KEY Review Unit 12 Test (Chp 20): Electrochemistry KEY

Name____[KEY]______Date:______Period:______

[KEY] Review Unit 12 Test (Chp 20): Electrochemistry [KEY]

Section I Multiple Choice

1. The loss of electrons by an element is called ______.

A) reduction

B) transduction

C) fractionation

D) oxidation LEO says GER

E) potentiation

Fe2S3 + 12 HNO3 ® 2 Fe(NO3)3 + 3 S + 6 NO2 + 6 H2O

2. Which substance oxidizes another substance in the reaction above?

A) Fe2S3 = gets reduced

B) S

C) NO2

D) HNO3 N gets reduced from +5 (in HNO3) to +4 (in NO2)

E) H2O

MnO4– + Br– ® MnO2 + Br2

3. As the reaction represented above proceeds, the oxidation number of manganese changes from…

A) +4 to 0

B) +7 to 0

C) +7 to +4

D) +8 to +4

E) +1 to +2

4. In which of the following species does phosphorus have the same oxidation number as it does in

H2PO4– ?

A) P4O6

B) P4O10

C) P8

D) HPO32–

E) H3PO3

5. Which one of the following reactions is a redox reaction?

A) H2O + NaCl ® NaOH + HCl

B) Pb2+ + 2 Cl– ® PbCl2

C) AgNO3 + HCl ® HNO3 + AgCl

D) NaOH + HCl ® NaCl + H2O

E) NONE of the above is a redox reaction.

…Pb + …H2O2 + …H+ ® …Pb4+ + …H2O

6. What is the coefficient of the peroxide molecule when the equation above is balanced in acidic solution?

A) 1

B) 2

C) 3

D) 4

E) 8

…Fe2+ + …O2 + …H+ ® …Fe3+ + …H2O

7. If 1 mole of O2 oxidizes Fe2+ in acidic solution according to the reaction represented above, how many

moles of Fe3+ ions can be formed?

A) 2

B) 3

C) 4

D) 5

E) 6

Questions 8-9 refer to a galvanic cell constructed using two half-cells and using the two half-reactions represented below.

Cu2+(aq) + e– ® Cu+(aq) Eo = 0.15 V

Pb2+(aq) + 2 e– ® Pb(s) Eo = –0.13 V

8. As the cell operates, ionic species that are found in the half-cell containing the anode include which of

the following?

I. Cu2+ II. Cu+ III. Pb2+

A) I only

B) II only

C) III only

D) I and III only

E) II and III only

9. What is the standard cell potential for the galvanic cell?

A) –0.28 V

B) –0.02 V

C) 0.02 V

D) 0.28 V

E) 0.43 V

M(s) + 2 Cr3+(aq) ® M2+(aq) + 2 Cr2+(aq) Eo = +0.15 V

Cr3+(aq) + e– ® Cr2+(aq) Eo = –0.41 V

10. According to the information above, what is the standard reduction potential for the half-reaction

M2+(aq) + 2 e– ® M(s) Eo = ?

A) –0.97 V

B) –0.56 V

C) –0.26 V

D) 0.26 V

E) 0.56 V

11. Which one of the following is the most effective choice to reduce another substance?

A) O2

B) Li

C) Ca

D) F2

E) H2

3 MnO4–(aq) + 24 H+(aq) + 5 Fe(s) ® 3 Mn2+(aq) + 5 Fe3+(aq) + 12 H2O(l)

12. True statements about the reaction represented above include which of the following?

I. MnO4–(aq) oxidizes another substance.

II. The oxidation state of hydrogen changes from +1 to 0.

III. Fe is oxidized.

A) I only

B) III only

C) II and III only

D) I and III only

E) I, II, and III

Fe2+ + 2 Na(s) ® Fe(s) + 2 Na+

13. If the ∆G° for the reaction above is +142 kJ/mol, which of the following correctly describes the

standard voltage, E°, and the equilibrium constant, K, for this reaction?

A) E° is positive and K > 1

B) E° is negative and K < 1

C) E° is positive and K < 1

D) E° is negative and K > 1

E) E° is zero and K = 1

14. A concentration cell using nickel electrodes as the anode and cathode was constructed, and the

observed voltage was found to be 1.00 volt instead of the standard cell potential, E°, of 0 volts. Which of the following could correctly account for this observation?

