More finals review please...
Ag^+ + 2e^- --> Ag E^nought = 0.800V
Sn^4+ + 2e^- --> SN^2+ E^nought + 0.154V
balanced equation:
2Ag^+ + Sn^4+ --> 2Ag + Sn^2+
E^nought cell = cathode - anode
I thought I heard say that I had to flip the "less negative" half reaction (here the Ag). So then do I also flip the sign on the voltage so that 0.800 becomes -0.800?
Then my E^nought equation becomes:
(-.800)-(.154)
and that can't be right because I want E^nought to be positive! Did I hear something wrong there?
Thank you!
I HAD to have actually heard "flip the less positive" didn't I? When I work it that way it seems to make sense.
The equation as you have it written isn't right. It doesn't balance by charge: +6 on the left and +2 on the right.
If you want the cell to be spontaneous as written then you write the more positive (less negative) REDUCTION first, then reverse the other one (and change the sign). So the equations are
Ag^+ + e ==> Ag Eo = 0.800
Sn^2+ ==> Sn^4+ + 2e Eo = -0.154
---------------------------------
Add equations. Add Eo.
I add the oxidation half equation to the reduction half equation to arrive at the Eocell.
If you "flip the less positive" (which is Sn in this case) you're doing the same thing. If you want Eocell you can do Eanode-Ecathode (both as reductions) = 0.800 -(+0.154) = 0.800-0.154 = ? and you arrive at the same number as if you add 0.800 + (-0.154)
Thank you for explaining! I think the balancing bit is my fault. I am getting tired and stupid. Time to go to bed and look at this closer in the morning.
I thank you for all your help this semester.
You are correct that when constructing the cell potential equation, you should consider the half-reaction with the less negative standard reduction potential (E°) as the cathode (reduction) and the other half-reaction as the anode (oxidation).
In your case, the half-reaction with Ag+ has a standard reduction potential of 0.800V, which is more positive than the Sn4+ half-reaction with a standard reduction potential of 0.154V. Therefore, the Ag+ half-reaction will be the cathode and the Sn4+ half-reaction will be the anode.
However, there is no need to flip the sign of the reduction potentials. The sign of the standard reduction potential already indicates the direction of the reaction spontaneously occurring at standard conditions. Therefore, you don't need to change the sign of 0.800V or 0.154V.
To calculate the standard cell potential (E°cell) of the overall reaction:
E°cell = E°cathode - E°anode
= 0.800V - 0.154V
= 0.646V
So, the standard cell potential (E°cell) for the balanced equation 2Ag+ + Sn4+ → 2Ag + Sn2+ is 0.646V.