1) The quantity of antimony in a sample can be determined by an oxidation-reduct

Question

1) The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 8.85-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3 (aq). The Sb3 (aq) is completely oxidized by 43.7 mL of a 0.115 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3^- (aq) + Sb^3+ (aq) yields Br^- (aq) + Sb^5+ (aq)

Calculate the amount of antimony in the sample and its percentage in the ore.

2) A galvanic (voltaic) cell consists of an electrode composed of iron in a 1.0 M iron(II) ion solution and another electrode composed of copper in a 1.0 M copper(I) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C. Standard reduction potentials can be found here.

Explanation / Answer

3Sb3+ + BrO3- + 6H+ ---------------------> 3Sb5+ + Br- + 3H2O

moles of BrO3 - = 0.115 x 43.7 /1000 = 5.02 x 10^-3

moles of BrO3- = 3 moles of Sb+3

Sb+3 moles = 1/3 x 5.02 x 10^-3

                     = 1.675 x 10^-3

molar mass of Sb = 121.8 g

mass of Sb+3 = 121.8 x 1.675 x 10^-3 = 0.204 g

antimony in the sample = 0.204 g

% of antimony in the sample = (0.204 / 8.85 ) x100

                                               = 2.3%

(2)

Fe+2 + 2e- ---------------------> Fe , E(o) = - 0.44V

Cu+   + e- -------------------> Cu , E(o) = +0.52 V

E(o)cell    =   E(o) reduction - E(o) oxidation

              = 0.52 - (-0.44)

                = 0.96 V

standard potential for this cell = 0.96 V

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