Question

Based on the following information, estimate the Ksp of silver bromide at 25°C.
Ag+(aq) + e– → Ag(s) E° = 0.80 V
AgBr(s) + e– → Ag(s) + Br–(aq) E° = 0.07 V


(1) 5 x 10–13
(2) 2 x 10–25
(3) 2 x 10–15
(4) 4 x 10–30
(5) 3 x 10–11

I understand the answer is 1, 5*10^-13 but I was wondering why you would subtract 0.07 V-0.8 V and not 0.8 V-0.07? That part confuses me.

Answer 1

Ag+(aq) + e– → Ag(s) E° = 0.80 V
AgBr(s) + e– → Ag(s) + Br–(aq) E° = 0.07 V

You want this.
AgBr(s) ==> Ag^+ + Br^- so that
Ksp = (Ag^+)(Br^-).

So reverse eqn 1 and add it to eqn 2 to get this.
Ag(s) + e– → Ag^+ Eo=-0.80
AgBr(s) + e– → Ag(s) + Br–(aq) 0.07

You get this.
AgBr(s) ==> Ag^+ + Br^- and the Ecell is -0.80+0.07 = -0.73 etc.or if you've been taught Ecell = E2-E1 it is 0.07-0.80 = -0.73

nEF = RTlnK