Consider the equation below: Cr$_2$O$_7 ^{2-}$ + 6Fe$^{2+}$ + 14H$^{+}$ $\to$ 2Cr$^{3+}$ + 6Fe$^{3+}$ + 7H$_2$O. The oxidation number of chromium changes from

1. Analyze the Question

We are asked to determine the change in the oxidation number of chromium (Cr) as it moves from the reactant side to the product side in the given redox equation: Cr$_2$O$_7^{2-}$ + 6Fe$^{2+}$ + 14H$^{+}$ $\to$ 2Cr$^{3+}$ + 6Fe$^{3+}$ + 7H$_2$O

2. Explain the Concept

To find the oxidation number of an element in a polyatomic ion or compound, we use these rules:

• The sum of oxidation numbers in a neutral compound is 0.
• The sum of oxidation numbers in an ion is equal to the charge of the ion.
• Oxygen (O) usually has an oxidation number of -2.

3. Show the Solution

Step 1: Find the oxidation number of Cr in the reactant (Cr$_2$O$_7^{2-}$) Let the oxidation number of Cr be $x$. Since there are two Cr atoms and seven O atoms, and the total charge is -2: $2(x) + 7(-2) = -2$ $2x - 14 = -2$ $2x = -2 + 14$ $2x = 12$ $x = +6$ So, the oxidation number of Cr in the reactant is +6.

Step 2: Find the oxidation number of Cr in the product (Cr$^{3+}$) In the product, chromium exists as a simple ion (Cr$^{3+}$). The oxidation number of a monatomic ion is simply its charge. Therefore, the oxidation number of Cr in the product is +3.

Step 3: Determine the change The oxidation number changes from +6 to +3.

4. Identify the Answer

Comparing our result to the options provided: A. +5 to +3 B. +6 to +3 C. -2 to +3 D. +7 to +3

The correct option is B.

5. Tip

When calculating oxidation numbers for complex ions like dichromate (Cr$_2$O$_7^{2-}$), always remember to set the equation equal to the charge of the ion (-2), not zero. If you set it to zero, you will get the wrong answer. Also, remember that in any redox reaction, the species that undergoes a decrease in oxidation number (like Cr here, from +6 to +3) is the one being reduced.