Y3+ + 3e- → Y E°red = 1.5 V Cell: 3 X + 2 Y3+ → 3 X Y E°cell = - E°red V E°cell = 𝑅𝑇 𝑛𝐹 ln𝐾 = 8.314𝐽 𝑚𝑜𝑙𝐾 298𝐾 6 𝑚𝑜𝑙 𝑒− 96,485 𝐶 𝑚𝑜𝑙𝑒− ln(1.2 x 1020) = - E°red V E°red = 1.3 V

27. Electrochemistry – ch 11 – Answers
11. Consider two electrodes connected by a wire. One side has M Fe2+/Fe (s) and the other side has 10 M Fe2+/Fe(s).
a. Assign the electrodes This is a concentration cell ⇒ the voltage is strictly due to the difference in concentration ⇒ at equilibrium the concentrations are equivalent ⇒ the side with lower concentration will have to increase and vice versa until the concentrations meet in the middle and there will no longer be a voltage aka equilibrium ⇒ so in order to increase the concentration on the lower side Fe needs to get oxidized making it the anode ⇒ to decrease the concentration on the other side the Fe2+ needs to be reduced making it the cathode
d. Calculate the cell voltage at 25 °C.
Anode: Fe → Fe2+ + 2e- E°ox = -0.44V
Cathode: Fe2+ + 2e- → Fe E°red = +0.44V
Cell: Fe(anode) + Fe2+(cathode) → Fe(cathode) + Fe2+ (anode) E°cell = 0
E = E° - 𝑅𝑇 𝑛𝐹 lnQ ⇒ E = 𝐽 𝑚𝑜𝑙𝐾 298𝐾 2 𝑚𝑜𝑙 𝑒− 96,485 𝐶 𝑚𝑜𝑙𝑒− ln ⇒ E = 0.15 V

28. Electrochemistry – ch 11 – Answers
12. What mass of Co (58.93 g/mol) will plate out from a solution of Co2+ when a current of 15 amps is applied for 1.15 hours? Co2+ + 2e- → Co ne- = It F = (15C/s)(1.15hr)(3600s/hr) (96,485C/mol e−) = mol e- (0.644 mol e-)(1mol Co/2mole-)(58.93g Co/mol)= 19 g Co

29. Electrochemistry – ch 11 – Answers
14. How long will it take (in min) to plate out 10.0 g of Bi (209 g/mol) from a solution of BiO+ using a current of 25.0 A?
BiO+ + 2H+ + 3e- → Bi + H2O
ne- = 𝐼𝑡 𝐹
10.0 g Bi x 1 𝑚𝑜𝑙 𝐵𝑖 209𝑔 x 3 𝑚𝑜𝑙 𝑒− 1 𝑚𝑜𝑙 𝐵𝑖 = ( 25𝐶 𝑠 )(𝑡) ( 96,485𝐶 𝑚𝑜𝑙𝑒 ) ⇒ t = 554 s or 9.23 min

30. Electrochemistry – ch 11 – Answers
15. What volume of gas at STP is produced from the electrolysis of water by a current of 3.5 amps in 15 minutes? 2H2O → 2H2 + O2 ne- = 𝐼𝑡 𝐹 = (3.5C/s)(15min)(60s/min) (96,485C/mol e−) ⇒ ne- = mol e- ( mol e-)( 3mol gas 4mol e− )(22.4 L/mol) = 0.55 L