5 a metal that is a better reducing agent to be

This preview shows page 2 - 5 out of 7 pages.

5. a metal that is a better reducing agent to be sacrificed. correct Explanation: 007 10.0 points Note: use a = 1 in this question. A compound decomposes with a half-life of 8.0 s and the half-life is independent of the concentration. How long does it take for the concentration to decrease to one-ninth of its initial value? 1. 32 s 2. 3.6 s 3. 72 s 4. 25 s correct 5. 64 s Explanation: 008 10.0 points Consider the reaction 2 Cu + (aq) Cu(s) + Cu 2+ (aq) . If the standard potentials of the Cu 2+ | Cu and Cu + | Cu couples are +0.34 and +0.52 V, respectively, calculate the value of E for the given reaction. 1. - 0.18 V 2. - 0.70 V 3. + 0.70 V
Image of page 2

Subscribe to view the full document.

Version 173 – Exam 4 302 – sutcliffe – (51060) 3 4. + 0.18 V correct 5. + 0.86 V Explanation: combine the 1/2 reactions in the following ways... 2 Cu+ + 2e- 2Cu +.52 V Cu Cu2+ 2e- -.34 V That gives an overall potential of +.18 V 009 10.0 points Given the following standard reduction po- tentials Fe 3+ + e Fe 2+ +0.771 V Cu 2+ + 2 e Cu +0.337 V Sn 2+ + 2 e Sn - 0 . 140 V Which of the following species would be the strongest oxidizing agent? 1. Fe 2+ 2. Cu 3. None of the species listed can act as an oxidizing agent. 4. Sn 2+ 5. Sn 6. Fe 3+ correct 7. Cu 2+ Explanation: 010 10.0 points What is the cathode in Ag(s) | Ag + (aq) || Fe 2+ (aq) | Fe(s) Ag + + e Ag E red = +0 . 80 Fe 2+ + 2 e Fe E red = - 0 . 44 and what type cell is it? 1. Ag(s) | Ag + (aq); a battery 2. Fe 2+ (aq) | Fe(s); an electrolytic cell cor- rect 3. Not enough information is provided. 4. Fe 2+ (aq) | Fe(s); a battery 5. Ag(s) | Ag + (aq); an electrolytic cell Explanation: The diagram A | B || C | D is read as follows: A B + ne (oxidation) C + me D (reduction) Since reduction occurs at the cathode, the cathode is Fe 2+ (aq) | Fe(s). To determine the cell type, calculate E cell: 2 Ag(s) 2 Ag + (aq) + 2 e E anode = - 0 . 80 V Fe 2+ + 2 e Fe E cathode = - 0 . 44 V 2 Ag(s) + Fe 2+ 2 Ag + (aq) + Fe E cell = - 1 . 24 V Since E cell is negative, the reaction is not spontaneous; potential has to be applied to the cell to enable this reaction to occur; i.e. , an electrolytic cell. 011 10.0 points The rate of formation of NO 2 (g) in the reac- tion 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) is 5 . 78 (mol NO 2 ) / L / s. What is the rate at which N 2 O 5 decomposes? 1. 5 . 78 (mol N 2 O 5 ) / L / s 2. 0 . 723 (mol N 2 O 5 ) / L / s 3. 11 . 6 (mol N 2 O 5 ) / L / s 4. 2 . 89 (mol N 2 O 5 ) / L / s correct 5. 1 . 45 (mol N 2 O 5 ) / L / s Explanation: 012 10.0 points
Image of page 3
Version 173 – Exam 4 302 – sutcliffe – (51060) 4 Three separate experiments were performed on the rate of the reaction 3 A 2 + 2 B 2 A 3 B The measured initial concentrations of A 2 (in moles per liter) are shown below along with the measured initial rates of formation of A 3 B (moles per liter per second). Initial Initial Initial Trial [A 2 ] 0 [B] 0 rate M M M/s 1 1 . 5 3 . 0 7 . 0 × 10 8 2 1 . 5 1 . 5 1 . 75 × 10 8 3 4 . 5 3 . 0 21 . 0 × 10 8 What is the order of the reaction?
Image of page 4

Subscribe to view the full document.

Image of page 5
You've reached the end of this preview.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern