10 mL of 3 M sulfuric acid and 10 mL of 3 M phosphoric acid are added into a 50

Question

10 mL of 3 M sulfuric acid and 10 mL of 3 M phosphoric acid are added into a 50 mL FeSO_4 7H_2O solution. The concentration of FeS0_4, 7H_2O is 5.6 grams per liter. This solution is then titrated with 0.20 M KMnO_4, and 9.95 mL of KMnO_4 are required to reach the endpoint. 50 mL of SnCl_2, 2H_O are titrated with 0.20 M KMnO_4. The concentration of SnCl_2 2H_2O is 2.83 grams per liter. The amount of KMnO_4 added was 9.7 mL. For each reaction, answer the following questions. Determine the number of moles of electrons lost when one mole of the reducing agent is oxidized. Determine which atom in the reducing agent is undergoing a change in oxidation number. Calculate the initial and final oxidation numbers of that atom. Determine the final products derived from the reducing agents. Write a balanced half-reaction for the oxidation reaction, and write an overall balanced redox reaction equation for the reaction.

Explanation / Answer

Experiment 1

3. when 1 mole of reducing agent Fe2+ is oxidized,

moles of electron = 1 x 1 = 1 mole of e-

2. Reducing agent atom : Fe2+

a. Initial oxidation number = +2

Final oxidation number = +3

3. Final product formed from the reducing agent : Fe2(HPO4)2(SO4)

4. Oxidation reaction

Fe2+ <==> Fe3+ + e-

Overall balanced redox reaction

2KMnO4 + 10H3PO4 + 10FeSO4 <==> 2MnSO4 + 5Fe2(SO4)3 + K2SO4 + 2H2SO4 + 8H2O

Experiment 2

1. when 1 mole of Sn2+ is oxidized = 2 x 1 = 2 mole of e- is lost

2. Initial oxidation number = +2

Final oxidation number = +4

3. Final product for reducing agent being oxidized is : Sn4+

4. Oxidation half cell

Sn2+ <==> Sn4+ + 2e-

Overall balanced redox reaction

5SnCl2 + 2KMnO4 + 16HCl <==> 5SnCl4 + 2MnCl2 + 2KCl + 8H2O

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