Determining iron(III) content in mixture of iron(II) and iron(III) [closed]

Question

Problem

Outline a plan of an experiment to determine the percentage of iron present as iron(III) in a solution containing $\ce{Fe^3+(aq)}$ and $\ce{Fe^2+(aq)}$ ions. You are provided with zinc, a standard solution of potassium dichromate(VI) and dilute sulfuric acid. Zinc can reduce $\ce{Fe^3+(aq)}$ to $\ce{Fe^2+(aq)}.$

Answer

1. Titrate measured volume solution against $\ce{K2Cr2O7}.$
2. Reduce same volume solution with zinc.
3. Filter off excess zinc.
4. Titrate total $\ce{Fe^n+}$ using $\ce{K2Cr2O7}.$ $$x(\ce{Fe^3+}) = \frac{c_2 - c_1}{c_2}\times 100\,\%$$

Questions

Firstly, why is the excess zinc filtered? $\ce{Fe^2+}$ has been oxidised to $\ce{Fe^3+}$ first by dichromate, and then we took a different aliquot and determined the amount of $\ce{Fe^3+}$ by reduction using zinc.

Secondly, the “titrate total $\ce{Fe^n+}$ using $\ce{K2Cr2O7}$” step confuses me. Would this be done in a volume that was the sum of the two aliquots taken previously? If so, does this mean I would have the amount of $\ce{Fe^2+}$ present in solution, yet the step using zinc would be made redundant?

Answer 1

Suppose your solution contains $a$ mol $\ce{Fe^{2+}}$ and $b$ mol $\ce{Fe^{3+}}$. Titration with $\ce{K2Cr2O7}$ gives you $a$. To obtain $b$ you first add some zinc powder, before titration by $\ce{K2Cr2O7}$. Zinc will reduce all the $\ce{Fe^{3+}}$ ions according to $$\ce{2 Fe^{3+} + Zn -> 2 Fe^{2+} + Zn^{2+}}$$ So the solution contains now some $\ce{Zn^{2+}}$ ions which do not interfere later on. But it contains an amount of $\ce{Fe^{2+}}$ equal to $a + b$. So titration of this second solution requires more $\ce{K2Cr2O7}$ than previous titration. This second titration by dichromate ions gives you $a+b$.

To summarize, first titration gives $a$. Second titration gives $a+b$. Substraction gives you $b$. And the problem is solved.

But the second titration must be done without any residual zinc metal, because zinc may also be oxidized by $\ce{K2Cr2O7}$ according to : $$\ce{ 3 Zn + Cr2O7^{2-} + 14 H^+ -> 2 Cr^{3+} + 3 Zn^{2+} + 7 H2O}$$ If you titrate the second solution without removing the zinc metal, you will need a huge amount of dichromate, just for destroying all the excess of zinc metal added.