Question

The cell potential of this cell is observed to be 0.445v Calculate the pH for the H+/H2 half cell
please explain the use of the half reaction in terms of the signs, positive and negative in the equations.

The cell potential of this electrochemical cell is observed to be 0.445 v. Calculate the pH in for the H+H2 half cell Zn (s)| Zn2+ (2.5 M) || H*(??)[H2 (g. 4.5 atm)|Pt(s)

Answer 1

224 +2e n E = 0.76 v Chell potential values ) 249 +2€ - He & FOU potential and cell eletrods. Now Coming to cell Anode (oxidation) 20(6) 02 the Catmale (recentions) 24ttre H2(g) Net (6) + 2 H+ Halg) + 212 Reaction in in CSRP) a= [m?+] [ Pne) E cell = Stenadared Redinha Potential SRP (Anode) SRP E cell = (Cathode) - 2.5* 4.51 E call = +0.760 0.0 - 0.76 V 0.445 VE Cut ): Cena o-(-0.76V) / E = E cell - 0.0591 log (a) 1945V = 0.36 v - 0.8591 Av. ( 26 ) 70-3152 t Biossa long 10 ( high ) 10-6588 = Moyse ( bolovu in CH+]= [ht] - 1.54 x 105 m. Now - s-dog (1.543) = 5-0.19005 10.6598 o - 2.24 * do M 10:25 10.6598 - 2-24 x 1 IpH = 4.809