Question

Calculate the cell potential, ? , for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (?∘red) . Pt(s)+Fe2+(aq)↽−−⇀Pt2+(aq)+Fe(s) [Fe2+]=0.0045 M[Pt2+]=0.022 M ?= V The cell is spontaneous. not spontaneous. Cu(s)+2Ag+(aq)↽−−⇀Cu2+(aq)+2Ag(s) [Cu2+]=0.013 M[Ag+]=0.013 M ?= V The cell is spontaneous. not spontaneous. Co2+(aq)+Ti3+(aq)↽−−⇀Co3+(aq)+Ti2+(aq) [Co2+] = 0.055 M[Co3+] = 0.020 M [Ti3+] = 0.0055 M[Ti2+] = 0.0090 M ?= V The cell is not spontaneous. spontaneous.

Answer 1

Pt(s) + Fe2+(aq)↽−−⇀Pt2+(aq)+Fe(s)

Eocell = Eored - Eooxidation

           = - 0.44 - (1.18)

           = - 1.62 V

Ecell = Eo - 0.05916 / 2 log [Pt2+ / Fe2+]

         = - 1.62 - 0.05196 / 2 log [0.022 / 0.0045]

E = -1.64 V

non spontaneous

Cu(s)+2Ag+(aq)↽−−⇀Cu2+(aq)+2Ag(s)

Eocell = 0.80 - 0.34 = 0.46 V

E = Eo - 0.05916 / 2 log [0.013 / 0.013^2]

E = 0.404 V

spontaneous

Co2+(aq)+Ti3+(aq)↽−−⇀Co3+(aq)+Ti2+(aq)

Ecell = - 0.85 - 1.92 = -2.77 V

Q = [0.020] [0.0090] / [0.055] [0.0055]

   = 0.595

E = Eo - 0.05916 / 1 log Q

   = -2.77 - 0.05916 / 1 log 0.595

E = - 2.76 V

non spontaneous