[yanna.banana]

Consider the following standard reduction potentials:
E°(A/B) = 0.60V
E°(B/C) = 1.25 V
Calculate the different species equilibrium concentrations when a 0.1M B solution is prepared. Assume both systems exchange one electron.
So i think A/B is being reduced and B/C is being oxidised.

I wrote logK= E°rd-E°ox/0.059 * n1n2 
which i got 1.25-0.6/0.059 *1 = -8.92
so then I guess  K=10^-8.92
Now is where I get confused:
my prof wrote
C=[ B]+[ B](takes off A) + [ B] (takes off C)
C=[ B]+[A]+[C]  = [ B] +2[A] --> [ B] = C-2[A]
then i really don't understand

Keq=[A]^2/[C-2[A]^2] = [A]^2/(C^2+4[A]^2-4C[A]

KC^2+4K[A]^2-4CK[A]-[A]^2 = 0
[A]^2[4K-1]-4[CK[A] + KC^2 = 0
[A] = 0.05M = [C]
Now we have A we can calculate B ?
I am just not following the logic here. I don't know if someone can link me to a video or a page where I can better understand? The problem is this class is not in my language (english) and so I am really struggling to follow. the answer she posted 1.54*10^-7 M

[mjc123]

I wrote logK= E°rd-E°ox/0.059 * n1n2 
which i got 1.25-0.6/0.059 *1 = -8.92

Elementary mistake! Brackets! You shouldn't be doing that at this level.

C=[ B]+[ B](takes off A) + [ B] (takes off C)
C=[ B]+[A]+[C]  = [ B] +2[A] --> [ B] = C-2[A]

I have no idea what this is about.
You start with only B, so A and C can only come from B.
2B  A + C
If you calculate K correctly it is very large, so assume almost complete reaction and let [A] = [C] = 0.05-x and [B ] = 2x
See where that gets you.