Question

Complete and balance the following equations. If no reaction occurs, write NR: (a) \(\mathrm{H}_{3} \mathrm{PO}_{4}(l)+\operatorname{NaI}(s) \longrightarrow\) (b) \(\mathrm{Cl}_{2}(g)+\mathrm{I}^{-}(a q) \longrightarrow\) (c) \(\mathrm{Br}_{2}(l)+\mathrm{Cl}^{-}(a q) \longrightarrow\) (d) \(\mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow\)

Short answer

NR; \(\mathrm{Cl}_{2}(g) + 2\mathrm{I}^{-}(aq) \longrightarrow 2\mathrm{Cl}^{-}(aq) + \mathrm{I}_{2}(s)\); NR; \(\mathrm{ClF}(g) + \mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g)\)

Step-by-step solution

- Analyze the first reaction

Let's consider the given reagents: ewline \textbf{(a)} \ \(\mathrm{H}_{3}\mathrm{PO}_{4}(l) + \operatorname{NaI}(s) \longrightarrow \). \(\mathrm{H}_{3}\mathrm{PO}_{4}\) is a phosphoric acid and \(\operatorname{NaI}\) is sodium iodide. Since NaI is a salt and phosphoric acid doesn't react with it, there is no reaction.

Conclusion for (a)

The reaction doesn’t occur: NR (No Reaction)

- Analyze the second reaction

Let's consider the second reaction: ewline \textbf{(b)} \(\mathrm{Cl}_{2}(g) + \mathrm{I}^{-}(aq) \longrightarrow \). In this case, chlorine gas can oxidize iodide ions to form iodine and chloride ions.

Balancing for (b)

The balanced equation for the reaction is: \[\mathrm{Cl}_{2}(g) + 2\mathrm{I}^{-}(aq) \longrightarrow 2\mathrm{Cl}^{-}(aq) + \mathrm{I}_{2}(s)\]

- Analyze the third reaction

Let's consider the third reaction: ewline \textbf{(c)} \(\mathrm{Br}_{2}(l) + \mathrm{Cl}^{-}(aq) \longrightarrow \). In this case, bromine cannot oxidize chloride ions because bromine is less reactive than chlorine. Thus, there is no reaction.

Conclusion for (c)

The reaction doesn’t occur: NR (No Reaction)

- Analyze the fourth reaction

Let's consider the last reaction: ewline \textbf{(d)} \(\mathrm{ClF}(g) + \mathrm{F}_{2}(g) \longrightarrow \). Chlorine monofluoride and fluorine gas may react to form chlorine trifluoride (\(\mathrm{ClF}_{3}(g)\)).

Balancing for (d)

The balanced equation for this reaction is: \[\mathrm{ClF}(g) + \mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g)\]

Key concepts

oxidation-reduction reactions

Oxidation-reduction reactions, or redox reactions, are chemical reactions where the oxidation states of atoms are changed. These reactions involve the transfer of electrons between chemical species. The substance that loses electrons is said to be oxidized, and the substance that gains electrons is said to be reduced.
For instance, in the reaction \textbf{(b)} \(\text{Cl}_{2}(g) + \text{I}^{-}(aq) \rightarrow\text{Cl}^{-}(aq) + \text{I}_{2}(s)\), chlorine gas (\text{Cl}_{2}) is reduced to chloride ions (\text{Cl}^{-}), and iodide ions (\text{I}^{-}) are oxidized to iodine (\text{I}_{2}).
Understanding who donates and accepts electrons helps in figuring out the changes in oxidation states.

• Oxidation: Loss of electrons
• Reduction: Gain of electrons
• Oxidizing agent: Gains electrons (is reduced)
• Reducing agent: Loses electrons (is oxidized)

reactivity series

The reactivity series is a list of elements organized by their ability to displace other elements from compounds. Highly reactive elements can displace less reactive elements from their compounds.
For example, in the given reactions:
\textbf{(c)} \(\text{Br}_{2}(l) + \text{Cl}^{-}(aq) \rightarrow\) shows that no reaction occurs because bromine (\text{Br}_{2}) is less reactive than chlorine (\text{Cl}^{-}). Therefore, bromine cannot oxidize chloride ions.
This concept is crucial in predicting whether a single displacement reaction will occur.

• More reactive elements displace less reactive ones
• Elements higher in the series will oxidize those below them

chemical reaction predictions

Predicting chemical reactions involves understanding the chemical nature of the reactants and the conditions under which they react. Not every combination of reactants will result in a reaction.
For example, in the given reactions:
\textbf{(a)} \(\text{H}_{3}\text{PO}_{4}(l) + \text{NaI}(s) \rightarrow \text{NR}\) shows no reaction because phosphoric acid (\text{H}_{3}\text{PO}_{4}) does not react with sodium iodide (\text{NaI}).
Several factors can influence predictions:

• Chemical reactivity (more reactive elements displace less reactive ones)
• Solubility rules (whether products are soluble or insoluble in water)
• Nature of reactants (some chemicals simply do not react with others)

chemical equation balancing

Balancing chemical equations is essential to ensure mass conservation in a reaction. It ensures that the number of each type of atom on both sides of the equation is the same.
For example, balancing the reaction \textbf{(d)} \(\text{ClF}(g) + \text{F}_{2}(g) \rightarrow \text{ClF}_{3}(g)\) involves ensuring that chlorine and fluorine atoms are equal on both sides of the equation.
Steps to balance equations:

• Write the unbalanced equation
• Count atoms of each element on both sides
• Add coefficients to balance the atoms (start with the most complex molecule)
• Repeat the counting and adjusting until balanced

For this reaction: Starting with 1 chlorine and 3 fluorine atoms on the product side, we adjust the coefficients to balance the equation:
\text{\text{ClF}(g) + \text{F}_{2}(g) \rightarrow \text{ClF}_{3}(g}}.