Question

Question 1)

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Pb(s) and Sn2+(aq) to give Sn(s) and Pb2+(aq) Use the reduction potential values for Sn2+(aq) of -0.14 V and for Pb2+(aq) of -0.13 V Question 2) What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Cr(s) and Cu2+(aq) to give Cu(s) and Cr3+(aq) Use the reduction potentials for Cr3+(aq) is -0.74 V and for Cu2+(aq) is +0.340 V. could someone please help me with these? thank you very much

Answer 1

1)

Pb(s) + Sn+2(aq) -------------------- Pb+2(aq) + Sn(s)

according to equation Pb undergoes oxidation and Sn undergoes reduction

oxidation reaction at anode          Pb(s) -------------------- Pb+2(aq) + 2e-

reduction reaction at cathode    Sn+2(aq) + 2e- ------------------ Sn(s)

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

                                                    Pb(s) + Sn+2(aq) ---------------- Pb+2(aq) + Sn(s)

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

E0 of Sn+2/Sn = - 0.14V                      E0 of Pb+2/Pb = - 0.13V

E0cell = E0 cathode - E0 anode

E0cell = - 0.14V - ( -0.13)

E0cell = - 0.01V

number of electrons transferred = n= 2

Faraday = F= 96500 coloumbs

$\Delta$ G0 = - nFE0cell

$\Delta$ G0 = - 2 x96500 x( -0.01)

$\Delta$ G0 = 1930 J = 1.930 KJ

$\Delta$ G0 = - RT lkK

R = 8.314x10^-3 KJ              T= 25C = 25+273= 298K

1.930 = - 8.314x10^-3 x 298x lnK

lnK = - 0.779

K= e^-0.779

K = 0.459

Equilibrium constant = 0.459

2)

Cr(s) + Cu+2(aq) -------------------- Cr+3(aq) + Cu(s)

oxidation at anode         [Cr(s) ------------------- Cr+3(aq)   + 3e-]x2

reduction at cathode     [ Cu+2(aq) + 2e ---------------- Cu(s)]x3

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

                             2 Cr(s) + 3 Cu+2(aq) ----------------- 2 Cr+3(aq) + 3 Cu(s)

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

E0 of Cr+3/Cr= - 0.74V                     E0 of Cu+2/Cu = +0.34V

E0cell = E0cathode - E0 anode

E0cell = - 0.74 - ( 0.34)

E0cell = - 1.08V

number of electrons transferred = n= 6

F= 96500 coloumbs

$\Delta$ G0 = - nFE0cell

$\Delta$ G0 = - 6 x96500 x ( - 1.08)

$\Delta$ G0 = - 625320 J

$\Delta$ G0 = - 625.32 KJ