Wednesday
May 22, 2013

Homework Help: Chemistry

Posted by Miaow on Sunday, May 20, 2012 at 10:17pm.

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!

• Chemistry - Miaow, Sunday, May 20, 2012 at 10:20pm

I HAD to have actually heard "flip the less positive" didn't I? When I work it that way it seems to make sense.

• Chemistry - DrBob222, Sunday, May 20, 2012 at 11:21pm

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
---------------------------------
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)

• Chemistry - Miaow, Sunday, May 20, 2012 at 11:32pm

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.

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