Iron(III) was reduced to iron(II) by chromium(III) in acidic solution according

Question

Iron(III) was reduced to iron(II) by chromium(III) in acidic solution according to the following unbalanced reaction: Cr^3+_(aq) + Fe^3+_(aq) rightarrow Fe^2+_(aq) + compound_(aq) The experiment requires 66.9 g of Fe^3+ to be completely reduced by 20.8 g of Cr^3+. Based on this information, what is the oxidation state of the chromium in the chromium oxide product? What is the possible chromium oxide product? CrO CrO_2 Cr_2O_3 Cr_2O^2+_4 Cr_2O^2-_7 Use the standard reduction potentials: to determine the net reaction and standard cell potential far the cell below

Explanation / Answer

66.9 g of Fe3+ = 66.9 g / 55.8 g/mol = 1.2 mol

20.8 g of Cr3+ = 20.8 g / 51.9 g/mol = 0.4 mol

So balanced equation should be

1 Cr3+ + 3 Fe3+ ----> 3 Fe2+ + 1 Cr6+

the only option e) Cr2O72- has Cr6+

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