The following values may be useful when solving this tutorial. In the activity,

Question

The following values may be useful when solving this tutorial. In the activity, click on the E degree _cell and K_eq quantities to observe how they are related. Use this relation to calculate K_eq for the following redox reaction that occurs in an electrochemical cell having two electrodes a cathode and an anode. The two half-reactions that occur in the cell are Cu^2 (aq) +2e rightarrow Cu (s) and Zn (s) rightarrow Zn^2 (aq) + 2e The net reaction is Cu^2+ (aq) + Zn (s) rightarrow Cu (s) + Zn^2 (aq) Use the given standard reduction potentials in your calculation as appropriate Express your answer numerically to three significant figures.

Explanation / Answer

from data table:

Eo(Zn2+/Zn(s)) = -0.763 V

Eo(Cu2+/Cu(s)) = 0.337 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Cu2+/Cu(s))

anode is (Zn2+/Zn(s))

The spontaneous chemical reaction taking place is

Cu2+ + Zn(s) --> Cu(s) + Zn2+

Eocell = Eocathode - Eoanode

= (0.337) - (-0.763)

= 1.100 V

here, number of electrons being transferred, n = 2

E = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.059

So, E = (0.059/n)*log Kc

1.100 = (0.059/2)*log Kc

log Kc = 37.288

Kc = 1.941*10^37

Answer: 1.941*10^37

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