The quantity of antimony in a sample can be determined by an oxidation-reduction

Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 5.97-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl and passed over a reducing agent so that all the antimony is in the form Sb3+. The Sb3+ is completely oxidized by 30.8 mL of a 0.110 M aqueous solution of KBrO3.

The unbalanced equation for the reaction is:
   BrO3- + Sb3+ -----> Br- + Sb5+

a) Calculate the potential of the solution after 15 mL of KBrO3 is added

b) Calculate the potential after 45.8 mL of KBrO3 is added

Explanation / Answer

a) Eo = Ecathode - Eanode

          = 1.44 - (-0.59)

          = 2.03 V

moles of BrO3- added = 0.110 M x 15 ml = 1.65 mmol

moles of BrO3- for complete oxidation = 0.110 M x 30.8 ml = 3.388 mmol

moles of Sb3+ present = 3.388 mmol

[Sb3+] remaining = (3.388 - 1.65)/15 = 0.116 M

[Sb5+] formed = 1.65/15 ml = 0.0.11 M

E = Eo - 0.0595/n log([Sn3+]/[Sn5+])^3

   = -0.59 - 0.0595/2 log(0.116/0.11)^3

   = -0.592 V

(b) after 45.8 ml KBrO3 added

moles of BrO3- added = 0.110 M x 45.8 ml = 5.04 mmol

Excess BrO3- = (5.04 - 3.388)/45.8 = 0.036 M

formed [Br-] = 3.388/45.8 = 0.074 M

E = Eo - 0.0595/n log([Br-]/[BrO3-])

   = 1.44 - 0.0595/6 log(0.074/0.036)

   = 1.437 V

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