Problem 1 - Chemistry Matter and Change

Question

Identify each of the following changes as either oxidation or reduction. Recall that e - is the symbol for an electron a. \(I_{2}+2 e^{-} \rightarrow 21^{-}\) c. \({Fe}^{2+} \rightarrow {Fe}^{3+}+{e}^{-}\) b. \(K \rightarrow K^{+}+e^{-}\) d. \({Ag}^{+}+{e}^{-} \rightarrow {Ag}\)

Step-by-Step Solution

Answer

a. Reduction b. Oxidation c. Oxidation d. Reduction

Step 1: Identify electron transfer in each reaction

For each of the given reactions, observe whether electrons are being gained or lost. a. \(I_{2}+2 e^{-} \rightarrow 2I^{-}\) In this reaction, the iodine molecule (\(I_{2}\)) accepts two electrons (\(2 e^-\)) to form two iodide ions (\(2I^{-}\)). b. \(K \rightarrow K^{+}+e^{-}\) In this reaction, a potassium atom (\(K\)) loses an electron (\(e^-\)) to become a potassium ion (\(K^+\)). c. \({Fe}^{2+} \rightarrow {Fe}^{3+}+{e}^{-}\) In this reaction, a ferrous ion (\({Fe}^{2+}\)) loses an electron (\(e^-\)) to become a ferric ion (\(Fe^{3+}\)). d. \({Ag}^{+}+{e}^{-} \rightarrow {Ag}\) In this reaction, a silver ion (\({Ag}^+\)) accepts an electron (\(e^-\)) to become a silver atom (\(Ag\)).

Step 2: Identify oxidation or reduction for each reaction

Now that we have identified electron transfer for each reaction, we can classify each reaction as oxidation or reduction. a. \(I_{2}+2 e^{-} \rightarrow 2I^{-}\) Since iodine molecules (\(I_{2}\)) are gaining electrons, this is a reduction reaction. b. \(K \rightarrow K^{+}+e^{-}\) Since the potassium atom (\(K\)) is losing an electron, this is an oxidation reaction. c. \({Fe}^{2+} \rightarrow {Fe}^{3+}+{e}^{-}\) Since the ferrous ion (\({Fe}^{2+}\)) is losing an electron, this is an oxidation reaction. d. \({Ag}^{+}+{e}^{-} \rightarrow {Ag}\) Since the silver ion (\({Ag}^+\)) is gaining an electron, this is a reduction reaction.

Key Concepts

Electron Transfer

In chemistry, electron transfer is an essential process involving the movement of electrons from one atom, molecule, or ion to another. This plays a fundamental role in oxidation-reduction reactions, often referred to as redox reactions. These reactions are pivotal in both biological systems and industrial applications. During electron transfer, there is a re-distribution of electrons that leads to the change in charge of the atoms involved.

• When electrons are lost by an atom, it is said to undergo oxidation.
• When an atom gains electrons, it is undergoing reduction.

This electron movement brings about the essential changes required to propel redox reactions forward. Understanding this transfer is key to mastering concepts related to oxidation and reduction.

Reduction

Reduction is half of the redox process where a chemical species gains electrons. This change is often accompanied by a decrease in oxidation state. Think of reduction as making the atom or ion more negative due to the addition of electrons.

• In the reaction \(I_{2}+2 e^{-} \rightarrow 2I^{-}\), the iodine molecule gains electrons and is reduced.
• Similarly, in \({Ag}^{+}+{e}^{-} \rightarrow {Ag}\), the silver ion \({Ag}^{+}\) gains an electron, converting it into metallic silver, \({Ag}\).

These examples highlight the fundamental process of electron gain during reduction. The mnemonic "OIL RIG" might help you remember: "Oxidation Is Loss, Reduction Is Gain" of electrons.

Oxidation

Oxidation is the complementary part of a reduction reaction, where a chemical species loses electrons and typically increases its oxidation state. You can remember oxidation as the "loss" of electrons, leading to a more positive oxidation state.

• In the reaction \(K \rightarrow K^{+}+e^{-}\), potassium loses an electron and is thus oxidized.
• Similarly, \({Fe}^{2+} \rightarrow {Fe}^{3+}+{e}^{-}\) shows how the ferrous ion \({Fe}^{2+}\) is oxidized to ferric ion \({Fe}^{3+}\) by losing an electron.

The term "oxidation" originally came from the process of adding oxygen; however, it now encompasses any electron loss, not just those involving oxygen.

Half-Reactions

Each redox reaction can be split into two half-reactions; one showing oxidation and the other depicting reduction. This helpful breakdown allows us to separately visualize the transfer of electrons.
Let's consider the complete redox reaction as a whole:

• The reduction half-reaction from \(I_{2}+2 e^{-} \rightarrow 2I^{-}\) shows the gain of electrons.
• On the flip side, the oxidation half-reaction \(K \rightarrow K^{+}+e^{-}\) demonstrates the loss of electrons.

Breaking it down into half-reactions helps in balancing complex redox equations, ensuring both mass and charge are balanced. It's a practical method for analyzing and understanding the flow of electrons in chemical reactions.