A voltaic cell consists of a copper electrode in a solution of Cu^2+ and a palla

Question

A voltaic cell consists of a copper electrode in a solution of Cu^2+ and a palladium electrode in a solution of Pd^2+ ions. The copper electrode is the anode. Write a balanced equation and cell notation to represent this cell. The standard cell potential (E degree cell) for this cell is found to be 0.609 V. What is the reduction potential for the Pd^2+/Pd half-reaction? Determine the equilibrium constant for this reaction at 25degree C. Consider a nonstandard cell with (Cu^2+] = 3.00 M and [Pd^2+] = 0.0500M. What is the cell voltage under these conditions?

Explanation / Answer

The reactions are Cu ------>Cu+2 +2e-   Eo= -0.34 V and Pd+2+2e- --------->Pd

Combined reaction is Pd+2+Cu---->Pd+Cu+2 , Eo= 0.609

EO= Ered + Eox

0.609 =Ered +Eox

0.609 =-0.340 +EOx

Eox= 0.609+0.340 =0.951 V

deltaG=-nFE =-RTlnK, n= 2 ( nubmer of eletrons transferred)

=-2*96500*0.609 = -8.314*298*lnK

lnK= 78

K= 6.77*1033

E =EO- 0.0596/n* logQ

E= 0.609-0.0596/2* log { [Cu+2]/[Pd+2]}

E=0.609-0.0596/2* log (3/0.05)=0.556V

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