Use the following Latimer Diagram to learn some descriptive chemistry of sulfur.

Question

Use the following Latimer Diagram to learn some descriptive chemistry of sulfur. Identify which, if any, species is(are) unstable WRT hydrogen gas evolution. Identify which, if any, species is(are) unstable WRT oxygen gas evolution. Identify which, if any, species is (are) unstable WRT. Write a balanced net ionic reaction for any one reaction satisfying "a" above. If none, Write NONE. Write a balanced net ionic reaction for any one reaction satisfying "b" above. If none, Write NONE. Write a balanced net ionic reaction for any one reaction satisfying "c" above. If none, Write NONE. Consider IE's (kJ/mol) of the Gr. 15 atoms: N (1400), P (1012), As 947), Sb (834), and Bi (703). Explain the expected order. Account for any deviation (s) from the expected order in terms of any applicability Network Ideas. What, if any, rationale beside the Network Idea can be invoked to account for and BiCI_3 being favored over SbCl_5 and BiCl_5, respectively? chi E (eV) of Gr. 15 elements: N (3.0), P (2.1), As (2.0), Sb (1.9), and Bi (1.9). What is the expected general trend for this periodic property? Account for these values in terms of any and all Network of Ideas. What is/are the reason (s) for the significant extent of catenation of C whereas that of Si is very limited? Account for this in terms of Network Ideas in light of the following data: C--C 356 kJ/mol Si--Si 226 kJ/mol C--C 413 kJ/mol Si = Si 298 kJ/mol C--0 356 kJ/mol Si--Si 368 kJ/mol

Explanation / Answer

a. If any of the couple in the diagram can reduce water, then H2 gas will evolve.

2H2O   +   2e   =   H2 (g) +   2OH- ; E0 = -0.829 V

If any of the couple has E0 < -0.829 V, then it can reduce water. As none of the couple has E0 < -0.829 V, so there is not any species which is unstable WRT hydrogen gas evolution.

b. If any of the couple can oxidise water, then O2 gas will evolve.

H2O = O2(g) + 4 H+ + 4e               ; E0 = 1.23 V

If any couple has E0 > 1.23 V, then it an oxidise water. From the Latimer diagram we see that :

E0S2O82-/(SO4)2- = 2.01 V > E0O2/H2O= 1.23 V

Also, for the couple, E0S2O82-/(S2O6)2- = (2.01+0.57)/2 = 1.29 > E0O2/H2O= 1.23 V

So S2O82- is unstable WRT oxygen gas evolution.

c. If the potential to the left of any species is less than the potential right to it, then the species will be unstable with respect to disproportionation. Here H2SO3 is is unstable with respect to disproportionation.

SO32- = S2O62- + S2O32-                  ;E0 = 0.88-(-0.08) = 0.96 >0

SO32- = S2O62- +   S                         ; E0= [(0.88+0.50)/2] - (-0.08) = 0.77 > 0

SO32- = S2O62- + H2S                      ; E0= [(0.88+0.50+0.14)/3] - (-0.08) = 0.587 > 0

d. NONE (As there is no such reaction)

e. Cathode:   2S2O82- + 4e = 4 SO42-

Anode:         H2O = O2(g) + 4 H+ + 4e

_____________________________________

Net Ionic reaction: 2S2O82- + H2O = 4 SO42- + O2(g) + 4 H+

f. Cathode: H2SO3 + 4e + 4H+ = S + 3H2O

Anode: 2 H2SO3 + 2H2O = 2SO42- + 4e + 8H+

_________________________________________

Net reaction:   3 H2SO3 = H2O + 4H+ + 2SO42- + S

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.