Question

Balance this redox reaction by the half reaction method by assuming an acidic solution. $$ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe} \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} $$

Short answer

\( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 2 \mathrm{Fe} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{Fe}^{3+} \)

Step-by-step solution

Identify Oxidation and Reduction Half-Reactions

In this redox reaction, we identify the changes in oxidation states. Chromium in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is reduced to \( \mathrm{Cr}^{3+} \), decreasing its oxidation state from +6 to +3. Iron \( \mathrm{Fe} \) is oxidized to \( \mathrm{Fe}^{3+} \), increasing its oxidation state from 0 to +3. Thus, we have two half-reactions: reduction (for chromium) and oxidation (for iron).

Write the Reduction Half-Reaction

The reduction half-reaction involves chromium:\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow 2 \mathrm{Cr}^{3+} \]Balance oxygen atoms by adding water:\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2}\mathrm{O} \]Next, balance the charge by adding electrons. \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) on reactant side has a charge of -2 which needs to balance with +6 from products with +12 from protons:\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2}\mathrm{O} \]

Write the Oxidation Half-Reaction

The oxidation half-reaction involves iron:\[ \mathrm{Fe} \rightarrow \mathrm{Fe}^{3+} \]Balance the charges by adding electrons:\[ \mathrm{Fe} \rightarrow \mathrm{Fe}^{3+} + 3 \mathrm{e}^{-} \]

Balance Electrons Between Half-Reactions

The reduction half-reaction uses 6 electrons, and the oxidation half-reaction releases 3 electrons. Multiply the oxidation half-reaction by 2 to have the same number of electrons in both half-reactions:\[ 2( \mathrm{Fe} \rightarrow \mathrm{Fe}^{3+} + 3 \mathrm{e}^{-}) \rightarrow 2\mathrm{Fe}\rightarrow 2\mathrm{Fe}^{3+} + 6 \mathrm{e}^{-} \]

Combine the Balanced Half-Reactions

Add the balanced reduction and oxidation half-reactions:\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{e}^{-} + 2 \mathrm{Fe} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{Fe}^{3+} + 6 \mathrm{e}^{-} \]Cancel out the electrons and simplify:\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 2 \mathrm{Fe} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O} + 2 \mathrm{Fe}^{3+} \]

Final Balanced Redox Reaction

The balanced redox reaction is now:\[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 2 \mathrm{Fe} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{Fe}^{3+} \]

Key concepts

Half Reaction Method

Redox reactions can seem complicated, but the half reaction method provides an effective tool for balancing them. The key is to separate the overall redox reaction into two distinct half-reactions: the oxidation half and the reduction half. Each of these half-reactions reflects the either gain or loss of electrons.

• **Oxidation half-reaction**: shows where a species loses electrons and increases its oxidation state.
• **Reduction half-reaction**: demonstrates where a species gains electrons, reducing its oxidation state.

By individually focusing on electron loss and gain, this method effectively compartmentalizes the complex interaction. Ultimately, combining the balanced half-reactions gives us a complete and balanced equation that accurately represents the redox process.

Oxidation States

Oxidation states, also known as oxidation numbers, are handy tools. They indicate the degree of oxidation of an atom in a substance. In a redox equation, these numbers change as reactions progress.
A prime example is chromium in the reaction: \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \to \mathrm{Cr}^{3+} \). Here, chromium's oxidation state transitions from +6 to +3, depicting reduction. Similarly, in iron's change from \( \mathrm{Fe} \to \mathrm{Fe}^{3+} \), its oxidation state increases from 0 to +3, showing oxidation.

• Assign oxidation states to each element in the reactants and products.
• Identify changes in oxidation states to determine which species are oxidized and which are reduced.

Grasping this concept is fundamental in identifying and balancing redox reactions.

Electron Balancing

Electron balancing is a vital component in balancing redox reactions. The main idea is that the total number of electrons lost during oxidation must equal the electrons gained during reduction. The half reaction method employs this concept in a systematic manner.
In balancing: - Chromium’s reduction requires 6 electrons. - Iron’s oxidation releases 3 electrons per atom.
To maintain balance, multiply the entire iron half-reaction by 2. Then, both half-reactions utilize or release a total of 6 electrons.
Using this method ensures that charge conservation remains consistent, which is crucial in achieving the final balanced equation.

Acidic Solution Redox Balancing

Balancing redox reactions in an acidic solution adds an additional layer. Here, hydrogen ions (\( \mathrm{H}^{+} \)) and water molecules often play a critical role in balancing atoms like oxygen and hydrogen:

• Add \( \mathrm{H}_{2} \mathrm{O} \) to balance oxygen atoms.
• Use \( \mathrm{H}^{+} \) ions to balance hydrogen atoms.

For the chromium half-reaction, adding 7 \( \mathrm{H}_{2} \mathrm{O} \) molecules balances the 14 oxygen atoms on the reactant side. Correspondingly, 14 \( \mathrm{H}^{+} \) ions ensure the hydrogen atoms are balanced.
This balancing process showcases how specific parameters change when working within an acidic medium, offering a methodical approach to maintaining equilibrium in complex redox systems.