Animated Solution for Chemistry - Basic Concepts in Chemistry: The volume, in mL, of $0.02\text{ M }{K}_{2}C{r}_2{O}_7$ solution required to react with $0.288\text{ g}$ of ferrous oxalate in acidic medium is …… . (Molar mass of $\text{Fe}=56{\text{ g mol}}^{-1}$)

The Dual Oxidation Trap

Imagine you are standing in a chemistry lab, performing a classic redox titration. In your conical flask, you have a precisely weighed amount of ferrous oxalate ($Fe{C}_2{O}_4$), and in the burette above, a standard solution of potassium dichromate ($K_{2}C{r}_2{O}_7$). The goal is simple: find the exact volume of the dichromate solution required to completely oxidize the ferrous oxalate. But beneath this simple setup lies a beautiful and often misunderstood concept of stoichiometry.

Law of Equivalence

To solve any titration problem without getting lost in balancing complex chemical equations, we rely on the Law of Equivalence. This law states that at the equivalence point of a reaction, the number of equivalents of the oxidizing agent perfectly matches the number of equivalents of the reducing agent.

Mathematically, this is expressed as:

We can expand this using the relationship between equivalents, molarity ($M$), volume ($V$), and the crucial n-factor ($n$):

The entire problem hinges on correctly identifying the n-factors for both reactants.

First, let's look at our oxidizing agent, potassium dichromate ($K_{2}C{r}_2{O}_7$). In an acidic medium, the dichromate ion ($C{r}_2{O}_7^{2-}$) gets reduced to chromium(III) ions ($C{r}^{3+}$). The oxidation state of each chromium atom drops from $+6$ to $+3$, which is a change of $3$ electrons. Since there are two chromium atoms in one dichromate ion, the total change is:

Now, let's analyze our reducing agent, ferrous oxalate ($Fe{C}_2{O}_4$). This is where most students fall into a trap! Ferrous oxalate is a special compound because both of its constituent ions undergo oxidation.

1. The ferrous ion ($F{e}^{2+}$) oxidizes to the ferric ion ($F{e}^{3+}$). This is a change of $1$ electron. 2. The oxalate ion ($C_2{O}_4^{2-}$) oxidizes to carbon dioxide gas ($C{O}_2$). Here, the oxidation state of carbon goes from $+3$ to $+4$. Since there are two carbon atoms, the change is $2\times 1=2$ electrons.

Therefore, the total n-factor for ferrous oxalate is the sum of these changes:

Before we substitute our values, we need the molar mass of ferrous oxalate ($Fe{C}_2{O}_4$).

Now, let's plug everything into our equivalence equation. Since we want the volume in milliliters, we will equate the milli-equivalents by multiplying the right side by $1000$:

Substituting the known values:

Let's simplify the math. On the left side, $0.02\times 6=0.12$. On the right side, $\frac{0.288}{144}=0.002$.

Finally, solving for $V$:

And there we have it! Exactly $50\text{ mL}$ of the potassium dichromate solution is required. The key takeaway here is to always be vigilant when dealing with compounds where multiple elements can change their oxidation states. Never rush the n-factor calculation!