Problem 66 - Chemistry Matter and Change

Question

Label each half-reaction as reduction or oxidation. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-}} \\ {\text { b. } \mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}} \\ {\text { c. } 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}} \\ {\text { d. } \mathrm{F}_{2} \rightarrow 2 \mathrm{F}^{-}+2 \mathrm{e}^{-}}\end{array} \end{equation}

Step-by-Step Solution

Answer

a: Oxidation, b: Reduction, c: Reduction, d: Oxidation.

Step 1: Understanding the Reaction Type

To determine if a reaction is a reduction or oxidation, remember that oxidation involves the loss of electrons, while reduction involves the gain of electrons. You can look at the changes in oxidation states or the movement of electrons.

Step 2: Determine Oxidation in Reaction a

In the reaction: \( \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \), iron (Fe) loses an electron as it changes from a 2+ to a 3+ oxidation state. This is an oxidation reaction.

Step 3: Determine Reduction in Reaction b

For the reaction: \( \mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \), manganese (Mn) gains electrons, indicating a reduction process.

Step 4: Determine Reduction in Reaction c

In the reaction: \( 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \), hydrogen gains electrons to form molecular hydrogen (H₂). Thus, it is a reduction reaction.

Step 5: Determine Oxidation in Reaction d

Looking at the reaction: \( \mathrm{F}_{2} \rightarrow 2 \mathrm{F}^{-}+2 \mathrm{e}^{-} \), each fluorine molecule splits into two fluoride ions, and electrons appear on the same side as the products indicating an oxidation process as electrons are released.

Key Concepts

Oxidation

Oxidation is a fundamental concept in chemistry that involves the loss of electrons from a substance. This can lead to an increase in oxidation state. To easily spot an oxidation reaction, look for the element where the positive charge increases, or electrons are on the product side of the reaction. For instance, in the reaction \( \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \), the iron ion (Fe) is losing an electron.
This indicates oxidation as it transforms from the +2 oxidation state to the +3 oxidation state. A convenient mnemonic to remember oxidation is "LEO goes GER": **LEO** stands for "Loss of Electrons is Oxidation." This helps you recognize that oxidation always involves losing electrons, not gaining.

Reduction

Reduction is the opposite of oxidation and involves the gain of electrons. This results in a decrease in oxidation state. To identify a reduction reaction, find the reactant gaining electrons or decreasing its positive charge. For example, in the reaction \( \mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \), manganese (Mn) is gaining five electrons.
Through this electron gain, the oxidation state of Mn changes from +7 in \( \mathrm{MnO}_{4}^{-} \) to +2 in \( \mathrm{Mn}^{2+} \), which qualifies as reduction. The mnemonic "LEO goes GER" continues to be helpful: **GER** stands for "Gain of Electrons is Reduction." This will help you identify reduction reactions quickly by pinpointing where electron gain occurs.

Electron Transfer

Electron transfer is the essence of redox reactions, comprising both oxidation and reduction processes. It represents the movement of electrons from one reactant to another. To determine electron transfer, observe both half-reactions where one substance loses electrons and the other gains them, demonstrating a balanced give-and-take system. For example, in reaction \( 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \), the hydrogen ions are accepting electrons to form molecular hydrogen. While in \( \mathrm{F}_{2} \rightarrow 2 \mathrm{F}^{-}+2 \mathrm{e}^{-} \), fluorine atoms are losing electrons, indicating electron transfer from F to form \( \mathrm{F}^{-} \) ions.Understanding electron transfer is critical to mastering redox reactions, which is why these reactions are often found in processes like battery operation and cellular respiration. Being able to identify electron flow helps in predicting reaction behavior. Look for where electrons appear in a reaction, as this marks the point of electron transfer.