Question

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. (a) \(\mathrm{P}_{4}(\mathrm{~s})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{PH}_{3}(\mathrm{~g})+\mathrm{HPO}_{2}^{-}(\mathrm{aq})\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{ClO}_{2}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}^{+}\)

Short answer

(a) \( \text{P}_4 \) is both the oxidizing and reducing agent; (b) \( \text{ClO}_3^- \) is oxidizing agent, \( \text{N}_2\text{H}_4 \) is reducing agent; (c) \( \text{Cl}_2\text{O}_7 \) is oxidizing agent, \( \text{H}_2\text{O}_2 \) is reducing agent.

Step-by-step solution

Step 1a: Write Unbalanced Equation for Reaction (a)

Start by considering the unbalanced reaction:\[ \text{P}_4(\text{s}) + \text{OH}^- (\text{aq}) \rightarrow \text{PH}_3(\text{g}) + \text{HPO}_2^- (\text{aq}) \]

Step 2a: Determine Oxidation States and Identify Redox Changes

Identify the change in oxidation states in the reaction. In \( \text{P}_4 \), phosphorus (\ (\text{P}) is in the zero oxidation state. In \( \text{PH}_3 \), \( \text{P} \) changes to -3. In \( \text{HPO}_2^- \), \( \text{P} \) changes to +1. This indicates the presence of both reduction and oxidation.

Step 3a: Write Half-Reactions

For the reduction (\( \text{P} \to \text{PH}_3 \)): \[ \text{P}_4 + 12\text{e}^- + 12\text{H}_2\text{O} \rightarrow 4\text{PH}_3 + 12\text{OH}^- \] For the oxidation (\( \text{P} \to \text{HPO}_2^- \)): \[ \text{P}_4 + 8\text{OH}^- \rightarrow 4\text{HPO}_2^- + 4\text{e}^- + 4\text{H}_2\text{O} \]

Step 4a: Balance Electrons and Add Half-Reactions

Multiply the oxidation half-reaction by 3 to balance electrons, then add both half-reactions to get:\[ \text{P}_4 + 12\text{OH}^- \rightarrow 4\text{PH}_3 + 4\text{HPO}_2^- \]

Step 5a: Identify Agents in Reaction (a)

The oxidizing agent is \( \text{P}_4 \) being reduced to \( \text{PH}_3 \), and the reducing agent is also \( \text{P}_4 \), being oxidized to \( \text{HPO}_2^- \).

Step 1b: Write Unbalanced Equation for Reaction (b)

Consider the unbalanced reaction:\[ \text{N}_2\text{H}_4(\text{l}) + \text{ClO}_3^- (\text{aq}) \rightarrow \text{NO}(\text{g}) + \text{Cl}^- (\text{g}) \]

Step 2b: Determine Oxidation States and Identify Redox Changes

In \( \text{N}_2\text{H}_4 \), the oxidation state of \( \text{N} \) is -2, changing to +2 in \( \text{NO} \). In \( \text{ClO}_3^- \), \( \text{Cl} \) changes from +5 to -1 in \( \text{Cl}^- \). This indicates oxidation and reduction steps.

Step 3b: Write Half-Reactions

For nitrogen oxidation (\( \text{N}_2\text{H}_4 \to \text{NO} \)): \[ \text{N}_2\text{H}_4 \rightarrow 2\text{NO} + 8\text{e}^- + 4\text{H}^+ \] For chlorine reduction (\( \text{ClO}_3^- \to \text{Cl}^- \)):\[ \text{ClO}_3^- + 6\text{e}^- + 6\text{H}^+ \rightarrow \text{Cl}^- + 3\text{H}_2\text{O} \]

Step 4b: Balance Electrons and Combine Half-Reactions

Multiply half-reactions to balance electrons and add: \[ 2\text{N}_2\text{H}_4 + 3\text{ClO}_3^- + 6\text{H}^+ \rightarrow 4\text{NO} + 3\text{Cl}^- + 6\text{H}_2\text{O} \]

Step 5b: Identify Agents in Reaction (b)

The oxidizing agent is \( \text{ClO}_3^- \) being reduced to \( \text{Cl}^- \), and the reducing agent is \( \text{N}_2\text{H}_4 \), being oxidized to \( \text{NO} \).

