Chemistry question- electrochem

Question

Consider the cell described below:

Zn/Zn^2(1.00M)//Cu^2(1.00 M)/Cu

Calculate the cell potential after the reaction has operated long enough for the Al^3 to have changed by 0.60 mol/L. Assume T=25 C

Answer 1

if you meant for us to use:

Zn/Zn^2(1.00M)//Cu^2(1.00 M)/Cu

& not to use:

"long enough for the Al^3"

then

Zn --> Zn+2  & 2e-  ... Eo = + 0.763 volts

Cu+2 & 2 e- --> Cu ... Eo = - 0.337 volts

cell potential ... Eo = 0.426 volts

Zn  & Cu+2 --> Zn+2  & Zn

was [1.00M] --> [1.00M]

is [0.40M] --->  [1.60M]

-------

nernst:

E = Eo - (0.0592/n) log Q

E = 0.426 volts - (0.0592/2) log [Zn+2] / [Cu+2]

E = 0.426 volts - (0.0296) log [1.6] / [0.4]

E = 0.426 volts - (0.0296) log 4

E = 0.426 volts - (0.0296) (0.602)

E = 0.426 volts - 0.018

your answer is: E = 0.408 volts

( references vary for Eo's you may have to use you text's values & its sig figs)