No.
The equation is ok, the voltage is not.
You can do this one of several ways. I prefer to add an oxidation half to a reduction half. If we do it that way, then
Ni to Ni^2+ is +.25 (you should confirm)
Ag^+ to Ag is +0.80 (confirm)
total is 1.05v
If you want to subtract E1-E2, then you keep the reactions as both oxidations or both reductions. You get the same answer either way (except for a change of sign).
Let's put both in oxidation form
Ni==> Ni^2+ E = + 0.25
Ag ==> Ag^+ E = -0.80
Ecell = E1-E2 = 0.25-(-0.80) = 1.05v
Do it with both in reduction form.
Ni^2+ ==> Ni Eo = -0.25v
Ag^+ ==> Ag Eo = +0.80
E1-E2 = -0.25 -(+0.80) = -1.05
The reason for the change of sign is that the first one above has the Ni as the anode as you have said but the lower one has the Ag as the anode and of course the cell won't go that way. Frankly, I don't like the E1-E2 method for that very reason. If I'm in a hurry I may forget to change the sign.
Calculate the E°cell for the following voltaic cell (use the standard electrode potentials given on page 829 of the textbook, or from some other source):
Ni(s) + 2 Ag+(aq) → Ni2+(aq) + 2Ag(s)
======================
I got -0.55 as my answer.
Ecell is Ecath-Eanod
The Ni part is oxidation/anode
The Ag is reduction/cathode
when i look up their values i get
-0.25-0.80
is this right?
1 answer
3 answers