Question

Balance the following redox rcactions by ion \(-\) clcctron method : (a) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}\) \((\mathrm{s})+\mathrm{I}_{2}(\mathrm{~s})\) (in basic medium) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}\) (aq) (in acidic solution) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Fe}^{2+}\) (aq) \(\rightarrow \mathrm{Fe}^{3+}\) (aq) \(+\mathrm{H}_{2} \mathrm{O}\) (l) (in acidic solution) (d) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (in acidic solution)

Short answer

Each reaction is balanced by separating into half-reactions, balancing atoms and charges, then combining them.

Step-by-step solution

Separate the half-reactions (a)

For reaction (a), write the oxidation and reduction half-reactions. The reduction half-reaction is \( \text{MnO}_4^- \rightarrow \text{MnO}_2 \). The oxidation half-reaction is \( \text{I}^- \rightarrow \text{I}_2 \).

Balance atoms in half-reactions (a)

Balance all atoms except H and O. For \( \text{I}^- \rightarrow \text{I}_2 \), write \( 2\text{I}^- \rightarrow \text{I}_2 \). For \( \text{MnO}_4^- \rightarrow \text{MnO}_2 \), \( \text{Mn} \) atoms are balanced.

Balance Oxygens and Hydrogens (a)

For \( \text{MnO}_4^- \rightarrow \text{MnO}_2 \), balance O by adding 2\( \text{H}_2\text{O} \) to the right side and balance H by adding 4\( \text{OH}^- \) to the left side.

Balance Charges by adding Electrons (a)

Add electrons to balance charges. For \( \text{MnO}_4^- + 4\text{OH}^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} + 3e^- \). For \( 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \).

Equalize Electrons (a)

Multiply the \( \text{I}^- \) reaction by 3 and \( \text{MnO}_4^- \) reaction by 2 to equalize electrons: \( 6\text{I}^- \rightarrow 3\text{I}_2 + 6e^- \) and \( 2\text{MnO}_4^- + 8\text{OH}^- + 6e^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} \).

Combine the half-reactions (a)

Combine and cancel electrons: \( 2\text{MnO}_4^- + 6\text{I}^- + 8\text{OH}^- \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 4\text{H}_2\text{O} \).

Balance Half-reactions (b)

Write the half-reactions: Reduction \( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \) and Oxidation \( \text{SO}_2 \rightarrow \text{HSO}_4^- \).

Balance Atoms and Charges (b)

Balance \( \text{MnO}_4^- \) by adding 8\( \text{H}^+ \) to make \( \text{Mn}^{2+} + 4\text{H}_2\text{O} \), and \( 5e^- \) to the left. For \( \text{SO}_2 \rightarrow \text{HSO}_4^- \), add \( \text{H}_2\text{O} \), \( 2\text{H}^+ \), and \( 2e^- \).

Equalize Electrons (b)

Multiply \( \text{SO}_2 \) reaction by 5 and \( \text{MnO}_4^- \) by 2: \( 10\text{MnO}_4^- + 80\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 20\text{H}_2\text{O} \) and \( 5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 5\text{HSO}_4^- + 5\text{H}^+ + 10e^- \).

Combine and Cancel Electrons (b)

Combine and simplify: \( 2\text{MnO}_4^- + 5\text{SO}_2 + 2\text{H}_2\text{O} + 10\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{HSO}_4^- \).

Write Half-reactions (c)

For reduction: \( \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} \). For oxidation: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \).

Balance Electrons (c)

Balance \( \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + 2e^- \) and \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \).

Equalize Electrons (c)

Multiply \( \text{Fe}^{2+} \) by 2 to get \( 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^- \).

Combine Reactions (c)

Add half-reactions: \( \text{H}_2\text{O}_2 + 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O} \).

Write and Balance Half-reactions (d)

Reduction is \( \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \), and oxidation is \( \text{SO}_2 \rightarrow \text{SO}_4^{2-} \).

Balance Electrons and Atoms (d)

Add 6\( \text{e}^- \) and 14\( \text{H}^+ \) to \( \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \rightarrow 2\text{Cr}^{3+} \), and 2\( \text{H}_2\text{O} \), 4\( \text{H}^+ \), and 2\( \text{e}^- \) to \( \text{SO}_2 \rightarrow \text{SO}_4^{2-} \).

Equalize Electrons (d)

Multiply \( \text{SO}_2 \) by 3 and use \( \text{Cr}_2\text{O}_7^{2-} \) as is. Combine both reactions: \( \text{Cr}_2\text{O}_7^{2-} + 3\text{SO}_2 + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + 7\text{H}_2\text{O} \).

Key concepts

Ion-Electron Method

The Ion-Electron Method is a systematic way to balance redox reactions, which involve the transfer of electrons between species. In redox reactions, one species undergoes oxidation (loses electrons) while the other undergoes reduction (gains electrons). The Ion-Electron Method simplifies balancing such reactions by separately focusing on oxidation and reduction processes.

Here's a general approach to using the Ion-Electron Method:

• Identify Redox Pairs: Determine which components act as the oxidizing and reducing agents.
• Separate Half-Reactions: Write individual oxidation and reduction reactions, highlighting electron transfers.
• Balance Atoms and Electrons: Ensure that each half-reaction balances both the number of atoms and the charges by adding electrons.
• Equalize Electrons: Multiply the half-reactions by appropriate coefficients to ensure the number of electrons lost in one reaction equals those gained in the other.
• Combine Half-Reactions: Merge the balanced half-reactions, canceling out the electrons.

This method not only helps in balancing equations but also provides insight into the electron exchange occurring in chemical reactions.

Half-Reactions

Half-reactions are individual components of a redox equation, representing either the oxidation or reduction change. They allow a clearer view of electron transfer and help in the precise balancing of redox reactions.

In any redox process:

• Oxidation Half-Reaction: This illustrates the species that loses electrons. For example, in the equation \[ \text{I}^- \rightarrow \text{I}_2 + 2e^- \], iodine ions lose electrons and form iodine molecules.
• Reduction Half-Reaction: This shows the species gaining electrons. As in the case of \[ \text{MnO}_4^- + 4e^- \rightarrow \text{MnO}_2 \], permanganate ions gain electrons and form manganese dioxide.

Remember, the sum of electrons lost in the oxidation half should equal the electrons gained in the reduction half. This equality ensures the complete redox equation will be balanced in terms of both mass and charge.

Balancing Chemical Equations

Balancing chemical equations is fundamental to chemistry, ensuring the conservation of mass where the number of atoms for each element is the same on both sides of the equation. This becomes particularly crucial in redox reactions, where both mass and charge must be balanced.

Here is how you can balance a redox reaction effectively:

• Write the Unbalanced Equation: Begin with a clear view of the reaction, identifying the species involved.
• Identify State and Medium: Note whether the reaction is occurring in an acidic or basic medium as it will impact balancing steps.
• Add Water, Hydrogen, and Hydroxide Ions as Needed: For reactions in acidic solutions, H₂O and H⁺ ions help balance oxygen and hydrogen atoms. In basic solutions, use OH^- instead.
• Balance Electrons Last: Add electrons where required to equalize the charge on both sides.
• Final Check: Reassess the equation to confirm all atoms and charges are identical on both sides, ensuring the law of conservation of mass and charge is satisfied.

By following these steps, you can accurately balance complex redox reactions and understand the underlying chemical changes.