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CHEM 104 chemistry ii Exam 4 Portage Learning
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Due No due date Points 100 Questions 10 Time Limit 120 Minutes
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Question 1 8 /^10 pts le.pdf) Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab This may be helpful throughout the exam. For the cell described by the following cell diagram and standard reduction potentials. Cd (s) | Cd+2^ (aq, 1 M) || Ni+2^ (aq, 1 M) | Ni (s) Cd+2^ + 2e-^ โ Cd E^0 = - 0.40 v Ni+2^ + 2e-^ โ Ni E^0 = - 0.23 v (1) The anode half reaction is (2) The cathode half reaction is (3) The overall cell reaction is (4) Show the calculation for the total cell potential
(5) State and explain whether the cell is voltaic or electrolytic Your Answer:
- anode half reaction is Cd(s) --> Cd^+2(aq) +2e-
- cathode half reaction is Ni ^+2 (aq) +2e ---------- > Ni (s)
- overall reaction is Cd (s) + Ni ^+2 (aq)------- > Cd+2(aq) + Ni (s)
- cell potentional =-0.40 - (-0.23) = - 0.17V
- Since E cell is negative, the reaction is nonspontaneous. This is a characteristic of an electrylytic cell.
This may be helpful throughout the exam.
In the following reaction, identify the following and explain your answers 2 C 8 H 18 + 25 O 2 โ 16 CO 2 + 18 H 2 O (1) the material undergoing oxidation (2) the material undergoing reduction (3) the oxidizing agent (4) the reducing agent Your Answer:
- Oxidatio number for C in C8H18 is - 4 and the oxidation number for C inCO2 is +4. So the oxidation number is increasing. C8H18 is undergoing oxidation
- oxidation number for O2 is 0 oxidation number for O in H2O is - 2, the oxidation number of oxygen is decreasing. oxidation number is decreasing. Oxygen is undergoing reducution.
- O2 is the oxidizing agent
- C8H18 is the reducing agent In the following reaction, identify the following and explain your answers (1) the material undergoing oxidation, (2) the material undergoing reduction, (3) the oxidizing agent and (4) the reducingagent. 2 C 8 H 18 + 25 O 2 โ 16 CO 2 + 18 H 2 O C 8 H 18 is undergoing oxidation (gain of O) to become CO 2 and H 2 OO 2 is undergoing reduction (gain of H) to become H 2 O O 2 is the oxidizing agent which causes oxidation of C 8 H 18 C 8 H 18 is the reducing agent which causes reduction of O 2
This may be helpful throughout the exam. For the voltaic cell described by the cell reaction below, (1) write the cell diagram, (2) identify which specie is being reduced, (3) which specie is being oxidized, (4) identify the anode and (5) identify the cathode.
Question 5 10 /^10 pts le.pdf) Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab Ba + Pb+2^ (0.10 M) โ Ba+2^ (0.20 M) + Pb Your Answer:
- cell diagram Ba (s) I Ba+2 (aq, 0.20M) I I Pb+2 (aq, 0.10 M)I Pb (s)
- reduced = Pb+2 is reduced to Pb
- oxidized = Ba is oxidized to Ba+
- anode = Ba
- cathode = Pb For the voltaic cell described by the cell reaction below, (1) write the cell diagram, (2) identify which specie is being reduced, (3) which specie is being oxidized, (4) identify the anode and (5) identify the cathode. Ba + Pb+2^ (0.10 M) โ Ba+2^ (0.20 M) + Pb (1) cell diagram 0.10 M) | Pb (s) Ba (s) | Ba+2^ (aq, 0.20 M) || Pb+2^ (aq, (2) specie being reduced Pb+2^ is reduced to Pb (3) specie being oxidized Ba is oxidized to Ba+ (4) anode Ba (5) cathode Pb
cell For the voltaic cell described by the following cell reaction: Mg + Sn+2^ (0.10 M) โ Mg+2^ (0.30 M) + SnMg+2^ + 2e-^ โ Mg E^0 = - 2.37 v Sn+2^ +^ 2e
- (^) โ Sn E (^0) = - 0.14 v (1) calculate the value of Q (2) using the given standard reduction potentials, show the calculationof the Ecell using the Nernst equation [Ecell = E^0 โ (0.0592/n x log Q)]. Your Answer:
- Value of Q Q = [Mg+2] /[Sn+2] = 0.30/0.10 = 3 = cathode- anode = - 0.14 - (-2.37) = 2.23 V
- = cell- (0.592/n x log Q) = 2.23 - (0.592 log 3 )/ 2 = 2.21V (1) calculate the value of Q Q = Product ion / Reactant ion = [Mg+2] / [Sn+2] = (0.30) / (0.10) = 3 (2) using the given standard reduction potentials, show the calculation of the Ecell using the Nernst equation [Ecell = E^0 cell โ (0.0592/n x log Q)]. E^0 cell = Ecathode - Eanode = - 0.14 - (- 2.37) = 2.23 v Ecell = 2.23 - (0.0592/2 x log 3) = 2.23 - (0.0592/2 x 0.477) = 2.23 - 0.014 = 2.216 v
Question 6 10 /^10 pts
- Which of the following is a naturally occurring solid inorganic compound that can have varying elemental composition? A. Mineral B. Ore
Question 7 7.5^ /^10 pts le.pdf) Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab C. Alloy D. Metallurgy E. Gangue Your Answer:
- D. Both A and B
- C Saturn
- C. 2H
- A. Mineral
- D
- C
- C
- A
D. Thermometallurgy E. None of the above
- Which of the following indicates that metals can be pulled into long strings? A. Malleable B. Luster C. Ductile D. Electrical conductivity E. Thermal conductivity
- Which of the following is not an alkali metal? A. Li B. Na C. K D. Both A and B E. All of the above are
- What gas is evolved when alkali metals react with water? A. Br 2 B. O 2 C. Cl 2 D. N 2 E. None of the above Your Answer:
- E. this is a metallurgical process
- C ductile
- E all the above are
- E. none of the above. it creates hydrogen
E. LiNO 3
- Which compound is the crucial ingredient in milk of magnesia?
A. CaCO 3 B. MgCl 2 C. Mg(OH) 2 D. Ca(OH) 2 E. CaO
- When magnesium burns in air, what is one compound produced? A. MgH 2 B. MgI 2 C. Mg 3 N 2 D. Mg 2 P 3 E. None of the above Your Answer:
- C. Na2CO
- A. KNO
- C. Mg(OH)
- E. Magneissum Oxide
- C
- A
- C
- C minus 2.5 point