A) The nickel anode was larger than the nickel cathode

B) The electrolyte at the anode was Ni(NO3)2, while the electrolyte at the cathode was NiSO4

C) The Ni2+ solution at the anode was less concentrated than the Ni2+ solution at the cathode

D) The solutions in the half-cells had different volumes

E) The Ni2+ solution at the cathode was less concentrated than the Ni2+ solution at the anode

15. Consider an electrochemical cell based on the reaction:

2 H+(aq) + Sn(s) ® Sn2+(aq) + H2(g)

Which of the following actions would change the measured cell potential?

A) increasing the [Sn2+] in the anode compartment

B) increasing the pH in the cathode compartment

C) lowering the pH in the cathode compartment

D) increasing the pressure of hydrogen gas in the cathode compartment

E) ALL of the above will change the measure cell potential.

16. In an electrolytic cell the sign of ∆G° is always ______and is the reaction is ______.

A) positive, favorable

B) negative, favorable

C) negative, favorable

D) positive, unfavorable

E) zero, at equilibrium

17. A steady electric current is passed through molten MgCl2 for exactly 1.00 hour, producing 243 g of

Mg metal. If the same current is passed through molten AlCl3 for 1.00 hour, the mass of Al metal

produced is closest to

A) 27.0 g

B) 54.0 g

C) 120. g

D) 180. g

E) 270. g

Questions 18-19 refer to an electrolytic cell that involves the electrolysis of CuCl2.

18. Which of the following occurs in the reaction?

A) Cu is reduced at the cathode.

B) Cl– is oxidized at the anode.

C) Copper is converted from the 0 oxidation state to the +2 oxidation state.

D) CuCl2 is reduced at the cathode.

E) Cl– acts as an oxidizing agent.

19. Which of the following expressions is correct for the maximum mass of copper, in grams, that could

be plated out by electrolyzing aqueous CuCl2 for 16 hours at a constant current of 3.0 amperes?

A) (16)(3,600)(3.0)(63.55)

(96,485)(2)

B) (16)(3,600)(3.0)(63.55)

(96,485)

C) (16)(3,600)(3.0)(63.55)(2)

(96,485)

D) (16)(60)(3.0)(96,485)

(63.55)(2)

E) (16)(60)(3.0)(96,485)(2)

(63.55)

20. If 0.060 faraday is passed through an electrolytic cell containing a solution of In3+ ions, the

maximum number of moles of In that could be deposited at the cathode is…

A) 0.010 mole

B) 0.020 mole

C) 0.030 mole

D) 0.060 mole

E) 0.18 mole

Questions 21-24 refer to the electrochemical cell and reaction below.

The spontaneous reaction that occurs when the cell above operates is:

2 Ag+ + Cd(s) ® 2 Ag(s) + Cd2+

(A) Voltage increases.

(B) Voltage decreases but remains positive.

(D) No change in voltage occurs.

(E) Direction of voltage change cannot be predicted without additional information.

Which of the above occurs for each of the following circumstances?

21. A 50-milliliter sample of a 2-molar Cd(NO3)2 solution is added to the left beaker.

(B) Voltage decreases but remains positive.

22. The silver electrode is made larger.

(D) No change in voltage occurs.

23. The salt bridge is replaced by a platinum wire.

24. Current is allowed to flow for 5 minutes.

(B) Voltage decreases but remains positive.

Questions 25-27 refer to galvanic cells made from different combinations of the three half–cells described below.

Half–cell 1: strip of Cu(s) in 1.00 M Cu(NO3)2(aq)

Half–cell 2: strip of Ag(s) in 1.00 M Ag(NO3)2(aq)

Half–cell 3: strip of Cr(s) in 1.00 M Cr(NO3)3(aq)

Galvanic Cell / Half–cells / Cell Reaction / E°cell (V)
A / 1 and 2 / Cu(s) + 2 Ag+(aq) ® Cu2+(aq) + 2 Ag(s) / 0.46
B / 1 and 3 / 2 Cr(s) + 3 Cu2+(aq) ® 2 Cr3+(aq) + 3 Cu(s) / ?
C / 2 and 3 / Cr(s) + 3 Ag+(aq) ® Cr3+(aq) + 3 Ag(s) / 1.54

25. What is the standard cell potential of galvanic cell B?

(A) 2.00 V

(B) 1.08 V

(D) –2.00 V

26. In galvanic cells A and B, which of the following takes place in half–cell 1 ?

(A) Reduction occurs in both cell A and cell B.

