The full equation you obtain from the set up. Read and write from left to right.
Sn==> Sn^2+ + 2e
Sn^4+ + 2e ==> Sn^2+
--------------------
Add the two half equations to obtain
Sn + Sn^4+ ==> Sn^2+ + Sn^2+ Eo = ?
Look up the Sn ==> Sn^2+ + 2e Eo value. You will find it in your table of reduction potentials as
Sn^2+ + 2e ==> Sn. My table shows -0.136 but my set is very old and you should use your value. Then reverse the sign to make it read +0.136 because your reaction is the reverse of what you looked up.
Then look up the other half equation. That one is a reduction potential written as a reduction and my table has +0.15. Again, you should use the value in your set.
Then add the oxidation half to the reduction half to find the Eo for the cell as you have it written.
That will be 0.136 + 0.15 = ? volts = Eo cell.
Predict the standard emf of each of the following galvanic cells.
(a) Sn(s)|Sn2+(aq)||Sn4+(aq),Sn2+(aq)|Pt(s)
(b) Pt(s)|Fe3+(aq),Fe2+(aq)||Ag+(aq)|Ag(s)
I don't want the answer, I want the procedure/methods of how to do these types of problems.
3 answers
I did the same thing for part b:
Flipped the E for the lesser value (Fe half reaction)
0.80 - 0.771 = 0.029, but it wasn't correct.
Flipped the E for the lesser value (Fe half reaction)
0.80 - 0.771 = 0.029, but it wasn't correct.
Did you get the right answer for part a? You should have.
I don't think you followed directions for part b. Why did you flip it? The way the cell is set up it is non-spontaneous. So
Fe^3+ + e ==> Fe^2+ Eo = 0.77
Ag ==> Ag^+ + e Eo = -.8
So Ecell should be 0.77+(-0.80) = -0.03 volts which means it will not go as written but will proceed in the opposite direction spontaneously.
I don't think you followed directions for part b. Why did you flip it? The way the cell is set up it is non-spontaneous. So
Fe^3+ + e ==> Fe^2+ Eo = 0.77
Ag ==> Ag^+ + e Eo = -.8
So Ecell should be 0.77+(-0.80) = -0.03 volts which means it will not go as written but will proceed in the opposite direction spontaneously.