Questions: reaction highlighted atom is being... ------------ oxidized reduced neither oxidized nor reduced CH₄(g)+H₂O(g) → CO(g)+3H₂(g) 0 0 H₂S(aq)+2NaOH(aq) → Na₂S(aq)+2H₂O(l) FeO(s)+CO(g) → Fe(s)+CO₂(g) 0 N₂(g)+3H₂(g) → 2NH₃(g)

reaction highlighted atom is being...
------------
oxidized reduced neither oxidized nor reduced
CH₄(g)+H₂O(g) → CO(g)+3H₂(g) 0 0
H₂S(aq)+2NaOH(aq) → Na₂S(aq)+2H₂O(l)
FeO(s)+CO(g) → Fe(s)+CO₂(g) 0
N₂(g)+3H₂(g) → 2NH₃(g)

Transcript text: \begin{tabular}{|c|c|c|c|} \hline reaction & \multicolumn{2}{|c|}{ highlighted atom is being... } \\ \cline { 2 - 5 } & oxidized & reduced \begin{tabular}{c} neither \\ oxidized nor \\ reduced \end{tabular} \\ \hline $\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$ & 0 & 0 & \\ \hline $\mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{NaOH}_{(a q)} \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{(a q)}+2 \mathrm{H}_{2} \mathrm{O}(l)$ & & \\ \hline $\mathrm{FeO}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)$ & & 0 & \\ \hline $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)$ & & & \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Analyze the First Reaction

The first reaction is: \[ \mathrm{CH}_{4}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightarrow \mathrm{CO}(g) + 3\mathrm{H}_{2}(g) \] We need to determine if the highlighted atom (carbon in this case) is being oxidized, reduced, or neither.

In $\mathrm{CH}_4$, the oxidation state of carbon is $-4$.
In $\mathrm{CO}$, the oxidation state of carbon is $+2$.

Since the oxidation state of carbon increases from $-4$ to $+2$, carbon is being oxidized.

Step 2: Analyze the Second Reaction

The second reaction is: \[ \mathrm{H}_{2}\mathrm{S}(aq) + 2\mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_{2}\mathrm{S}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) \] We need to determine if the highlighted atom (sulfur in this case) is being oxidized, reduced, or neither.

In $\mathrm{H}_2\mathrm{S}$, the oxidation state of sulfur is $-2$.
In $\mathrm{Na}_2\mathrm{S}$, the oxidation state of sulfur is still $-2$.

Since the oxidation state of sulfur does not change, sulfur is neither oxidized nor reduced.

Step 3: Analyze the Third Reaction

The third reaction is: \[ \mathrm{FeO}(s) + \mathrm{CO}(g) \rightarrow \mathrm{Fe}(s) + \mathrm{CO}_2(g) \] We need to determine if the highlighted atom (iron in this case) is being oxidized, reduced, or neither.

In $\mathrm{FeO}$, the oxidation state of iron is $+2$.
In $\mathrm{Fe}$, the oxidation state of iron is $0$.

Since the oxidation state of iron decreases from $+2$ to $0$, iron is being reduced.

Final Answer

\[ \begin{array}{|c|c|c|c|} \hline \text{reaction} & \multicolumn{2}{|c|}{\text{highlighted atom is being...}} \\ \cline{2-4} & \text{oxidized} & \text{reduced} & \text{neither oxidized nor reduced} \\ \hline \mathrm{CH}_{4}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightarrow \mathrm{CO}(g) + 3\mathrm{H}_{2}(g) & 1 & 0 & 0 \\ \hline \mathrm{H}_{2}\mathrm{S}(aq) + 2\mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_{2}\mathrm{S}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) & 0 & 0 & 1 \\ \hline \mathrm{FeO}(s) + \mathrm{CO}(g) \rightarrow \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) & 0 & 1 & 0 \\ \hline \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g) & & & \\ \hline \end{array} \]

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