Question

1)1. You are given the following cell at 25°C: Pb(s) | Pb2+(aq, 0.60M) || Sn2+(aq, 0.1 M), Sn4+(aq, 0.15 M) |Pt(s) Write the half reactions for the cell and using Appendix D in your text book, calculate the E°cell and the Ecell

oxidation half reaction = Pb_(s) ------------> Pb⁺²_(aq) + 2e^-

reduction half reaction = Sn⁺⁴_(aq) + 2e^- --------> Sn⁺²_(aq)

E⁰_cell = E⁰_{Sn⁺⁴/sn⁺²}   E⁰_Pb+2/Pb

E⁰_cell = 0.15 - (-0.126)

E⁰_cell = 0.276 V ( i'm not sure if it correct or not)

2) For the cell in #1 , compare the spontaneity of the cell as it is written to the cell under standard conditions.

3) From reading through the lab, describe the process of how you could determine if the reactions are driven by enthalpy or entropy. (Hint: how can you determine the sign of DS as you might do in this lab?)

Answer 1

oxidation half reaction = Pb_(s) ------------> Pb⁺²_(aq) + 2e^-        E0 = -0.126v

reduction half reaction = Sn⁺⁴_(aq) + 2e^- --------> Sn⁺²_(aq)                   E0 = 0.15v

                                        ------------------------------------------------------------------------

                                         Pb(s) + Sn^4+ (aq) ---------> Pb^2+ (aq) + Sn^2+ (aq)   E0cell = 0.025v

                      n = 2

    G⁰    = -nE0cell*F

                = -2*0.025*96500   = -4825J

   G⁰ < 0 it is spontaneous reaction

   Ecell   = E0 cell - 0.0592/n logQ

              = 0.025 -0.0592/2 log[Pb^2+][Sn^2+]/[Sn^4+]

               = 0.025 -0.0296log0.6*0.1/0.15

             = 0.025-0.0296log0.4

           = 0.025-0.0296*-0.3949   = 0.0367v

S >0