Electrochemistry Chapter 11 E-mail: [email protected]@gmail.com Web-site:

ElectrochemistryChapter 11E-mail: [email protected]

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Electrochemistry – ch 11

1. Consider the following reaction:

a. What substance is getting reduced?

b. How many moles of electrons are transferred?

2 VO2+ + 4 H+ + Cd → 2 VO2+ + 2 H2O + Cd2+


2. Balance the following reaction under basic conditions:

Fe(CN)64- + Ce4+ → Ce(OH)3 + Fe(OH)3 + CO3

2- + NO3-


3. Which is the strongest oxidizing agent?

a. Mn2+

b. Br-

c. Br2

d. Ag+


4. Which is the strongest reducing agent?

a. Na+

b. Al

c. Zn2+

d. F-

e. Mn


5. 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


6. 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?


7. 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 HCl solution?

d. Will Zn dissolve in an HCl solution?

e. What can oxidize Al but not Fe?

f. What can reduce Ag+ but not I2?


8. Consider the following reaction at 75 °C.Pb2+

(aq) + 2 Cr2+(aq) Pb(s) + 2 Cr3+

(aq) a. Calculate the standard potentialb. Calculate the standard change in free energy c. Calculate the equilibrium constant d. If the above cell is allowed to operate spontaneously, will the voltage increase, decrease or remain the same?e. If the above cell is allowed to operate spontaneously, what will

happen to the concentration and mass at the electrodes?f. Calculate the initial cell voltage for the above reaction if the initial concentrations are [Pb2+] = 0.25 M, [Cr2+] = 0.20M and [Cr3+] = 0.005 M.


9. Consider the following cell: Cu (s) | Cu2+

(aq)(0.001 M) || Br2(l) | Br-(aq) (? M), Pt(s)

a. Assign the electrodes.b. Determine the direction of electron flow.c. Describe the flow of ions in the salt bridged. Calculate the standard potential.e. At 25 °C the measured cell voltage is 0.975 V. Calculate the concentration of the bromide ion.


10. 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 0.6 M at 25 °C.


11. Consider two electrodes connected by a wire. One side has 0.0001 M Fe2+/Fe (s) and the other side has 10 M Fe2+/Fe(s).a. Assign the electrodes b. Determine the direction of electron flowc. Calculate the standard cell voltage at 25 °C. d. Calculate the cell voltage at 25 °C. e. What are the concentrations at equilibrium?


12. What mass of Co forms from a solution of Co2+ when a current of 15 amps is applied for 1.15 hours?


13. A solution of iron chloride underwent electrolysis for 2 hrs at 10 amps yielding 20.84 g of iron. What is the oxidation state of the iron in the iron chloride?


14. How long will it take (in min) to plate out 10.0 g of Bi from a solution of BiO+ using a current of 25.0 A?


15. What volume of gas at STP is produced from the electrolysis of water by a current of 3.5 amps in 15 minutes?

1. Consider the following reaction:

a. What substance is getting reduced? VO2+

b. How many moles of electrons are transferred? 2

2 VO2+ + 4 H+ + Cd → 2 VO2+ + 2 H2O + Cd2+

Electrochemistry – ch 11 – Answers

+5 +4

Oxidation number for V is decreasing ⇒ Getting reduced

Electrochemistry – ch 11 – Answers 2. Balance the following reaction under basic conditions:

Fe(CN)64- + Ce4+ → Ce(OH)3 + Fe(OH)3 + CO3

2- + NO3-

Red ½: Ce4+ + 3H2O Ce(OH)3 + 3H+

Ox ½: Fe(CN)64– + 39H2O Fe(OH)3 + 6CO3

2– +6NO3– + 75H+ + 61e–

Acidic: Fe(CN)64– + 61 Ce4+ + 222H2O Fe(OH)3 + 6CO3

2– +6NO3– + 258H+ + 61Ce(OH)3

Basic: Fe(CN)64– + 61 Ce4+ + 258 OH– Fe(OH)3 + 6CO3

2– +6NO3– + 36H2O + 61Ce(OH)3

3. Which is the strongest oxidizing agent?

a. Mn2+

b. Br-

c. Br2 d. Ag+

4. Which is the strongest reducing agent?

a. Na+

b. Al

c. Zn2+

d. F-

e. Mn


Electrochemistry – ch 11 – Answers 5. 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

Electrochemistry – ch 11 – Answers 6. 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?

Au

Au3+(aq) Ag+(aq)

Ag

Cathode Anode

Cathode ½: Au3+ + 3e– Au E° = 1.5 VAnode 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.

