# Redox titrations : Oxidation of metallic Iron by Potassium Dichromate

0.31 g of an alloy of Fe + Cu was dissolved in excess of dilute $$\ce{H2SO4}$$ and the solution was made upto 100 ml. 20 ml of this solution required 3 ml of $$\frac{N}{10} \ce{K2Cr2}\ce{O7}$$ solution for the oxidation. The percentage purity{ in closest value} of Fe in wire is ____.

I am approaching the problem with the following method. Let us assume weight of Fe in alloy to be x grams. Now my law of chemical equivalence we know that no. of equivalents of Fe is equal to the no. of equivalents of $$\ce{K2Cr2}\ce{O7}$$. To find the no. of equivalents of Fe we need the corresponding chemical equation which I tried to find on Google but in vain. I just want to know the oxidation state Fe would acquire. Then I can easily answer this problem by substituting 'n'(no. of electrons lost by Fe) in the equivalence equation for this particular problem : $$\frac{1000xn}{280} = \frac1{10}$$. From here I can easily calculate the percentage purity of Fe.

• Google cannot find these calculations, because such calculations are no longer taught so probably no one has webpages on this topic. They are obsolete (although some chemical industries still use it). Search normality calculations or see solved examples in your textbook which is still using normality. Sep 25 at 19:54
• You should be able to formulate the chemical equation of the reaction on your own, knowing the basics of chemistry of iron and chromium. Sep 25 at 19:55
• There is no problem in doing the mathematics. I just can't find the Reaction involved here. I will once again try to find it in my textbook. Thank you guys. Sep 25 at 19:58
• That was my point. You should not need to search for it. Hint: Cr: +VI->+III Fe: 0->+III Sep 25 at 19:59
• The beauty of normality calculations is that you do not need a chemical equation for this problem. Plug and chug in C1V1=C2V2. C1 is Fe conc. in normality, V1 is volume of iron sample, C2 is concentration of K2Cr2O7 in normality and V2 is the burette volume. Sep 25 at 19:59

Let's reason with molar units. Expressed in molar units, $$\ce{K2Cr2O7}$$ N/$$10$$ is $$\ce{K2Cr2O7}$$ M/$$60$$.
$$3$$ mL $$\ce{K2Cr2O7}$$ N/$$10$$ contains $$0.003/10$$ equivalent.
In molar units $$0.003/10$$ equivalent is $$0.003/60$$ mol $$\ce{K2Cr2O7}$$ = $$0.05$$ mmol $$\ce{K2Cr2O7}$$.
The equation of the titration becomes : $$\ce{Cr2O7^{2-} + 6 Fe^{2+} + 14 H+ -> 2 Cr^{3+} + 6 Fe^{3+} + 7 H2O}$$ Because of the coefficient $$6$$ in front of $$\ce{Fe^{2+}}$$ in the equation, $$0.05$$ mmol $$\ce{Cr2O7^{2-}}$$ has reacted with $$6$$ times more $$\ce{Fe^{2+}}$$ = $$0.3$$ mmol $$\ce{Fe^{2+}}$$. This amount of $$\ce{Fe^{2+}}$$ comes from the same amount of metallic Iron, i.e. $$\ce{0.3}$$ mmole $$\ce{Fe}$$, because the reaction of iron with a diluted acid is $$\ce{Fe + 2 H+ -> Fe^{2+} + H2}$$ But the initial solution of $$\ce{Fe^{2+}}$$ had a volume of $$100$$ mL, and only $$20$$ mL was used for titration. It means that the initial metallic sample contained $$5$$ times more iron than $$0.3$$ mmol, which is $$1.5$$ mmol $$\ce{Fe}$$. As $$1.5$$ mmol $$\ce{Fe}$$ weighs $$1.5·56$$ mg = $$84$$ mg, it means that the proportion of iron in the initial sample is $$\frac{0.084}{0.31} = 27.1$$%