Question

SO₄^-2 + 8H⁺ + 6e^- <-> S_(s) + 4H₂O

Log K = 36.2

[SO₄^-2] = 10^-2

Solve for the relationship between Eh and pH, which would be the boundary between the two species on an Eh-pH diagram. Using the Nernst equation Eh=Emeasured-Ereference - where Eh is the electromotive force of any reaction measured relative to the standard hydrogen electrode ( Eref = 244.4 mV)

Answer 1

or a general reaction, aA + bB = c(+ dD - S4=84 + RT du a no - ufk= -nfeo + RT hea {= 4 - RI na Q- reactant quoti ent = aC a no C MF пе - 6 = 6 - RI mai ar replacing activities by concentration, on 2,303 RT 2- E-aus der er en toute (une 2013080] lun= 8.3 ne At 25°C, 090sti v 2,303 R1 = 2303 x P314 x 298 gouro => TF-fº- Dolly love (0306074 € Can 0 is called reelust equation Rxn Soy2 +out+ bé e-) S (s) + 4 40 Lop k = 36.2 [soy?]= 102 fry = 294.YmV = 012 44V1mV= los v] cs Scanned with CamScanner

[=&º- of one simb Csawit n=6 £= f- 00 991 lange multe cases - @ Loy K = 362 At quilibrium, cell stope work uif, t=0 fº=009 long Layk - ne 3612 x 090311 F (n=6) 0 6=0-356 V – ☺ Put ③ in eqni @) . E = 0.356 – 0.03€! by {4*3*8 (160) E- 01356-008985 [ Log lutje + leglar] [Lagab= Laga + ] logh J =) € = 0.356-0.60985(-8 log [4 +3 + 2] (lyate-blaga] . As ph=-Laght] E = 0.356 - 0.68 985 (8 pk +2] E = 0.356 - 090197 - 010788 p4 E = 0.3363-000788 pH As th= Enneasured - Ereference Eu = 6.33634 00 0788 pu) -0.0444 v (980.899y] - TEH = 0.0919V - 0.0788 px]@ the epith desnibe the relation between tu & pH Cs Cah'scanner