Question

Identify the oxidizing and reducing agents in the following reactions: (a) \(\operatorname{Sn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (b) \(2 \mathrm{H}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{~F} \mathrm{e}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

a) \(\operatorname{Sn}(s)\): reducing agent, \(\mathrm{H}^{+}(aq)\): oxidizing agent. b) \(\mathrm{Fe}^{2+}(aq)\): reducing agent, \(\mathrm{H}_{2}\mathrm{O}_{2}(aq)\): oxidizing agent.

Step-by-step solution

- Assign Oxidation States (Reaction a)

Identify the oxidation states of each element in the reactants and products.For \(\operatorname{Sn}(s)\): Oxidation state = 0For \(\mathrm{H}^{+}(aq)\): Oxidation state = +1For \(\mathrm{Sn}^{2+}(aq)\): Oxidation state = +2For \(\mathrm{H}_{2}(g)\): Oxidation state = 0

- Determine Changes in Oxidation States (Reaction a)

Identify which elements' oxidation states have changed.\(\operatorname{Sn}(s)\) changes from 0 to +2 (oxidation)\(\mathrm{H}^{+}(aq)\) changes from +1 to 0 (reduction)

- Identify Oxidizing and Reducing Agents (Reaction a)

The element that is oxidized is the reducing agent, and the element that is reduced is the oxidizing agent.\(\operatorname{Sn}(s)\): reducing agent (it loses electrons)\(\mathrm{H}^{+}(aq)\): oxidizing agent (it gains electrons)

- Assign Oxidation States (Reaction b)

Identify the oxidation states of each element in the reactants and products.For \(\mathrm{H}^{+}(aq)\): Oxidation state = +1For \(\mathrm{H}_{2} \mathrm{O}_{2}(aq)\): Oxidation state of O = -1For \(\mathrm{Fe}^{2+}(aq)\): Oxidation state = +2For \(\mathrm{Fe}^{3+}(aq)\): Oxidation state = +3For \(\mathrm{H}_2\mathrm{O}(l)\): Oxidation state of O = -2

- Determine Changes in Oxidation States (Reaction b)

Identify which elements' oxidation states have changed.\(\mathrm{Fe}^{2+}(aq)\) changes from +2 to +3 (oxidation)\(\mathrm{H}_{2} \mathrm{O}_{2}(aq)\) changes from -1 (each O) to -2 (each O in \(\mathrm{H}_{2}\mathrm{O}(l)\)) (reduction)

- Identify Oxidizing and Reducing Agents (Reaction b)

The element that is oxidized is the reducing agent, and the element that is reduced is the oxidizing agent.\(\mathrm{Fe}^{2+}(aq)\): reducing agent (it loses electrons)\(\mathrm{H}_{2}\mathrm{O}_{2}(aq)\): oxidizing agent (it gains electrons)

Key concepts

Oxidation State

The oxidation state, also known as oxidation number, helps determine the degree of oxidation of an atom in a chemical compound. It represents the total number of electrons that an atom either gains or loses to form a chemical bond. For example, in the reaction \( \mathrm{Sn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{H}_{2}(g) \), the oxidation state of tin (\( \operatorname{Sn}(s) \)) changes from 0 to +2, and the hydrogen ion (\( \mathrm{H}^{+}(aq) \)) changes from +1 to 0. These changes in oxidation states indicate whether an element is oxidized or reduced.
Understanding oxidation states allows us to identify how electrons are transferred in a chemical reaction, which helps figure out the roles of different elements in a reaction.

Oxidizing Agent

An oxidizing agent, or oxidant, is a substance that gains electrons and is reduced in a chemical reaction. It causes another substance to become oxidized by taking electrons from it. For example, in \( \mathrm{Sn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{H}_{2}(g) \), the hydrogen ion (\( \mathrm{H}^{+}(aq) \)) is the oxidizing agent because it gains electrons (is reduced), changing its oxidation state from +1 to 0.
In the reaction \( 2 \mathrm{H}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \), hydrogen peroxide (\( \mathrm{H}_{2} \mathrm{O}_{2}(aq) \)) is the oxidizing agent as it gains electrons, reducing each oxygen from -1 to -2.
Identifying oxidizing agents helps in understanding how reactions proceed and which substances facilitate electron transfer.

Reducing Agent

A reducing agent, or reductant, is a substance that loses electrons and is oxidized in a chemical reaction. It causes another substance to become reduced by donating electrons to it. For instance, in \( \mathrm{Sn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{H}_{2}(g) \), tin (\( \operatorname{Sn}(s) \)) is the reducing agent because it loses electrons (is oxidized), changing its oxidation state from 0 to +2.
Similarly, in the reaction \( 2 \mathrm{H}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \), ferrous ion (\( \mathrm{Fe}^{2+}(aq) \)) acts as the reducing agent as it loses electrons and oxidizes from +2 to +3.
Understanding reducing agents is crucial as they drive the reduction half of redox reactions.

Electron Transfer

Electron transfer is the movement of electrons from one element or molecule to another in a chemical reaction. This transfer is fundamental in redox reactions (oxidation-reduction reactions). When an atom or molecule gains electrons, it is reduced, and when it loses electrons, it is oxidized. For example, in \( \mathrm{Sn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{H}_{2}(g) \), electrons are transferred from tin to hydrogen ions. Tin loses electrons (is oxidized), while hydrogen ions gain those electrons (are reduced).
Electron transfer is key to many processes, such as cellular respiration and photosynthesis in biology, and corrosion and energy production in industry. Understanding how electrons move between reactants and products helps explain reaction mechanisms and energy changes during the reactions.