Question

help with these please

A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode PHlatm) immersed in a solution of unknown [H]. If the cell potential is 0.181 V, what is the pH of the unknown solution at 298 K? pH

Answer 1

Dear Student, Kindly refer to the solution of your first two questions. Feel free to ask if you face difficulty in understanding any step of the solution attached so that we can assist you in a better way.

Deax ltudent, ② As per the conditione mentioned the representation of cell is as followe :- pt(s) 1H (lath) 14+ (x mall) |/ ht (lmall) Il lato let The reactions involved - at anode: Hallam 24 (& mol/L) + 2é aut cathode: 24 t + dě Hllath) (1 mol/L) loverall Reaction e Hallatm) + 2H+ (imall) 24t (amall) + 4 (lath) since the too half cells are identical I Ecele - Red - E x = 0 Thus, acc. to Nernst equation, Ecell cell n u t 1² bu n Pred 20-0089/ bog Luty** feel = -0.059) long [h* 32 Ecell = -0.0891 x 2 log Eht 47 -0.0591 pro Р.Ј.0

[pH = -logi Ecele = 0.0591 pH Thus, pH = Ecel 0.osal pH = 0.181 0.0591 pH = 3.06 @ As per the conditions peovided, cell diagoaum. can be written as: - ptcod Halt atm) | H+(amol/L) |) ** (Imoll) 14. Clatan) + (g) Reactiones- at anode: H (latm) 24+ (x mol/ the at cethode: art (Imall) + 2e - Clathr) overall: HClathr) +24+ (mall) 24+ (amoll) +4(laten) I since the two half cell reactions are identical sole Eoed - ko x = 0 As pee Neenst Eq. Ecell - Feell - 0.0591 log [HT] Pria [++] ²po2 P.T.O.

be 0-oogallong Dut 3! Free - Bora x log Chtj? Belle -0.059) va log. Lht] free = 0.0591 PH PHE-log [ht]) pH = £ceb 0.0591 pH = 0.186 0.0591 PH= 3015 Best wishes!!