Questions: For each reaction in the table below, write the chemical formulae of any reactants that will be oxidized in the second column of the table. Write the chemical formulae of any reactants that will be reduced in the third column. reaction reactants oxidized reactants reduced --------- 2 Cl2(g) + Pb(s) → PbCl4(s) I2(s) + Cu(s) → CuI2(s) S8(s) + 8 Mg(s) → 8 MgS(s)

For each reaction in the table below, write the chemical formulae of any reactants that will be oxidized in the second column of the table. Write the chemical formulae of any reactants that will be reduced in the third column.

reaction reactants oxidized reactants reduced
---------
2 Cl2(g) + Pb(s) → PbCl4(s)
I2(s) + Cu(s) → CuI2(s)
S8(s) + 8 Mg(s) → 8 MgS(s)

Transcript text: For each reaction in the table below, write the chemical formulae of any reactants that will be oxidized in the second column of the table. Write the chemical formulae of any reactants that will be reduced in the third column. \begin{tabular}{|c|c|c|} \hline reaction & \begin{tabular}{c} reactants \\ oxidized \end{tabular} & \begin{tabular}{c} reactants \\ reduced \end{tabular} \\ \hline $2 \mathrm{Cl}_{2}(g)+\mathrm{Pb}(s) \rightarrow \mathrm{PbCl}_{4}(s)$ & $\square$ & $\square$ \\ \hline $\mathrm{I}_{2}(s)+\mathrm{Cu}(s) \rightarrow \mathrm{CuI}_{2}(s)$ & $\square$ & $\square$ \\ \hline $\mathrm{S}_{8}(s)+8 \mathrm{Mg}(s) \rightarrow 8 \mathrm{MgS}(s)$ & $\square$ & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Identify Oxidation and Reduction in Reactions

In a chemical reaction, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. To determine which reactants are oxidized or reduced, we need to analyze the changes in oxidation states of the elements involved in the reactions.

Step 2: Analyze Reaction 1

Reaction: $2 \mathrm{Cl}_{2}(g) + \mathrm{Pb}(s) \rightarrow \mathrm{PbCl}_{4}(s)$

Oxidation: Lead ($\mathrm{Pb}$) goes from an oxidation state of 0 to +4 in $\mathrm{PbCl}_{4}$, indicating it is oxidized.
Reduction: Chlorine ($\mathrm{Cl}_{2}$) goes from an oxidation state of 0 to -1 in $\mathrm{PbCl}_{4}$, indicating it is reduced.

Step 3: Analyze Reaction 2

Reaction: $\mathrm{I}_{2}(s) + \mathrm{Cu}(s) \rightarrow \mathrm{CuI}_{2}(s)$

Oxidation: Copper ($\mathrm{Cu}$) goes from an oxidation state of 0 to +2 in $\mathrm{CuI}_{2}$, indicating it is oxidized.
Reduction: Iodine ($\mathrm{I}_{2}$) goes from an oxidation state of 0 to -1 in $\mathrm{CuI}_{2}$, indicating it is reduced.

Step 4: Analyze Reaction 3

Reaction: $\mathrm{S}_{8}(s) + 8 \mathrm{Mg}(s) \rightarrow 8 \mathrm{MgS}(s)$

Oxidation: Magnesium ($\mathrm{Mg}$) goes from an oxidation state of 0 to +2 in $\mathrm{MgS}$, indicating it is oxidized.
Reduction: Sulfur ($\mathrm{S}_{8}$) goes from an oxidation state of 0 to -2 in $\mathrm{MgS}$, indicating it is reduced.

Final Answer

\[ \begin{array}{|c|c|c|} \hline \text{reaction} & \begin{array}{c} \text{reactants} \\ \text{oxidized} \end{array} & \begin{array}{c} \text{reactants} \\ \text{reduced} \end{array} \\ \hline 2 \mathrm{Cl}_{2}(g)+\mathrm{Pb}(s) \rightarrow \mathrm{PbCl}_{4}(s) & \boxed{\mathrm{Pb}} & \boxed{\mathrm{Cl}_{2}} \\ \hline \mathrm{I}_{2}(s)+\mathrm{Cu}(s) \rightarrow \mathrm{CuI}_{2}(s) & \boxed{\mathrm{Cu}} & \boxed{\mathrm{I}_{2}} \\ \hline \mathrm{S}_{8}(s)+8 \mathrm{Mg}(s) \rightarrow 8 \mathrm{MgS}(s) & \boxed{\mathrm{Mg}} & \boxed{\mathrm{S}_{8}} \\ \hline \end{array} \]

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