Questions: Identifying oxidized and reduced reactants in a metal-nonmetal reaction 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 --------- O2(g) + 2 Cu(s) → 2 CuO(s) 8 Ni(s) + S8(s) → 8 NiS(s) Cr(s) + 2 Cl2(g) → CrCl4(s)

Identifying oxidized and reduced reactants in a metal-nonmetal reaction

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
---------
O2(g) + 2 Cu(s) → 2 CuO(s)
8 Ni(s) + S8(s) → 8 NiS(s)
Cr(s) + 2 Cl2(g) → CrCl4(s)

Transcript text: Identifying oxidized and reduced reactants in a metal-nonmetal reaction 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 $\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{Cu}(s) \rightarrow 2 \mathrm{CuO}(s)$ & $\square$ & $\square$ \\ \hline $8 \mathrm{Ni}(s)+\mathrm{S}_{8}(s) \rightarrow 8 \mathrm{NiS}(s)$ & $\square$ & $\square$ \\ \hline $\mathrm{Cr}(s)+2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CrCl}_{4}(s)$ & $\square$ & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Identify the Oxidized and Reduced Reactants in the First Reaction

The first reaction is: \[ \mathrm{O}_{2}(\mathrm{~g}) + 2 \mathrm{Cu}(s) \rightarrow 2 \mathrm{CuO}(s) \]

Oxidation: Copper ($\mathrm{Cu}$) is oxidized because it loses electrons to form $\mathrm{CuO}$.
Reduction: Oxygen ($\mathrm{O}_2$) is reduced because it gains electrons to form $\mathrm{CuO}$.

Step 2: Identify the Oxidized and Reduced Reactants in the Second Reaction

The second reaction is: \[ 8 \mathrm{Ni}(s) + \mathrm{S}_{8}(s) \rightarrow 8 \mathrm{NiS}(s) \]

Oxidation: Nickel ($\mathrm{Ni}$) is oxidized because it loses electrons to form $\mathrm{NiS}$.
Reduction: Sulfur ($\mathrm{S}_8$) is reduced because it gains electrons to form $\mathrm{NiS}$.

Step 3: Identify the Oxidized and Reduced Reactants in the Third Reaction

The third reaction is: \[ \mathrm{Cr}(s) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CrCl}_{4}(s) \]

Oxidation: Chromium ($\mathrm{Cr}$) is oxidized because it loses electrons to form $\mathrm{CrCl}_4$.
Reduction: Chlorine ($\mathrm{Cl}_2$) is reduced because it gains electrons to form $\mathrm{CrCl}_4$.

Final Answer

\[ \begin{array}{|c|c|c|} \hline \text{reaction} & \text{reactants oxidized} & \text{reactants reduced} \\ \hline \mathrm{O}_{2}(\mathrm{~g}) + 2 \mathrm{Cu}(s) \rightarrow 2 \mathrm{CuO}(s) & \boxed{\mathrm{Cu}} & \boxed{\mathrm{O}_2} \\ \hline 8 \mathrm{Ni}(s) + \mathrm{S}_{8}(s) \rightarrow 8 \mathrm{NiS}(s) & \boxed{\mathrm{Ni}} & \boxed{\mathrm{S}_8} \\ \hline \mathrm{Cr}(s) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CrCl}_{4}(s) & \boxed{\mathrm{Cr}} & \boxed{\mathrm{Cl}_2} \\ \hline \end{array} \]

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >

Thank you for your feedback.

Copied!