Chemistry pre-Lab 12

Question

a. Balance the following reaction in an acidic solution.

Fe2+ + Cr2O7^2- → Fe3+ + Cr3+

b. When a 0.80 g sample of an unknown Fe2+ salt is titrated with 0.0285 M K2Cr2O7, it is found that 59.2 mL of this dichromate solution is required to reach the endpoint. Calculate the percent iron in the unknown using the balanced equation from part 1/a. Show work!

Answer 1

Fe²⁺ + Cr₂O₇^2- → Fe³⁺ + Cr³⁺ .. unbalanced redox equation

Fe²⁺ goes to Fe³⁺ so it has been oxidized

Cr goes from Cr⁶⁺ to Cr³⁺ so it has been reduced

Using the half reaction method, we have the following:

Oxidation half reaction:

Fe²⁺ ==> Fe³⁺ .. unbalanced

Fe²⁺ ==> Fe³⁺ + e- .. balanced for mass and charge = balanced equation

Reduction half reaction:

Cr₂O₇^2- ==> Cr³⁺ .. unbalanced

Cr₂O₇^2- ==> 2Cr³⁺ .. balanced for Cr

Cr₂O₇^2- ==> 2Cr³⁺ + 7H₂O .. balanced for Cr and O

Cr₂O₇^2- + 14H⁺ ==> 2Cr³⁺ + 7H₂O .. balanced for Cr, O and H (by using acid, H⁺)

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O .. balanced for Cr, O, H and charge = balanced equation

Since oxidation half reaction has only 1 electron and reduction has 6, we will multiply oxidation by 6

6Fe²⁺ ==> 6Fe³⁺ + 6e- .. balanced oxidation

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O .. balanced reduction

Adding them together and combining/canceling like terms, we end up with ...

6Fe²⁺ + Cr₂O₇^2- + 14H⁺ ==> 6Fe³⁺ + 2Cr³⁺ + 7H₂O .. balanced redox equation

moles of K₂Cr₂O₇ used = 0.0285 mol / L x 0.0592 L = 1.687x10^-3 mols

moles Fe²⁺ present = 1.687x10^-3 mols K₂Cr₂O₇ x 6 mols Fe²⁺ / 1 mol K₂Cr₂O₇ = 0.0101 mols Fe²⁺

grams Fe²⁺ present = 0.0101 mols Fe²⁺ x 55.85 g/mol = 0.565 g

% Fe in unknown = 0.565 g / 0.80 g (x100%) = 70.7% = 71% (2 sig. figs.)