Question

1. Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K.

A. Fe(s)+Ni2+(aq)?Fe2+(aq)+Ni(s)

B.Co(s)+2H+(aq)?Co2+(aq)+H2(g)

C.10Br?(aq)+2MnO?4(aq)+16H+(aq)?2Mn2+(aq)+8H2O(l)+5Br2(l)

2. If the equilibrium constant for a two-electron redox reaction at 298 K is 1.8×10^?4, calculate the corresponding ?G? and E?cel under standard conditions.

2A. Express your answer using two significant figures. Delta G=kJ

2B. Express your answer using two significant figures. E cell= V

Answer 1

1)

A   Fe(s) + Ni²⁺(aq)               Fe²⁺(aq) + Ni(s)

Fe     Fe²⁺ + 2e^-     E= -0.45 V

Ni²⁺ + 2e^-      Ni    E = -0.26

E^ocell = Ecathode – Eanode = -0.26 –(-0.45) = 0.19 V

We have,

E^ocell = (RT/nF) x lnKeq

0.19 = [(8.314 x 298)/(2 x 96485)] x lnKeq

Keq = 2.6 x 10⁵

B   Co(s) + 2H⁺(aq)               Co²⁺(aq) + H₂(g)

Co    Co²⁺ + 2e^-     E= -0.28 V

2H⁺ + 2e^-     H₂    E = 0

E^ocell = Ecathode – Eanode = 0 –(-0.28) = 0.19 V

We have,

E^ocell = (RT/nF) x lnKeq

0.28 = [(8.314 x 298)/(2 x 96485)] x lnKeq

Keq = 2.3 x 10⁹

C   10Br^-(aq) + 2MnO₄^-(aq) +16 H⁺               2Mn²⁺(aq) + 8H₂O(l) + 5Br₂(l)

MnO4^-(aq) + 8H⁺(aq) + 5e^-     Mn²⁺(aq) + 4H2O(l)      E= 1.51 V

5Br-                    5Br + 5e^-                      E = 1.065 V

E^ocell = Ecathode – Eanode = 1.51 -1.065 = 0.445 V

We have,

E^ocell = (RT/nF) x lnKeq

0.445 = [(8.314 x 298)/(2 x 96485)] x lnKeq

Keq = 1.8 x 10⁷⁵

2)

We have,

G^o = -RT lnK = - 8.314 J/mol K x 298 K x ln (1.8 x 10-4) = 21.36 kJ

Also,

G^o= - nFEo

Here, n=2

Therefore,

G^o = -2 x 96500 x E^o

21.363 x1000 J = -2 x 96484 x E^o

E^o= -0.11 V