A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JoannaBlackwelder

  • one year ago

Half cell problem help: number 15 please

  • This Question is Closed
  1. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @.Sam. @nincompoop

  2. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I get A. Can anyone confirm, please?

  3. JFraser
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you have each half-cell potential?

  4. Rushwr
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    U need to have the potentials. E cell = Ecathose - Eanode

  5. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes. To find the overall potential of this cell, you need to have the standard potentials of each. It goes from left to right.

  6. cuanchi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I didn't get any of the answers. What half reactions did you choose? Cr -> Cr+3 oxidation Mn+2 ->Mn reduction

    1 Attachment
  7. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I used those half reactions you did, but my chart said that the Mn one had a red potential of -1.04 V

  8. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So, I technically got 1.78, which I figured was pretty close to A

  9. cuanchi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I don't know where did you got your cell potentials This is my calculation. Cr(s) -> Cr+3(aq) + 2 e- oxidation 0.740 V Mn+2 (aq) +2e- ->Mn (s) reduction -1.180V E= -0.44 V it is not spontaneous

  10. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok, so you can also separate it by left and right cell. and do E\(\sf _{total} = E_{left}-E_{right}\)

  11. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where E = E\(\sf ^o\) + \(\frac{0.0831}{n}\)ln\(\sf \frac{prod}{react}\) n = numer of electrons being transferes, here it's 1.

  12. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow, thanks so much everyone!

  13. cuanchi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @abb0t , sorry to tell you but the formula is Ecell=Eright−Eleft is ALLWAYS final minus initial, and the number of electrons being transferred in these reactions are 2e-. Respect to the other equation E = Eo + (0.0831/n) x ln ([prod]/[react]) it is not necessary .In this case the two concentrations are the same and the ln 1 =0 then E=Eo and besides the question in the problem is not asking for E but Eo.

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.