Please help me. Thanks In this step, you add 16 M HNO_3 to the precipitate of St

Question

Please help me. Thanks In this step, you add 16 M HNO_3 to the precipitate of Step 5. Nitric acid is a strong oxidizing agent. As a result, the sulfide ions in the precipitate are oxidized to solid sulfur, and the nitrate ions in HNO_3 are reduced to NO_2(g), a toxic brown gas. This step must be conducted in the hood. Nitric acid should be added until all of the initial precipitate has dissolved. First, consider the reaction of HNO_3 with PbS, the precipitate formed with Pb^2+. Write a balanced oxidation half-reaction. The reactant should be PbS(s), and the products should include Pb^2+ (aq) and S(s). Write a balanced reduction half-reaction. The reactants should include HNO_3 (aq), and the products should include NO_2(g). Using the method of half-reactions, write a complete balanced chemical equation for the reaction between PbS(s) and HNO_3 (aq). Repeat parts (a), (b) and (c) for the sulfide precipitates of Bi^3+, Cu^2+ and Cd^2+ (Use the back of the page for more space if necessary.) Once this step is complete, what will be present in the aqueous phase, and what will be present in the solid phase?

Explanation / Answer

(a): Oxidation - half reaction: PbS(s)   ----- > Pb2+(aq) + S(s) + 2e-  

(b) Reduction - half reaction: HNO3(aq) + H+(aq) + e-  ----- > NO2(g) + H2O(l)

(c) By multiplying the reduction - half reaction by 2 and then adding will cancel out the electron and we get the complete balanced reaction.

PbS(s)   ----- > Pb2+(aq) + S(s) + 2e-  

[HNO3(aq) + H+(aq) + e-  ----- > NO2(g) + H2O(l)] x 2

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PbS(s) + 2HNO3(aq) + 2H+(aq)  ------ >  Pb2+(aq) + S(s) + 2NO2(g) + 2H2O(l) (answer)

(d): Bi3+:

Oxidation - half reaction: Bi2S3(s)   ----- > 2Bi3+(aq) + 3S(s) + 6e-  

Reduction - half reaction: HNO3(aq) + H+(aq) + e-  ----- > NO2(g) + H2O(l)

Complete balanced reaction: Bi2S3(s) + 6HNO3(aq) + 6H+(aq) ---- > 2Bi3+(aq) + 3S(s) + 6NO2(g) + 6H2O(l)

Cu2+:

Oxidation - half reaction: CuS(s)   ----- > Cu2+(aq) + S(s) + 2e-  

Reduction - half reaction: HNO3(aq) + H+(aq) + e-  ----- > NO2(g) + H2O(l)

Complete balanced reaction: CuS(s) + 2HNO3(aq) + 2H+(aq)  ------ > Cu2+(aq) + S(s) + 2NO2(g) + 2H2O(l)

Cd2+:

Oxidation - half reaction: CdS(s)   ----- > Cd2+(aq) + S(s) + 2e-  

Reduction - half reaction: HNO3(aq) + H+(aq) + e-  ----- > NO2(g) + H2O(l)

Complete balanced reaction: CdS(s) + 2HNO3(aq) + 2H+(aq)  ------ > Cd2+(aq) + S(s) + 2NO2(g) + 2H2O(l)

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