Problem 17 - Chemistry The Science in Context

Question

Complete and balance the partial chemical equation below, using the appropriate half-reactions in Table \(A 6.1\) for acidic solutions. $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}^{3+}(a q)$$

Step-by-Step Solution

Answer

Answer: The oxidation half-reaction is: $$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^{-}$$ and the reduction half-reaction is: $$\mathrm{Cr}_{2} \mathrm{O}_{7}(2-) + 14 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$$. The balanced chemical equation is: $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+ 14\, \mathrm{H}^{+} + 6\, \mathrm{Fe}^{2+}(a q) \rightarrow 2\, \mathrm{Cr}^{3+}(a q) + 7\, \mathrm{H}_{2} \mathrm{O} + 6\, \mathrm{Fe}^{3+}(a q)$$.

Step 1: Determine the oxidation and reduction half-reactions

Given the partial chemical equation, we can determine the half-reactions involving Cr and Fe. The Cr is changing from \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\), while the Fe is changing from \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\). The chromium is being reduced, while the iron is being oxidized.

Step 2: Identify the half-reactions from Table A 6.1

According to Table A 6.1 for acidic solutions, the half-reactions are: Reduction half-reaction: $$\mathrm{Cr}_{2} \mathrm{O}_{7}(2-) + 14 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$$ Oxidation half-reaction: $$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^{-}$$

Step 3: Balance the number of electrons in both half-reactions

The reduction half-reaction involves the transfer of 6 electrons, while the oxidation half-reaction involves the transfer of 1 electron. In order to balance the number of electrons in both half-reactions, we need to multiply the oxidation half-reaction by 6: $$6\, \mathrm{Fe}^{2+} \rightarrow 6\, \mathrm{Fe}^{3+} + 6\, \mathrm{e}^{-}$$ Now, both half-reactions involve the transfer of 6 electrons.

Step 4: Combine the half-reactions to form the balanced chemical equation

After having balanced the number of electrons, we can now add the two half-reactions to form the overall balanced chemical equation: $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\,\mathrm{H}^{+} + 6\,\mathrm{e}^{-} + 6\,\mathrm{Fe}^{2+} \rightarrow 2\, \mathrm{Cr}^{3+} + 7\,\mathrm{H}_{2} \mathrm{O} + 6\,\mathrm{Fe}^{3+} + 6\,\mathrm{e}^{-}$$ The balanced chemical equation is: $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+ 14\, \mathrm{H}^{+} + 6\, \mathrm{Fe}^{2+}(a q) \rightarrow 2\, \mathrm{Cr}^{3+}(a q) + 7\, \mathrm{H}_{2} \mathrm{O} + 6\, \mathrm{Fe}^{3+}(a q)$$

Key Concepts

Oxidation-Reduction Reactions

When we talk about oxidation-reduction reactions, also known as redox reactions, we are referring to chemical processes where electrons are transferred between two substances.
These reactions involve a change in the oxidation state of the involved species:

• Oxidation refers to the loss of electrons. This means an increase in oxidation state. For example, in our equation, iron (\( \mathrm{Fe}^{2+} \)) is oxidized to \( \mathrm{Fe}^{3+} \).
• Reduction involves the gain of electrons. This results in a decrease in oxidation state. In this exercise, chromium in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is reduced to \( \mathrm{Cr}^{3+} \).

Identifying which species are oxidized and which are reduced is critical for balancing redox reactions. This identification enables us to use methods like the half-reaction method effectively.

Half-Reaction Method

The half-reaction method is a straightforward technique for balancing redox reactions, particularly useful in complex ionic equations. It involves separately balancing the oxidation and reduction parts of the reaction equation.
Here's how it generally works:

• First, break down the equation into two half-reactions: one for oxidation and one for reduction. In our example, the oxidation half-reaction is \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^{-} \), and the reduction half-reaction is \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O} \).
• Next, balance each half-reaction in terms of atoms and charge. This typically means ensuring the number of ions and elements, particularly oxygen and hydrogen in acidic or basic solutions, are balanced.
• Then, multiply each half-reaction by the necessary coefficients to ensure that the electrons gained in reduction equal the electrons lost in oxidation. In the solution, the oxidation half-reaction is multiplied by 6 to balance the electron exchange.
• Finally, recombine the balanced half-reactions to form a single, balanced redox equation.

This method simplifies the process of balancing equations by dealing separately with the changes in electron transfer, ensuring accuracy and clarity.

Acidic Solutions

In the context of balancing redox reactions, the characteristics of the solution—whether acidic or basic—affect the balancing process especially for the half-reactions involving oxygen and hydrogen.
For acidic solutions, as detailed in this exercise:

• We use \( \mathrm{H}^{+} \) ions to balance hydrogen atoms in the equation. In the reduction half-reaction here, 14 \( \mathrm{H}^{+} \) ions are used to balance the equation adequately.
• The involvement of water molecules (\( \mathrm{H}_{2} \mathrm{O} \)) often balances oxygen atoms. In the reduction half, 7 \( \mathrm{H}_{2} \mathrm{O} \) molecules are added to achieve balance.

Being aware of the solution type allows us to apply the correct species for balancing purposes, ensuring both mass and charge are preserved in the final balanced equation.