Zn(s) ----------------------> Zn^2+ (aq) + 2e^- E0 = 0.733v Cu^2+ (aq) + 2e^- --

Question

Zn(s) ----------------------> Zn^2+ (aq) + 2e^-                E0 = 0.733v

Cu^2+ (aq) + 2e^- ---------> Cu(s)                                E0 = 0.367v

-----------------------------------------------------------------------------------------------

Zn(s) + Cu^2+ (aq) ----------> Zn^2+ (aq) + Cu(s)        E0cell = 1.1v

    n = 2

Ecell    = E0cell - 0.0592/n logQ

            = 1.1-0.0592/2 log[Zn^2+]/[Cu^2+]

            = 1.1-0.0296log0.3/0.1

             = 1.1-0.0296*0.4771

               = 1.086V >>>>answer

7.

      K   = 2.303/t log[A0]/[A]

    [A]   = 1/4

   K    = 1.05*10^-4 sec^-

1.05*10^-4    = 2.303/t log1/1/4

1.05*10^-4   = 2.303/tlog4

1.05*10^-4    = 2.303*0.6020/t

t                     = 2.303*0.6020/1.05*10^-4    = 13203.86sec

Explanation / Answer

6. Calculate Ecel for the following voltaic cell reaction, Zn(s) + Cu?"(0.100 M) Cu(s) + Zn2+(0.300 M), Eeuanc.- 0.367 v, Emman 8 is radioactive, with k = 1.05x104 s, How long would it take to reduce the concentration to o quarter? 7, F-1 8. Give signs (+, -, or 0) ofH, AS, aG, and wpv for the condensation: H2O(g) H00 at 250

Related Questions

Navigate

• Browse (All)
• Browse Z
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.