{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chhw11 - valdez(vv689 Homework#11 Holcombe(52460 This...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
valdez (vv689) – Homework #11 – Holcombe – (52460) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The equilibrium constant for the reaction 2 Hg( ) + 2 Cl (aq) + Ni 2+ (aq) Ni(s) + Hg 2 Cl 2 (s) is 5.6 × 10 20 at 25 C. Calculate the value of E for a cell utilizing this reaction. 1. + 1.14 V 2. + 0.57 V 3. - 0.57 V correct 4. - 0.25 V 5. - 1.14 V Explanation: 002 10.0 points The standard voltage of the cell Ag(s) | AgBr(s) | Br (aq) || Ag + (aq) | Ag(s) is +0.73 V at 25 C. Calculate the equilibrium constant for the cell reaction. 1. 2 . 0 × 10 15 2. 4 . 6 × 10 13 3. 2 . 2 × 10 12 correct 4. 5 . 1 × 10 14 5. 3 . 9 × 10 29 Explanation: 003 10.0 points Calculate the reduction potential for the Zn 2+ | Zn electrode if the Zn 2+ concentra- tion is 4 × 10 3 M. The standard potential for Zn 2+ + 2 e Zn is - 0 . 7628 V. 1. - 0 . 833 V correct 2. - 1 . 808 V 3. - 1 . 666 V 4. - 0 . 904 V Explanation: [Zn 2+ ] = 4 × 10 3 M E = E 0 - parenleftbigg 0 . 0592 n parenrightbigg log Q = - 0 . 7628 V - parenleftbigg 0 . 0592 2 parenrightbigg log parenleftbigg 1 4 × 10
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ 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