Question

1. (a) Develop a pE/pH expression for S⁰/H₂S given that [H₂S] = 1x10^-2M

S⁰ + 2H⁺ + 2e ↔ H₂S, E⁰ = 0.141 V

(b) Determine the E⁰(w) for the half reaction.

please answer ASAP will give immediate thumbs up!

Answer 1

ox. bH - 0.059 + E'$/bs - 15/1125 | As.- LoqW"): ") 0.059 ox. BP - 0.059 +0.141- Esilts 0.059. px 1H= .0.2-ES/ H2S 0.059. :: pH: 0.2 – Esilns 0.059. (6) Has molecule can easily dissociate is 1 H₂S Ž 2H+ +5= of C 20 c. d's wnanfration of 165 = 1x1074m), .; con confratim Of HT - 2x1072m). $- wglu*] =- wg2x1074) = - wg(2) - logiese = 2-wg2 1.698 Froom the above caniation, we can en calentate Es/has.. - Now given that. ph 10698 : 1.698 = 0.2 - Esitys - 0.059 - 0.2 – Eslins 3 1698X0.059 - 0-100 - ES/ H23 - 10.2-0.1) = 0.100 volti ; Ęsti s = 0.1 voru a cs Scanned with CamScanner

Let consiаue a reaction: Mtn +ne ² M The Nerenst eanatim for the above reactionis Ecell = Ecell - RT IN [M] - Mn] = Eceu - 23036T log on thing = Ecel - 0.059 og LM 1 at 25°c. (a) now consider the follning reaction 30+ 2H+ +26H H25 Here given that 2. Ecevi 0.141 v [ H25] = 1x 10-2 M. [3] . I Goitis in elemental state Now Nerenst eanation is Eggs : Egjiths - 0.05 wg has ng ... (Esrags – Eeyms) - - 0059 vogliono slike) = -log[155] + q log [te) or (E5/125 - Es/its) [As [3] : 1] 0.0295 * £3/125 Esilks 2 log10 +2 wglat 0.0295 es Eslins - Esiks 1+ loghat 0.059 -lugli) = 1- nay Scan be 0.059 – Esllbs + Ešliks CamScanner - 0.059 371ks 0.059-