Electro Chemistry Practice Exam ElectroChemistry...

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Electro Chemistry Practice Exam ElectroChemistry Name (last)________________ (First)______________________ Read all questions before you start. Show all work and explain your answers to receive full credit . Report all numerical answers to the proper number of significant figures. By signing your signature above you agree that you have worked alone and neither give nor received help from any source. Keep your eyes on your own paper at all times System LENGTH: VOLUME MASS Temperature English: 1 ft = 12 in 1 mile = 5280 ft 1 yd = 3 ft 1 gal = 4 qt 1 qt = 57.75 in 3 1 lb = 16 oz 1 ton = 2000 lb T F = 1.8T C + 32 SI- English: 1 in = 2.54 cm 1 mi = 1.609 km 1 L = 1.057 qt 1 qt = 0.946 L 1 lb = 453.6 g 1 oz = 28.35 g T C = (T F 32) 1.8 Misc. info 1 mole = 6.02•10 23 R = 8.314 J/ mol•K or .08206 L•atm/mol•K Cell concentration and the Nernst equation = (R T / n F) ln Keq = (0.0592/ n) log Keq E = - (0.0592 / n) log Q 2 H2O (l) + 2e- H2 (g) + 2 OH- (aq) E° = - 0.83 V O2(g) + 4H+ (aq) + 2e- 2 H2O (l) E° = +1.23V Constants R = 8.314 J / mol•K F = 96,500 C / mol e- 1 V = 1 J/ C

Transcript of Electro Chemistry Practice Exam ElectroChemistry...

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Electro Chemistry

Practice Exam ElectroChemistry Name (last)________________(First)______________________

Read all questions before you start. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper number of significant figures. By signing your signature above you agree that you have worked alone and neither give nor received help from any source. Keep your eyes on your own paper at all times

System LENGTH: VOLUME MASS Temperature

English: 1 ft = 12 in

1 mile = 5280 ft

1 yd = 3 ft

1 gal = 4 qt

1 qt = 57.75 in3

1 lb = 16 oz

1 ton = 2000 lb

T F = 1.8T C + 32

SI-

English:

1 in = 2.54 cm

1 mi = 1.609 km

1 L = 1.057 qt

1 qt = 0.946 L

1 lb = 453.6 g

1 oz = 28.35 g T C =

(TF−32)1.8

Misc. info 1 mole = 6.02•1023 R = 8.314 J/ mol•K or .08206 L•atm/mol•K

Cell concentration and the Nernst equation

E° = (R T / n F) ln Keq

E° = (0.0592/ n) log Keq E = E° - (0.0592 / n) log Q

2 H2O (l) + 2e- H2 (g) + 2 OH- (aq) E° = - 0.83 V

O2(g) + 4H+ (aq) + 2e- 2 H2O (l) E° = +1.23V

Constants

R = 8.314 J / mol•K

F = 96,500 C / mol e-

1 V = 1 J/ C

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Multiple -choice. Show work for partial credit.

1 What is the E°cell for the voltaic cell: Zr(s) | Zr4+ (1M) || Cu2+ (1M) | Cu(s) a) 1.10 V b) 1.87 V c) 1.19 V d) –1.19 V

2 In the reaction Ca(s) + O2 CaO, which is the reducing agent?

a) Ca b) O2 c) CaO d) None

3 What type of cell is nonspontaneous?

a) Voltaic b) Galvanic c) Electrolytic d) All

4 How many electrons are involved in the following half-reaction when it is balanced? ClO-(aq) Cl2

a) 0 b) 1 c) 2 d) 4 5 The chemical reaction for a two compartment cell, separated with a NaCl salt-bridge, has the following line notation: Cu(s) |

Cu2+(aq) || Cu+(aq) | Cu(s). Which change will increase the voltage of the cell?

a) Decrease the [Cu+] at the cathode b) Double size of Cu electrodes at anode and cathode

c) Increase the [Cu+2] at the anode d) none 6 Which reaction is the most spontaneous?

a) Al(s) | Cr3+(aq) || Al3+(aq) | Cr(s) b) Al(s) | Al3+(aq) || Cr3+(aq) | Cr(s)

c) Al(s) | Cr(s) || Al3+(aq) | Cr3+(aq) d) Cr(s) | Cr3+(aq) || Al(s) | Al3+(aq) 7 In electrolysis, oxidation takes place at the

a) Cathode b) Surface of the electrolyte c) Salt bridge d) Anode 8 Which one atom has the largest change in its oxidation state in the following reaction?

