Problem 52 - Chemistry The Central Science In Si Units

Question

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions. (a) \(\mathrm{P}_{4}(s)+10 \mathrm{HClO}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+10 \mathrm{HCl}(a q) $$ (b) \(\mathrm{Br}_{2}(l)+2 \mathrm{~K}(s) \longrightarrow 2 \mathrm{KBr}(s)\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)\) (d) \(\mathrm{ZnCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+\) $$ 2 \mathrm{NaCl}(a q) $$

Step-by-Step Solution

Answer

Reactions (a), (b), and (c) are redox reactions; reaction (d) is a precipitation reaction.

Step 1: Analyze Reaction (a)

The reaction \( \mathrm{P}_{4}(s)+10 \mathrm{HClO}(aq)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(aq) + 10 \mathrm{HCl}(aq) \) involves chlorine going from \( +1 \) in \( \mathrm{HClO} \) to \( -1 \) in \( \mathrm{HCl} \) (reduction), while phosphorus is oxidized from \( 0 \) in \( \mathrm{P}_{4} \) to \( +5 \) in \( \mathrm{H}_{3} \mathrm{PO}_{4} \). Thus, it is a redox reaction.

Step 2: Analyze Reaction (b)

In the reaction \( \mathrm{Br}_{2}(l)+2 \mathrm{K}(s) \rightarrow 2 \mathrm{KBr}(s) \), potassium is oxidized from \( 0 \) to \( +1 \), and bromine is reduced from \( 0 \) to \( -1 \). This is a redox reaction because there is a transfer of electrons.

Step 3: Analyze Reaction (c)

The reaction \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{H}_{2}\mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \) shows carbon in ethanol oxidized from \( -2 \) to \( +4 \) in \( \mathrm{CO}_{2} \), and oxygen reduced from \( 0 \) to \( -2 \) in water. This indicates a redox reaction.

Step 4: Analyze Reaction (d)

The reaction \( \mathrm{ZnCl}_{2}(aq)+2 \mathrm{NaOH}(aq) \rightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{NaCl}(aq) \) does not involve changes in oxidation states. It is a precipitation reaction because \( \mathrm{Zn}(\mathrm{OH})_{2} \) precipitates out of solution.

Key Concepts

Oxidation

Oxidation is a key concept in redox reactions. It involves the process where an element loses electrons. This loss of electrons increases its oxidation state. For instance, in the reaction \( \mathrm{Br}_{2}(l)+2 \mathrm{~K}(s) \longrightarrow 2 \mathrm{KBr}(s) \), potassium goes from an oxidation state of \(0\) to \(+1\). This increase signifies that potassium is being oxidized.
To identify oxidation in a reaction, follow these steps:

• Assign oxidation states to all elements in the reactants and products.
• Look for any increase in oxidation state to spot which element is oxidized.

A simple memory aid is "OIL RIG" where "Oxidation Is Loss" of electrons. Keeping this in mind helps to understand its role in various chemical processes.

Reduction

Reduction is the counterpart of oxidation in redox reactions. It happens when an element gains electrons, which decreases its oxidation state. An example is seen in the same reaction \( \mathrm{Br}_{2}(l)+2 \mathrm{~K}(s) \longrightarrow 2 \mathrm{KBr}(s) \), where bromine is reduced from \(0\) to \(-1\). This decrease indicates bromine is undergoing reduction.
To determine reduction in a reaction, you should:

• Identify the elements involved in the reaction.
• Examine their oxidation states before and after the reaction.
• A decrease in oxidation state shows that reduction has occurred.

"Reduction Is Gain" of electrons is another part of the "OIL RIG" acronym, emphasizing the transfer of electrons that defines reduction.

Precipitation Reactions

Precipitation reactions differ from redox reactions in that they involve the formation of a solid from two aqueous solutions. No transfer of electrons takes place. For example, in the reaction \( \mathrm{ZnCl}_{2}(aq)+2 \mathrm{NaOH}(aq) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{NaCl}(aq) \), \( \mathrm{ZnOH}_{2} \) forms as a precipitate. It's a classic precipitation reaction because the product is not soluble in water.
To identify these reactions:

• Check if a solid is formed from the liquid reactants.
• Look for insoluble compounds identified in solubility rules.

Understanding the nature of solutions and precipitates helps in identifying the type of reaction and balancing chemical equations.

Oxidation States

Oxidation states, also known as oxidation numbers, help determine how electrons are distributed among atoms in a molecule. They play a crucial role in identifying redox reactions. An oxidation state is the hypothetical charge of an atom if all bonds to atom were 100% ionic. It aids in tracking electron transfer by providing a numeric representation.
Here's how to assign oxidation states:

• Elements in their elemental form have an oxidation state of \(0\).
• For a simple monatomic ion, the oxidation state is equal to the charge on the ion.
• In compounds, hydrogen is usually \(+1\) and oxygen \(-2\), except in peroxides where oxygen is \(-1\).
• The sum of oxidation states for all atoms in a neutral compound must be zero.
• In polyatomic ions, the sum must equal the charge of the ion.

By assigning oxidation states, you can easily determine which species are oxidized or reduced in a reaction.