Question

Calculate the reduction potential of the following solutions:
a- 1 M CuSO4
b- 1 M FeSO4
c- 0.1 M CuSO4

I'm now the same answer as you. :( This is how I did it can you please check.

a- Cu(s)---> Cu(aq) + 2e = -0.34 V
SO4^-2(aq) + 4H(aq) + 2e --> H2SO3(aq) + H2O(l)= 0.20V
Cu(s) + SO4^-2(aq) + 4H(aq) ---> Cu^+2(aq) + H2SO3(aq) + H2O(l)

Ecell= Ecat -Eanode = 0.20-(-0.34)=0.54V

and I did the same for the other two and my answers were:

b- -0.25V

c- 0.451V (but for this one I used the equation : Ecell= Ecell - (0.0592V/n)logQ

I'm really trying but I don't if I'm doing it right.

Answer 1

I'm unsure of what you are doing. If you want the potential of 1M CuSO4, I think that is 0.34v (since 1M is standard that is the number that comes from the table). Same thing for FeSO4 since that is 1 M. It appears to me that you are trying to make a voltaic cell out of just one metal. The SO4^2- is a spectator ion. The reduction potential for 0.1M CuSO4 is
E = Eocell - (0.0592/n)log(reduced)/(oxidized)
E = 0.34 - (0.0592/2)log(Cu/Cu^2+)
E = 0.34 - (0.0296)log(1/0.1)
E = 0.34 - (0.0296)(1)
E = 0.34 - 0.0296 = 0.31 (I'm remembering that I wrote 0.34+0.0296 = 0.37 before. Check this carefully. If I did that it's because I multiplied by -1 and that is wrong.