(B) Oxidation occurs in both cell A and cell B.

(D) Oxidation occurs in cell A, and reduction occurs in and cell B.

27. If the half-cell containing 1.00 M Ag(NO3)2(aq) in galvanic cells A and C is replaced with a half-cell

containing 0.10 M Ag(NO3)2(aq), what will be the effect on the cell voltage of the two galvanic cells?

(A) The voltage will increase in both cells.

(B) The voltage will decrease in both cells.

(D) The voltage will decrease in cell A and increase in cell C.

Section II

Free Response Questions

YOU MAY USE YOUR CALCULATOR FOR SECTION II

Directions: Read each question carefully and write your response in the space provided following each question. Your responses to these questions will be scored on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Specific answers are preferable to broad, diffuse responses. For calculations, clearly show the method used and the steps involved in arriving at your answers. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not.

1. An unknown metal M forms a soluble compound, M(NO3)2.

(a) A solution of M(NO3)2 is electrolyzed. When a constant current of 2.50 amperes is

applied for 35.0 minutes, 3.06 grams of the metal M is deposited. Calculate the molar

mass of M and identify the metal.

35.0 min x 60 s x 2.50 C x 1 mol e– x 1 mol M = 0.0272 mol M 3.06 g M = 112 g/mol = Cd

1 min 1 s 96,485 C 2 mol e– 0.0272 mol M

(b) The metal identified in (a) is used with zinc to construct a galvanic cell shown below.

Write the net ionic equation for the cell reaction and calculate the cell potential, E°.

Half-Reaction / Eo (V)
M2+(aq) + 2e– ® M(s) / –0.40
Zn2+(aq) + 2 e– ® Zn(s) / –0.76

RED: Cd2+ + 2e– ® Cd E°red = –0.40 V

OX: Zn ® Zn2+ + 2e– E°ox = +0.76 V

Cd2+ + Zn ® Cd + Zn2+ E° = E°red + E°ox

+0.36 V = (–0.40) + (+0.76)

DG° is – because the galvanic cell described is a favorable reaction under standard reactions with

E° that is + which produces a voltage.

(d) How would the cell potential, Ecell , differ from the cell shown in (b) if the initial concentration of ZnSO4 is 0.10-molar, but the concentration of the M(NO3)2 solution remains unchanged.

Q = [Zn2+] Q = (0.10) .

[Cd2+] (1.0)

Q 1, therefore E Eo because there is more reactant so the reaction would shift right at a faster rate.

2. An electrochemical cell consists of a tin electrode in an acidic solution of 1.0 molar Sn2+ connected by a salt bridge to a second compartment with a silver electrode in an acidic solution of 1.0 molar Ag+. The two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below.

Half-Reaction / Eo (V)
Ag+(aq) + e– ® Ag(s) / +0.80
Sn2+(aq) + 2 e– ® Sn(s) / –0.14

(a) Write the equation for the half–cell reaction occurring at the anode.

RED: Ag+ + e– ® Ag

OX: Sn ® Sn2+ + 2e– (anode reaction)

(b) Write the balanced chemical equation for the overall favorable cell reaction that

occurs when the circuit is complete. Calculate the standard voltage, E°, for this cell.

RED: Ag+ + e– ® Ag E°red = +0.80 V

2 Ag+ + 2 e– ® 2 Ag

OX: Sn ® Sn2+ + 2 e– E°ox = +0.14 V E° = E°red + E°ox

2 Ag+ + Sn ® 2 Ag + Sn2+ +0.94 V = (0.80) + (+0.14)

DG° = –nFE°

= –(2)(96,485)(0.94) = –180,000 J/molrxn or –180 kJ/molrxn

(per molrxn b/c 2 mol e– per molrxn)

(d) Calculate the equilibrium constant for this cell reaction at 298 K.

ΔGo = −RT ln K ln K = ΔGo Keq = e–DGo/ RT

−RT

e–(–180,000/(8.314 x 298)) = 3.57 ´ 1031

(e) A cell similar to the one described above is constructed with solutions that have initial

concentrations of 1.0 molar Sn2+ and 0.020 molar Ag+. How would the cell potential, Ecell , differ from the standard cell potential, E°cell ? Justify your answer in terms of LeChatelier’s principle.

Q = [Sn2+] Q = (1.0) .

[Ag+]2 (0.020)2

Q > 1, therefore E < Eo because there is more product so the reaction would shift left at a faster rate.