Question

Calculate ?G?rxn and K for each of the following reactions.

Part A

The reaction of Cr2+(aq) with Cr2O2?7(aq) in acid solution to form Cr3+(aq).
Calculate ?G?rxn.

Part B

Calculate K.

Express your answer using two significant figures.

Part C

The reaction of Cr3+(aq) and Cr(s) to form Cr2+(aq). [The reduction potential of Cr2+(aq) to Cr(s) is -0.91 V.]
Calculate ?G?rxn.

Express your answer using two significant figures.

Part D

Calculate K.

Express your answer using one significant figure.

Answer 1

Part 1)

6Cr²⁺ + Cr₂O₇^2- + 14H⁺ 8Cr³⁺ + 7H₂O

E^ocell = 1.33 - (-0.5) = 1.83 V

We have, Nernst equation,

G^o = -nFE^o = -6 x 96485 x 1.83 = -1059405.3 J/mol = -1059.4 kJ/mol

G^o = -1059.4 kJ/mol

At 25 ^oC

G^orxn = -RTlnK

lnK = G^o/RT = -(-1059.4KJ/mol) /(8.314 x 298) = 427.6

Therefore, K = 5.2 x 10¹⁸⁵

Part 2)

Reduction half is

Cr³⁺ (aq) + e^-          Cr²+ (aq) E^o(red) = -0.5 V

Oxidation half is

Cr(s)                       Cr²⁺(aq) + 2e^-          Eo(ox) = -(-0.91 V) = 0.91

Combining the oxidation half and reduction half,

2Cr³⁺(aq) + Cr(s)                        3Cr²⁺(aq) Eo(cell) = 0.91 V -0.50 V = 0.41 V

G^o = -nFE^o = -2 x 96485 x 0.41 = -79117.7 J/mol = -79.11 kJ/mol

G^o = -79 kJ/mol

G^orxn = -RTlnK

lnK = G^o/RT = -(-79.11KJ/mol) /(8.314 x 298) = 0.0319

Therefore, K = 1.0324 = 1.0    ( 2 significant figure)