Compare your results between the KMnO_4. solution you prepared and the solution

Question

Compare your results between the KMnO_4. solution you prepared and the solution provided by the lab. Are they similar? If not, what do you think the reason is for the difference? What is the purpose of dissolving the iron supplement pill in H_2SO_4, and why is phosphoric acid added later on during the titration? Balance the following redox equation: Fe^2+(aq) + Cr_2O_7^2-(aq) + H^+ rightarrow Fe^3+(aq) + Cr^3+(aq) + H_2O. A solution of iron (II) sulfate was prepared by dissolving 10.00g of FeSO_4*7H_2O (FW = 277.9) in water and diluting up to a total volume of 250 mL. The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidized to ron(III) ions. A 25.0 mL sample of this partially oxidized solution required 23.70 mL of 0.0100 M potassium dichromate K_2Cr_2O_7 solution for complete reaction in the presence of dilute sulfuric acid. Calculate the percentage of iron(II) ions that had been oxidized by the air. You need to use the balanced redox reaction in question 3, in order to complete your calculations.

Explanation / Answer

1) Need data to answer the question.

2) H2SO4 is added to dissolve the solid supplement pill so that all the iron in the sample (Fe0) is converted to Fe2+. H2SO4 oxidizes Fe0 to Fe2+.

In this reaction, KMnO4 oxidizes Fe2+ to Fe3+. Fe3+ is yellow in colour. The colour change of the titration is indicated by the appearance of pink/purple end point at the equivalence point. Fe3+ can mask the pink colour at the end point of the titration and hence Fe3+ is complexed with phosphoric acid to form a white compound, [Fe(HPO4)]2+.

3) Fe2+ (aq) + Cr2O72- (aq) + H+ (aq) -------> Fe3+ (aq) + Cr3+ (aq) + H2O (l)

Change in oxidation number of Fe = 3 – 2 = 1

Change in oxidation number of Cr = 6 – 3 = 3

Total change in oxidation number for 2 atoms of Cr = 6

To balance the oxidation numbers of Fe and Cr, multiply Fe by 6 and Cr by 1. The reaction is

6 Fe2+ + Cr2O72- + H+ -------> 6 Fe3+ + 2 Cr3+ + 7 H2O

Balance the number of H atoms on both sides by multiplying the left by 14 and write

6 Fe2+ (aq) + Cr2O72- (aq) + 14 H+ (aq) -------> 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (l) (balanced)

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