Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1. Using the standard reduction potentials listed in Appendix E in the textbook,

ID: 1059428 • Letter: 1

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

Explanation / Answer

1. Equilibrium constant

A. Eo = Ecathode - Eanode

          = -0.25 - (-0.44)

          = 0.19 V

nFEo = RTlnKeq

2 x 96485 x 0.19 = 8.314 x 298 lnKeq

Keq = 2.67 x 10^6

B. Eo = Ecathode - Eanode

          = 0 - (-0.28)

          = 0.28 V

nFEo = RTlnKeq

2 x 96485 x 0.28 = 8.314 x 298 lnKeq

Keq = 2.96 x 10^9

C. Eo = Ecathode - Eanode

          = 1.51 - 1.066

          = 0.444 V

nFEo = RTlnKeq

10 x 96485 x 0.444 = 8.314 x 298 lnKeq

Keq = 1.24 x 10^75

---

2. Keq = 1.8 x 10^-4

dGo = -RTlnKeq

        = -8.314 x 298 ln(1.8 x 10^-4)

        = 21.36 kJ

nFEo = RTlnKeq

2 x 96485 Eo = 21363

Eo = 0.11 V

---