Question

Indicate the coefficient in front of \(\mathrm{H}_{2} \mathrm{O}_{2}\) when the following redox equation is balanced in an acidic medium. \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{aq})\) (a) 4 (b) 3 (c) 2 (d) 1

Short answer

The coefficient in front of \( \mathrm{H}_2\mathrm{O}_2 \) is (d) 1.

Step-by-step solution

Assign Oxidation Numbers

First, assign oxidation numbers to each element in the reaction. For \( \mathrm{H}_2\mathrm{O}_2 \), hydrogen is always +1, and oxygen is -1 in peroxides such as \( \mathrm{H}_2\mathrm{O}_2 \). In \( \mathrm{Cl}^- \), chlorine is -1. In \( \mathrm{H}_2\mathrm{O} \), oxygen is -2, and hydrogen is +1. Finally, in \( \mathrm{Cl}_2 \), chlorine is 0.

Identify the Changes in Oxidation Numbers

Note the changes in oxidation states: \( \mathrm{O}_2^{2-} \) changes from -1 in \( \mathrm{H}_2\mathrm{O}_2 \) to -2 in \( \mathrm{H}_2\mathrm{O} \), indicating reduction. Chlorine changes from -1 in \( \mathrm{Cl}^- \) to 0 in \( \mathrm{Cl}_2 \), indicating oxidation.

Write Half-Reactions

Separate the reaction into two half-reactions. The reduction half-reaction is \( \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} \), and the oxidation half-reaction is \( \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 \).

Balance Atoms Other Than O and H

Balance the elements other than oxygen and hydrogen for each half-reaction. For the reduction half-reaction, both sides have one O. For the oxidation half-reaction, two Cl- ions produce one \( \mathrm{Cl}_2 \) molecule, so consider \( 2 \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 \).

Balance Oxygen Atoms

Balance the oxygen atoms using water. The reduction half-reaction is already balanced with respect to oxygen.

Balance Hydrogen Atoms Using \( \mathrm{H}^+ \) Ions

Add \( \mathrm{H}^+ \) ions to balance the hydrogen atoms. The reduction half-reaction \( \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} \) becomes \( \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ \rightarrow 2\mathrm{H}_2\mathrm{O} \).

Balance Charges with Electrons

Add electrons to balance the charge on each side of the half-reactions. Reduction half-reaction: \( \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \). Oxidation half-reaction: \( 2\mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2e^- \).

Balance the Electrons by Equalizing Electron Numbers

The electrons are already balanced at 2 for both half-reactions, so no further adjustments are necessary.

Combine the Balanced Half-Reactions

Combine the two half-reactions: \( \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2\mathrm{Cl}^- \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{Cl}_2 \).

Verify the Overall Balance

Check that both sides of the final reaction have the same number of each element and the same charge. Both sides have 2 oxygen, 2 hydrogen, and 2 chlorine atoms with a net charge of 0.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the conservation of mass. This means the number of each type of atom must be the same on both sides of the equation. In redox reactions, balancing can be trickier because you must consider both mass and charge.
To balance a redox equation:

• First, separate the reaction into two half-reactions, one for oxidation and one for reduction.
• Balance all atoms except oxygen and hydrogen.
• Balance the oxygen atoms by adding water molecules.
• Balance the hydrogen atoms by adding hydrogen ions (\(\mathrm{H}^+\)).
• Finally, balance the charges by adding electrons (\(e^-\)).

Once each half-reaction is balanced, combine them, ensuring the electrons cancel out. This results in a balanced equation for both mass and charge. Balancing equations is vital for understanding chemical reactivity and stoichiometry.

Oxidation Numbers

Oxidation numbers are a useful way to keep track of electrons in chemical reactions. They help identify which atoms are being oxidized and which are being reduced. In simple terms, oxidation involves losing electrons, while reduction involves gaining them.
Assigning oxidation numbers follows specific rules:

• Elements in their natural state have an oxidation number of 0. For instance, \(\mathrm{Cl}_2\) is 0.
• The oxidation number for a monatomic ion is equal to its charge; for example, \(\mathrm{Cl}^-\) is -1.
• Hydrogen is usually +1, while oxygen is typically -2.
• Sum of oxidation numbers in a compound should equal its charge.

Understanding oxidation numbers is key to identifying transfer of electrons in redox reactions, aiding in balancing them accurately.

Half-Reactions

Half-reactions are a breakdown of a redox reaction into its oxidation and reduction components. This method simplifies the process of examining which species lose electrons (oxidation) and which gain them (reduction).
Each half-reaction shows:

• Only the species involved in one of the processes, either oxidation or reduction.
• How atoms and charges are balanced separately for each process.

Consider the half-reactions for \(\mathrm{H}_2\mathrm{O}_2\) and \(\mathrm{Cl}^-\):

In the reduction half-reaction, \(\mathrm{H}_2\mathrm{O}_2\) is reduced to \(\mathrm{H}_2\mathrm{O}\), while in the oxidation half-reaction, \(\mathrm{Cl}^-\) is oxidized to \(\mathrm{Cl}_2\). By balancing these half-reactions, we can clearly see the electron flow and verify the number of electrons lost equals those gained.
The half-reaction method is crucial for a clear understanding of the electron transfer process and is essential for balancing redox equations in complex scenarios.