Fe2S3(s) + 12 HNO3(aq) 2 Fe(NO3)3(aq) + 3S(s) + 6NO2(g) + 6 H2O(l) a) Fe b) S c) N d) All right atom

9 Calculate the solubility product of CdCO3 (s) using the following standard reduction potential:

CdCO3 (s) + 2e- Cd(s) + CO32- (aq) E° = -2.37 V

Cd2+ (aq) + 2e- Cd(s) E° = -0.40 V

a) 3•10-12 b) 2.80 • 10-67 c) 3.58 • 1066 d) 2 • 10-6 10 Use the Standard Reduction table to rank the following metals in order of decreasing reducing strength: Al, Zn, Cu.

a) Al > Zn > Cu b) Zn > Cu > Al c) Cu > Zn > Al d) Al > Cu > Zn

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11 Solve the following: (show your work) i) Balance the overall reaction and calculate E°cell:

Cl2 (g) Cl- (aq) + ClO3- (aq) E° cell = ?

5 ( 2e- + Cl2 2 Cl - ) 1.358 V

6 H2O + Cl2 2 ClO3- + 12 H+ + 10 e- -1.47 V

Net: 6 H2O + 6Cl2 10 Cl- + 2 ClO3- + 12 H+ E° = - 0.112 V

ii) Balance the overall reaction and calculate the E° for U3+/ U:

U(s)| U3+

(aq) || Cu2+(aq) | Cu(s) E° cell = + 2.13 V

2e- + Cu+2 Cu 0.337 V

U U3+ + 3 e- 1.793 V

Net: 3Cu+2 + 2U 2 U3+ + 3Cu E° = 2.13 V

12 Alcohol levels in blood can be determined by a redox titration with potassium dichromate according to the redox equation: C2H5OH

(aq) + Cr2O72-(aq) CO2(g) + Cr23+(aq) What is the blood alcohol level in mass percent if 8.76 mL of 0.04988M K2Cr2O7 is required for titration of 10.002 g sample of

blood?

13 Give the oxidation number for the atoms in brackets [ ] from the following molecules and ions:

i) ClF4- Oxidation state of [Cl] ii) Cl2 Oxidation state of [Cl] v) H2O2 Oxidation state of [O] iv) Mg3N2 Oxidation state of [N] vi) KAuCl4 Oxidation state of [Au]

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14 Use the ion-electron method to balance each of the redox reactions below. After balancing the equations, answer the following questions for each reaction: a) In the half reaction you wrote, which species is undergoing oxidation and which is undergoing reduction? b) What are the possible oxidation state of chlorine in the reactions (i & ii) below?

i) Cl2 (aq) → Cl-(aq) + ClO3- (aq) (basic conditions)

ii) ClO-(aq) + CrO2 -(aq) → Cl-(aq) + CrO42- (aq) (basic conditions)

15 A sample containing H3AsO3 (0.196 g) is dissolved in acid solution and titrated with 0.501 M I3- solution to the end point. In this

redox reaction the H3AsO3 is converted to H3AsO4 and the triiodide ion is reduced to iodide ion. How many milliliters of I3- is

required in this titration? (H3AsO3 MW = 126.0 g/mol)

16 In a solution containing Ag+, Mg+2, Cu+2 and Fe+2 (each 1.0 M) rank the order in which these ions will be reduced to their elemental state. (Easiest to hardest).

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17 For the reaction Al3+(aq) + 3Cr2+(aq) → 3Cr3+(aq) + Al(s) i) Find E°, ΔG° and the equilibrium constant for the reaction (T = 298K, F = 96500 C/mol e-). ii) Will this reaction proceed spontaneously in the forward or reverse direction?

iii) What is the voltage (E) if the conc in this reaction is [Al3+] = 0.5 M, [Cr2+] = 0.1 M, and [Cr3+] = 1.0 x 10-4 M

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18 (18 pts) Sketch the following cell, indicating the direction of the flow of electrons in the external circuit and the motion of ions in

the salt bridge. [ Pb (s) | Pb+2(0.10M) || HCl (0.10M), H2 (g) | Pt(s) ], T = 298 K i) Write the half-reaction at the cathode.

Cathode (Reduction): 2H+ + 2e- H2 (g) E° = 0.000V

ii) Write the half-reaction at the anode.

Anode (Oxidation): Pb(s) Pb2+

(aq) + 2e- E° = 0.126V

iii) What is the charge on the electrode labeled "anode" and that on the "cathode"?

Anode (-), Cathode (+)

iv) Which is the electrode that electrons leave the cell and the electrode that

electrons enter the cell?

Electrons leave the Anode (Oxidation reaction is the electron source).

Electrons enters the Cathode (Reduction reaction is where electrons are needed). v) Calculate Ecell for the reaction.

vi) Calculate ΔG for the reaction and state the sign of ΔG.

vii) If the cell were carried out at standard state condition, what is the equilibrium

constant for this reaction?

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