Step 1c: Write Unbalanced Equation for Reaction (c)

Consider the unbalanced reaction:\[ \text{Cl}_2\text{O}_7 (\text{g}) + \text{H}_2\text{O}_2 (\text{aq}) \rightarrow \text{ClO}_2^- (\text{aq}) + \text{O}_2 (\text{g}) + \text{H}^+ \]

Step 2c: Determine Oxidation States and Identify Redox Changes

In \( \text{Cl}_2\text{O}_7 \), \( \text{Cl} \) is +7, changing to +3 in \( \text{ClO}_2^- \). In \( \text{H}_2\text{O}_2 \), \( \text{O} \) goes from -1 to 0 in \( \text{O}_2 \). This indicates oxidation and reduction.

Step 3c: Write Half-Reactions

For chlorine reduction (\( \text{Cl}_2\text{O}_7 \to \text{ClO}_2^- \)):\[ \text{Cl}_2\text{O}_7 + 14\text{e}^- + 8\text{H}^+ \rightarrow 2\text{ClO}_2^- + 7\text{H}_2\text{O} \]For oxygen oxidation (\( \text{H}_2\text{O}_2 \to \text{O}_2 \)):\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{e}^- + 2\text{H}^+ \]

Step 4c: Balance Electrons and Combine Half-Reactions

Multiply the peroxide half-reaction by 7 to balance electrons:\[ \text{Cl}_2\text{O}_7 + 7\text{H}_2\text{O}_2 \rightarrow 2\text{ClO}_2^- + 7\text{O}_2 + 2\text{H}^+ \]

Step 5c: Identify Agents in Reaction (c)

The oxidizing agent is \( \text{Cl}_2\text{O}_7 \) being reduced to \( \text{ClO}_2^- \), and the reducing agent is \( \text{H}_2\text{O}_2 \), being oxidized to \( \text{O}_2 \).

Key concepts

Ion-Electron Method

The Ion-Electron Method, also known as the half-reaction method, is a technique used to balance redox reactions. This approach involves separating the oxidation and reduction processes into two half-reactions. These half-reactions show the electron exchange clearly, making it easier to balance complex redox equations.
To start, we identify each reactant and product and determine their respective oxidation states. Then we write separate equations for the oxidation and reduction halves, showing how electrons are transferred. Each half-reaction is balanced for mass and charge. Finally, the half-reactions are combined, ensuring that the electrons lost and gained are equal. This method provides a step-by-step approach to achieve a balanced redox equation in both acidic and basic media, crucial for understanding chemical reactions deeply.

Oxidation State

Oxidation states, sometimes called oxidation numbers, help us track how many electrons are lost or gained during a chemical reaction. An atom's oxidation state is a fictional charge that it might have if all bonds were considered ionic. Understanding changes in oxidation states allows us to identify redox partners and reactions.
An increase in the oxidation state indicates oxidation (loss of electrons), while a decrease indicates reduction (gain of electrons). These changes are central to identifying and understanding redox processes. For example, metals in compounds usually have positive oxidation states, whereas nonmetals may have negative states. By writing out oxidation states, complex reactions become more comprehensible, facilitating the balancing of redox equations.

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry, translating how reactants convert to products without losing or gaining any matter. In redox reactions, balancing involves ensuring both mass and charge are conserved.
When using the ion-electron method, we first balance atoms and then charges. This dual process ensures that atoms are equal on both sides and that electrons lost in oxidation equal those gained in reduction. At times, it may involve adding water (in aqueous solutions) or hydroxide ions (in basic conditions) to maintain balance. Successful balancing confirms that both the laws of conservation of mass and charge are upheld, giving a true picture of the chemical change taking place.

Oxidizing Agent

In redox reactions, the oxidizing agent is the substance that oxidizes another by accepting electrons. The oxidizing agent undergoes reduction itself, because it gains electrons.
An important part of identifying an oxidizing agent is recognizing that it is often associated with high oxidation states, which may come down through the reaction as it gains electrons. For instance, in a reaction, if a compound such as \( ext{ClO}_3^- \) is reduced to \( ext{Cl}^- \), this compound serves as the oxidizing agent because it facilitates the oxidation of another species while being reduced itself. This function is vital in many industrial, biochemical, and environmental processes.

Reducing Agent

The reducing agent in a redox reaction facilitates the reduction of another substance by losing electrons itself. As a result, the reducing agent is oxidized during the reaction.
Typically, a reducing agent is characterized by its low oxidation state, which it increases by losing electrons. In a reaction like the conversion of \( ext{N}_2 ext{H}_4 \) to \( ext{NO} \), \( ext{N}_2 ext{H}_4 \) acts as the reducing agent because it donates electrons, undergoing oxidation in the process. Understanding reducing agents is crucial not only for identifying them in reactions but also for harnessing these reactions in processes like energy generation and material synthesis.