Question

What are the oxidizing and reducing agents in the following redox reactions? (a) \(5 \mathrm{SO}_{3}^{2-}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \longrightarrow\) \(5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\) (b) \(2 \mathrm{NO}_{2}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+} \longrightarrow\) \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+2 \mathrm{H}_{2} \mathrm{O}\)

Short answer

The oxidizing and reducing agents are (a) Oxidizing agent MnO4^-, Reducing agent SO3^2-, (b) Oxidizing agent NO2, Reducing agent H2, (c) Oxidizing agent H2O2, Reducing agent [Fe(CN)6]^4-.

Step-by-step solution

Identify the Oxidation States

Firstly, identify the oxidation states of the species in the reactants and products. (a) For \(5 \mathrm{SO}_{3}^{2-}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \longrightarrow 5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\), in SO3^2-, S is in +6 oxidation state and in SO4^2-, it's also +6, so sulfur doesn't change. Mn in MnO4^- is +7 and in Mn^2+ it's +2; Therefore, Mn is reduced. (b) For \(2 \mathrm{NO}_{2}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), N in NO2 is +4 and in NH3 it's -3; thus N is reduced. H in H2 is in 0 state and in H2O it's +1; thus H is oxidized. (c) For \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+} \longrightarrow 2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+2 \mathrm{H}_{2} \mathrm{O}\), in [Fe(CN)6]^4-, Fe is in +2 state and in [Fe(CN)6]^3-, Fe is is +3; Fe is therefore oxidized. H2O2 is 0 state and in H2O it's +1; H is hence reduced.

Assign the Oxidizing and Reducing Agents

Next, assign the oxidizing and reducing agents based on the ions which have been oxidized or reduced. (a) MnO4^- is the oxidizing agent (it is reduced), and SO3^2- is the reducing agent (it is oxidized). (b) NO2 is the oxidizing agent (it is reduced), and H2 is the reducing agent (it is oxidized). (c) [Fe(CN)6]^4- is the reducing agent (it is oxidized), and H2O2 is the oxidizing agent (it is reduced).

Key concepts

Oxidizing Agents

In a redox reaction, an oxidizing agent, also known as an oxidant, is a substance that accepts electrons from another substance. When an oxidizing agent gains electrons, it becomes reduced. This means the oxidizing agent helps another substance to oxidize by accepting electrons. It's like the teammate that takes the ball in a relay run, allowing the other to run faster and win the oxidation race.

An example can be seen in the reaction: \(5 \mathrm{SO}_{3}^{2-}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \rightarrow 5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\). Here, \(\mathrm{MnO}_{4}^{-}\) acts as the oxidizing agent because it accepts electrons and is reduced to \(\mathrm{Mn}^{2+}\). This acceptance of electrons is crucial for the completion of the redox reaction.

Other common oxidizing agents include oxygen, chlorine, and hydrogen peroxide.

Reducing Agents

Reducing agents are the heroes that offer electrons to another species during a redox reaction. By donating electrons, reducing agents allow other substances to be reduced while they themselves are oxidized. Think of a reducing agent as someone handing over a baton in a relay race, allowing their partner to keep going, in this case, get reduced.

Take the reaction \(2 \mathrm{NO}_{2}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). Here, \(\mathrm{H}_{2}\) acts as the reducing agent. It donates electrons to \(\mathrm{NO}_{2}\), helping it reduce to \(\mathrm{NH}_{3}\). As a result, \(\mathrm{H}_{2}\) itself is oxidized.

Potassium, lithium, and hydrazine are examples of strong reducing agents.

Oxidation States

Oxidation states, also known as oxidation numbers, are a handy tool for keeping track of electrons in elements during chemical reactions. They are assigned to atoms to predict how electrons are distributed in a molecule. It's similar to giving each atom a 'score' that shows how many electrons it gained or lost.

To determine what happens during a reaction, we compare the oxidation states of an element in the reactants and the products. For example, in \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+} \rightarrow 2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+2 \mathrm{H}_{2} \mathrm{O}\), iron's oxidation state changes from +2 to +3, indicating oxidation.

Understanding oxidation states helps in balancing equations, deciding the nature of compounds, and deducing electron flow in reactions.