Zn(s) ----------------------> Zn^2+ (aq) + 2e^- E0 = 0.733v Cu^2+ (aq) + 2e^- --
ID: 700064 • Letter: Z
Question
Zn(s) ----------------------> Zn^2+ (aq) + 2e^- E0 = 0.733v
Cu^2+ (aq) + 2e^- ---------> Cu(s) E0 = 0.367v
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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 Ecell for the following voltaic cell reaction, Zn(s) + Cu2+(0.100 M) Cu(s) + Zn2"(0.300 M), Ealcu= 0.367 v. Ennanzn Znl)n-0.733 v F-1 8 is radioactive, with k = 1.05x10-4 s. How long would it take to reduce the concentration to on 7. quarter?
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