Question

Determine whether each of the following half-reactions represents an oxidation or a reduction. Supply the correct number of electrons to the appropriate side to balance the cquation. (a) \(\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}^{1+}\) (b) \(\mathrm{F}_{2} \rightarrow 2 \mathrm{~F}^{-}\) (c) \(2 \mathrm{IO}_{4}^{-}+16 \mathrm{H}^{+} \longrightarrow \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{Mn} \longrightarrow \mathrm{Mn}^{2+}\)

Short answer

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

Step-by-step solution

Understand the terms

Oxidation involves the loss of electrons, while reduction involves the gain of electrons. A common mnemonic is 'OIL RIG' – Oxidation Is Loss, Reduction Is Gain.

Analyze Reaction (a)

For the half-reaction \(\text{Cu}^{2+} \rightarrow \text{Cu}^{1+}\), determine the number of electrons needed to balance the charge. In this case, \(\text{Cu}^{2+} \rightarrow \text{Cu}^{1+} + e^{-}\). Since the reactant is losing electrons, this is an oxidation reaction.

Analyze Reaction (b)

For the half-reaction \(\text{F}_{2} \rightarrow 2\text{~F}^{-}\), determine the number of electrons required to balance the charge. We have \(\text{F}_{2} + 2e^{-} \rightarrow 2 \text{~F}^{-}\). Since electrons are being gained, this is a reduction reaction.

Analyze Reaction (c)

For the half-reaction \(2 \text{IO}_{4}^{-} + 16 \text{H}^{+} \rightarrow \text{I}_{2} + 8 \text{H}_{2} \text{O}\), determine the number of electrons. Balancing both charges and atoms: \(2 \text{IO}_{4}^{-} + 16 \text{H}^{+} + 10 e^{-} \rightarrow \text{I}_{2} + 8 \text{H}_{2} \text{O}\). Electrons are added to the reactants indicating a reduction.

Analyze Reaction (d)

For the half-reaction \(\text{Mn} \rightarrow \text{Mn}^{2+}\), add the correct number of electrons to balance the charge \(\text{Mn} \rightarrow \text{Mn}^{2+} + 2e^{-}\). Since electrons are lost, this is an oxidation reaction.

Key concepts

half-reactions

Half-reactions are a way to split the overall reaction into two parts: oxidation and reduction. Each half-reaction shows either the loss or gain of electrons. This method helps in understanding and balancing redox reactions more effectively. In the analysis of the given exercise, we separated each chemical change, observing how electrons are transferred during the process. Let's take a closer look at each half-reaction:

(a) \(\text{Cu}^{2+} \rightarrow \text{Cu}^{1+}\) is broken down to illustrate copper's electron loss or gain.
(b) \(\text{F}_{2} \rightarrow 2 \text{~F}^{-}\) shows the electron gain by fluorine.
(c) \(2 \text{IO}_{4}^{-} + 16 \text{H}^{+} \rightarrow \text{I}_{2} + 8 \text{H}_{2} \text{O}\) is analyzed for electron changes in iodine and hydrogen.
(d) \(\text{Mn} \rightarrow \text{Mn}^{2+}\) highlights manganese's oxidation.
Understanding each half-reaction allows us to identify the process of oxidation and reduction more clearly.

electron loss and gain

Electron loss and gain are the core of redox reactions. Oxidation is when an atom or molecule loses electrons, making it more positively charged. Reduction is when an atom or molecule gains electrons, making it more negatively charged. The mnemonic 'OIL RIG' can help you remember: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). In the given exercise:

* In part (a) \(\text{Cu}^{2+} \rightarrow \text{Cu}^{1+} + e^{-}\), copper loses an electron, hence it is oxidized.
* In part (b) \(\text{F}_{2} + 2e^{-} \rightarrow 2 \text{~F}^{-}\), fluorine gains electrons, causing reduction.
* In part (c) \(2 \text{IO}_{4}^{-} + 16 \text{H}^{+} + 10 e^{-} \rightarrow \text{I}_{2} + 8 \text{H}_{2} \text{O}\), electrons are gained indicating a reduction.
* In part (d) \(\text{Mn} \rightarrow \text{Mn}^{2+} + 2 e^{-}\), manganese loses electrons, showing oxidation.
Understanding how electrons are transferred provides the foundation for identifying and balancing chemical reactions.

balancing chemical equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation. In redox reactions, it's also crucial to balance the number of electrons lost and gained. Follow these steps to balance equations for the given exercise:

1. Write the unbalanced half-reactions.
2. Balance all elements except hydrogen and oxygen.
3. Balance oxygen atoms by adding \(\text{H}_{2} \text{O}\).
4. Balance hydrogen atoms by adding \(\text{H}^{+}\).
5. Balance the charge by adding electrons.
For example, in reaction (c) \(2 \text{IO}_{4}^{-} + 16 \text{H}^{+} \rightarrow \text{I}_{2} + 8 \text{H}_{2} \text{O}\), the equation is balanced by adding 10 electrons to the reactant side. Balancing ensures that mass and charge are conserved, maintaining the principle of matter conservation.

redox reactions

Redox reactions are chemical processes where oxidation and reduction occur simultaneously. One species loses electrons (oxidized) while another gains electrons (reduced). These reactions are integral in various fields like biochemistry, industrial processes, and environmental science.

To identify redox reactions in the exercise:

* In part (a), the conversion of \(\text{Cu}^{2+} \rightarrow \text{Cu}^{1+}\) is an oxidation, as electrons are lost.
* In part (b), \(\text{F}_{2} \rightarrow 2 \text{~F}^{-}\) represents a reduction, as electrons are gained.
* For part (c), \(2 \text{IO}_{4}^{-} + 16 \text{H}^{+} + 10 e^{-} \rightarrow \text{I}_{2} + 8 \text{H}_{2} \text{O}\) involves the reduction of iodate ions.
* In part (d), \(\text{Mn} \rightarrow \text{Mn}^{2+} + 2 e^{-}\), manganese undergoes oxidation.
Identifying these processes helps in understanding electron transfer mechanisms and energy changes in chemical reactions.