Question

In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced: a. \(2 \mathrm{Li}(s)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{LiF}(s)\) b. \(\mathrm{Cl}_{2}(g)+2 \mathrm{KI}(a q) \longrightarrow 2 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\) c. \(2 \mathrm{Al}(s)+3 \mathrm{Sn}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \operatorname{Sn}(s)\) d. \(\mathrm{Fe}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{Cu}(s)\)

Short answer

a. Li is oxidized, F is reduced b. I is oxidized, Cl is reduced c. Al is oxidized, Sn is reduced d. Fe is oxidized, Cu is reduced

Step-by-step solution

- Identify Reactants and Products

For each reaction provided, list all reactants and products clearly. This helps in identifying which species undergo oxidation and reduction.

- Determine Oxidation Numbers

Assign oxidation numbers to each atom in the reactants and products. This will help determine which atoms gain and lose electrons.

- Identify Oxidation and Reduction

Compare the oxidation numbers from reactants to products. The atom whose oxidation number increases is oxidized (loses electrons), and the atom whose oxidation number decreases is reduced (gains electrons).

- Apply to Reaction (a)

For the reaction: \(2 \text{Li}(s) + \text{F}_2(g) \rightarrow 2 \text{LiF}(s)\), - Li: Oxidation number changes from 0 to +1 (Oxidized)- F: Oxidation number changes from 0 to -1 (Reduced)

- Apply to Reaction (b)

For the reaction: \(\text{Cl}_2(g) + 2 \text{KI}(aq) \rightarrow 2 \text{KCl}(aq) + \text{I}_2(s)\), - I: Oxidation number changes from -1 to 0 (Oxidized)- Cl: Oxidation number changes from 0 to -1 (Reduced)

- Apply to Reaction (c)

For the reaction: \(2 \text{Al}(s) + 3 \text{Sn}^{2+}(aq) \rightarrow 2 \text{Al}^{3+}(aq) + 3 \text{Sn}(s)\), - Al: Oxidation number changes from 0 to +3 (Oxidized)- Sn: Oxidation number changes from +2 to 0 (Reduced)

- Apply to Reaction (d)

For the reaction: \(\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)\), - Fe: Oxidation number changes from 0 to +2 (Oxidized)- Cu: Oxidation number changes from +2 to 0 (Reduced)

Key concepts

oxidation number

The oxidation number, or oxidation state, is a key concept in understanding redox reactions. It represents the hypothetical charge of an atom if all bonds were ionic. Here's a simple rule to remember: In their elemental form, atoms have an oxidation state of 0.

For example, in \[\text{F}_2(g)\], both fluorine atoms have an oxidation state of 0.

Some basic rules for determining oxidation numbers:

• Oxygen usually has an oxidation number of -2.
• Hydrogen typically has an oxidation number of +1.
• Group 1 elements (like Li, Na) have an oxidation number of +1.
• Group 2 elements (like Mg, Ca) have an oxidation number of +2.
• Fluorine always has an oxidation number of -1.

Analyzing these rules, you can determine oxidation numbers for complex molecules, crucial for identifying which elements are oxidized and which are reduced.

electron transfer

Electron transfer is central to oxidation-reduction (redox) reactions. When a substance loses electrons, it undergoes oxidation. Conversely, when a substance gains electrons, it undergoes reduction.

To visualize this, consider the reaction \[\text{2Li}(s) + \text{F}_2(g) \longrightarrow 2\text{LiF}(s)\]. Lithium atoms lose electrons, which means lithium is oxidized. Fluorine atoms gain electrons, meaning fluorine is reduced.

• Lithium (Li) goes from oxidation number 0 to +1, losing one electron per atom.
• Fluorine (F) goes from oxidation number 0 to -1, gaining one electron per atom.

Seeing how electrons transfer between reactants helps in understanding the 'flow' of a redox reaction.

redox reactions

Redox reactions involve simultaneous oxidation and reduction. These processes are sealed to the idea that \[\text{Oxidation}\] always accompanies \[\text{Reduction}\]. One of the reactants gives up electrons while the other gains them.

For example, in the following reaction:
\text{\text{Fe}}(s) + \text{\text{CuSO}}_4(aq) \rightarrow \text{\text{FeSO}}_4(aq) + \text{\text{Cu}}(s)
Iron (Fe) is oxidized from 0 to +2, while copper (Cu) is reduced from +2 to 0. This reaction shows iron's loss of electrons (oxidation) and copper's gain of electrons (reduction).

A handy mnemonic: OIL RIG - \text{Oxidation Is Loss}, \text{Reduction Is Gain}.

chemical equations

Chemical equations concisely present a chemical reaction with reactants and products. They must be balanced, meaning the number of atoms for each element is the same before and after the reaction.

Consider the balanced equation:
\[2\text{Li}(s) + \text{F}_2(g) \rightarrow 2\text{LiF}(s)\]
Here, we see:

• 2 lithium atoms react with 1 fluorine molecule (consisting of 2 fluorine atoms).
• The products are 2 units of lithium fluoride (LiF) where each LiF contains 1 lithium and 1 fluorine atom.

Properly balancing equations ensures the law of conservation of mass is upheld. Thus, closing loops on redox reactions helps critically analyze both electron transfer and chemical equilibrium.