91Ó°ÊÓ

A redox titration happens when the analyte and the titrant undergo an oxidation–reduction process. The endpoint is frequently detected using an indicator, much as it is in acid–base titrations. Oxalic acid titrated against potassium permanganate in acid medium is an example of redox titration.

In Step 1, chromium above +3 state was used to oxidize $F{e}^{2+}$. By looking at the first reaction in step 1 , we can see that the ratio of moles of $C{r}^{4+}$and $F{e}^{4+}$is 1:1

$$\begin{gathered} c\left(F{e}^{2+}\right)=2.786mM=2.786×{10}^{-3}M \\ V\left(F{e}^{2+}\right)=0.498mL=0.498×{10}^{-3}L \\ m\left(crystal\right)=0.4375g \end{gathered}$$

Number of moles of $F{e}^{2+}$used in step is:

$$\begin{gathered} n\left(F{e}^{2+}\right)=c\left(F{e}^{2+}\right)×V\left(F{e}^{2+}\right) \\ n\left(F{e}^{2+}\right)=0.498×{10}^{-3}L×2.786×{10}^{-3}M \\ n\left(F{e}^{2+}\right)=1.387×{10}^{-6}mol \end{gathered}$$

Hence the number of moles of $F{e}^{2+}$is $1.387×{10}^{-6}mol$.

In Step 2 , the total chromium content was determined by using the same solution. The only difference is the ratio of moles. Each chromium reacts with three $F{e}^{2+}$.

$$\begin{gathered} c\left(F{e}^{2+}\right)=2.786mM=2.786×{10}^{-3}M \\ V\left(F{e}^{2+}\right)=0.703mL=0.703×{10}^{-3}L \\ m\left(crystal\right)=0.1566g \end{gathered}$$

The average oxidation number is:

Average oxidation number= role="math" localid="1664877813554" $$\begin{gathered} 3+\frac{n(Crabove3+)}{n(totalCr)} \\ 3+\frac{3.17×{10}^{-6}mol}{4.17×{10}^{-6}mol} \end{gathered}$$

= 3+0.76

=3.76

Hence the average oxidation number is

Total micrograms of chromium per gram of crystal can be calculated by using total moles and molar mass of chromium:

$$\begin{gathered} m(totalCrincrystal)=n(totalCr)â‹\dots M\left(Cr\right) \\ m(totalCrincrystal)=4.17×{10}^{-6}molâ‹\dots 51.996g/mot \\ m(totalCrincrystal)=2.17×{10}^{-4}g=217\mathrm{μ}g \end{gathered}$$

Hence the total amount of chromium is $217\mathrm{μ}g$.