Dichromate and ferrous ions react in acidic solution to form chromic and ferric

Question

Dichromate and ferrous ions react in acidic solution to form chromic and ferric ions, respectively. If 1.285 grams of iron(II) bisulfate dissolved in sulfuric acid solution requires 35.78 mL of sodium dichromate solution for complete titration, what is the molarity of the sodium dichromate solution? Show the balanced net ionic and complete chemical equations for this reaction.

102 .. come of the 103 QUESTIONS FOR RE-DOX EXP. NAME 1. Dichromate and ferrous ions react in acidic solution to form chromic and ferric ions, respectively. If 1.285 grams of iron(iI) bisulfate dissolved in sulfuric acid solution requires 35.78 ml of sodium dichromate solution for complete titration, what is the molarity of the sodium dichromate solution? Show the balanced net ionic and complete chemical equations for this reaction. 2. The sodium dichromate solution from problem 1 was used to titrate a solution made by dissolving 3.500 g of a pure ferrous salt in sulfuric acid dichromate solution. Calculate the percent by mass of iron in the pure sal g of a pure ferrous salt in sulfuric acid. The titration required 47.22 ml of the sodium t. The net ionic equation here is the same as in problem 1. You will not be able to write a balanced molecular equation for this because the anion in the ferrous salt was not specified in this problem.

Explanation / Answer

(1) write out the two half-reactions

Redcution Half reaction (Cr2O7)2- + 14H+ + 6e- = 2Cr3+ + 7H2O

Oxidation half reaction Fe2+ = Fe3+ + e-

The number of electrons must be equal on both sides, so the second equation must be multiplied by 6. Also in acidic conditions balance the equation by adding H+ to whatever side needed and half of the number of H+ on the other side of reation. Therefore the net equation is written as follows :

(Cr2O7)2- + 14H+ + 6e- +6Fe2+ = 2Cr3+ + 7H2O +6Fe3+ + 6e-

The net ionic equation is then

(Cr2O7)2- + 14H+ +6Fe2+ = 2Cr3+ + 7H2O +6Fe3+

Now the molarity of Na2Cr2O7 = ?

According to the net equation the number of moles of Na2Cr2O7 is equal to (number of moles of FeSO4) / 6 ,so lets evaluate the number of moles of FeSO4 first as follows :

n = weight/ Molecular weight

n(FeSO4) = 1.285 / 152 = 0.00845 mol

Now divide the n(FeSO4) /6 = n(Na2Cr2O7)

n(Na2Cr2O7) = 0.00140899122 mol

Hence molarity = n(Na2Cr2O7) / volume of solution in litres

Cm(Na2Cr2O7) = 0.00140899122 / 0.03578 = 0.0339 M

(2)   n(Na2Cr2O7) = 0.04722 * 0.0339 = 0.0016 mol
n(FeSO4) = 0.0016 * 6 = 0.0096 mol
m(Fe) = M(Fe) / M(FeSO4) * m
m(Fe) = 56/152 * 3.5 = 1.288 g
%(Fe) = 1.288 / 3.5 * 100 % = 36.8 %

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