Question

17. Given the following two half-cell reactions and their half-cell potentials:

Fe³⁺ + e --> Fe²⁺ Eo = 0.68 V

Cr₂O₇^2- + 14H⁺ + 6e^– --> 2 Cr³⁺ + 7H₂O E° = 1.33 V

a. Write the balanced full-cell reaction for a spontaneous reaction and determine the Eo cell ([H+ ] = 1.0 M):

Balanced full-cell reaction:

Eo cell: Answer: 0.65 V (How they got that?)

b) What’s the equilibrium constant at room temperature ?

Answer: K = 8.66 x 10⁶⁵

How they got that?

Answer 1

From the reactions, note that the most positive will REDUCE, and the least positive will be OXIDIZED

the oxidized will be Fe3+ ; since E is low; we must invert the reactants/products and the E°value

Cr2O72- + 14H+ + 6e– --> 2 Cr3+ + 7H2O E° = 1.33 V

Fe2+ --> Fe3+ + e- Eo = -0.68 V

E = E°red + E°ox = 1.33-0.68 = 0.65 V

the balanced equation

Cr2O72- + 14H+ + 6e– + 6Fe2+ ---> 2 Cr3+ + 7H2O + 6Fe3+ + 6e-

cancel common terms

Cr2O72- + 14H+ + 6Fe2+ ---> 2 Cr3+ + 7H2O + 6Fe3+

b)

for K

E°cell = (RT)/(nF)*lnK

K = exp(n*F/(RT)*E°cell)

K = exp((6*96500)/(8.314*298)*0.65)

K = exp(151.9027) = 9.3433*10^65