Electrochemistry – ch 11 – Answers 7. 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 Fe? Cr3+, Zn2+, H2O and Mn2+

f. What can reduce Ag+ but not I2? Hg, Fe2+, H2O2 and MnO42-

8. 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 = 0.37Vb. Calculate the standard change in free energy ΔGº =-nFEº

ΔGº =-(2mol e-)(96,485C/mol e-)(0.37J/C) = -71.4 kJc. Calculate the equilibrium constant ΔGº =-RTlnK

-71.4kJ=(0.008314J/molK)(348K)lnK ⇒ K=5.2x1010

d. If the above cell is allowed to operate spontaneously, will the voltage increase, decrease or remain the same? Rxns spontaneously go to equilibrium where E=0

e. If the above cell is allowed to operate spontaneously, what will happen to the concentrations at the electrodes?

The concentrations of the reactants will decrease and the products increase


Continue to next slide…

Electrochemistry – ch 11 – Answers f. Calculate the initial cell voltage for the above reaction if the initial

concentrations are [Pb2+] = 0.25 M, [Cr2+] = 0.20M and [Cr3+] = 0.005 M. Since the concentrations are not 1M these are

nonstandard conditions ⇒ E = E° - ⇒E = 0.37V - ln = 0.46

Electrochemistry – ch 11 – Answers 9. Consider the following cell:

Cu (s) | Cu2+(aq)(0.001 M) || Br2(l) | Br-

(aq) (? M), Pt(s) a. Assign the electrodes. Left side ⇒ anode and right ⇒ cathodeb. Determine the direction of electron flow and the direction of the current. Electrons flow from anode to cathode and vice versa for the currentc. Describe the flow of ions in the salt bridge anions flow to the anode and cations to cathode

d. Calculate the standard potential. Eoxº = -0.34 V and Eredº = 1.09 V ⇒ Ecellº = 0.75 V


Electrochemistry – ch 11 – Answers e. At 25 °C the measured cell voltage is 0.975 V. Calculate the concentration of the bromide ion. E = E° - ⇒0.975 = 0.75V - ln(0.001)[Br-]2 [Br⇒ -] = 0.005

Electrochemistry – ch 11 – Answers 10. 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 0.6 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° - ⇒E = 1.53V - ln = 1.5 V

3Pb2+ 2Al3+

1M 1M

-3/2(0.6)

+0.6

0.1 1.6

Electrochemistry – ch 11 – Answers 11. Consider two electrodes connected by a wire. One side has 0.0001 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 cathodeb. Determine the direction of electron flow anode to cathode or low conc to high conc


Electrochemistry – ch 11 – Answers d. Calculate the cell voltage at 25 °C.

Anode: Fe Fe2+ + 2e- E°ox = -0.44V

Cathode: Fe2+ + 2e- Fe E°red = +0.44VCell: Fe(anode) + Fe2+(cathode) Fe(cathode) + Fe2+ (anode)

E°cell = 0e. What are the concentrations at equilibrium? at equilibrium E = 0 and Q = K and for a concentration cell [anode]=[cathode]

They will be equal at the midpoint ⇒ [Fe2 ]= = 5M

Electrochemistry – ch 11 – Answers 12. What mass of Co forms 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-) = 0.644 mol e-

(0.644 mol e-)(1mol Co/2mole-)(58.933g Co/mol)= 19 g Co

Electrochemistry – ch 11 – Answers 13. A solution of iron chloride underwent electrolysis for 2 hrs at 10 amps

yielding 20.84 g of iron. What is the oxidation state of the iron in the iron chloride? FeX+ + Xe- Fe So we need to figure out how many moles of electrons per mol of Fe

Fe ⇒ (20.84g)/(55.85g/mol) = 0.373 mol Fe

e- ⇒ ne- = ⇒ ne- =(10C/s)(2 hr)(3600s/hr)/(96,485C/mol e-) = 0.746 mol e-

0.373 mol Fe:0.746 mol e- ⇒ 1 mol Fe : 2 mol e- ⇒ Fe2+

Electrochemistry – ch 11 – Answers 14. How long will it take (in min) to plate out 10.0 g of Bi from a

solution of BiO+ using a current of 25.0 A?BiO+ + 2H+ + 3e- Bi + H2O

ne- =

10.0 g Bi xx = t = 554 s or 9.23 min⇒

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- = = (3.5C/s)(15min)(60s/min)/(96,485C/mol e-)

ne- = 0.0326 mol e-

(0.0326 mol e-)(3mol gas/4mol e-)(22.4 L/mol) = 0.55 L

Electrochemistry Chapter 11 E-mail: [email protected]@gmail.com Web-site:

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