Question

for each chemical reaction listed in the table below, decide whether the highlighted atom is being oxidized or reduced. reaction highlighted atom is being... oxidized reduced neither oxidized nor reduced feo(s)+co(g)→fe(s)+co₂(g) co(g)+h₂o(g)→co₂(g)+h₂(g) h₂s(aq)+2naoh(aq)→na₂s(aq)+2h₂o(l) 2cro₄²⁻(aq)+2h₃o⁺(aq)→cr₂o₇²⁻(aq)+3h₂o(l)

Explanation:

Step1: Determine oxidation states

Oxidation state rules: In a compound, oxygen usually has an oxidation state of - 2, hydrogen + 1, and in a free - element the oxidation state is 0.

Step2: Analyze the first reaction

For the reaction $\text{FeO}(s)+\text{CO}(g)
ightarrow\text{Fe}(s)+\text{CO}_2(g)$. In $\text{FeO}$, let the oxidation state of $\text{Fe}$ be $x$. Since oxygen has an oxidation state of - 2, $x+( - 2)=0$, so $x = + 2$. In $\text{Fe}(s)$, the oxidation state of $\text{Fe}$ is 0. The oxidation state of $\text{Fe}$ decreases from + 2 to 0, so $\text{Fe}$ is reduced.

Step3: Analyze the second reaction

For the reaction $\text{CO}(g)+\text{H}_2\text{O}(g)
ightarrow\text{CO}_2(g)+\text{H}_2(g)$. In $\text{CO}$, let the oxidation state of $\text{C}$ be $y$. Since oxygen has an oxidation state of - 2, $y+( - 2)=0$, so $y = + 2$. In $\text{CO}_2$, let the oxidation state of $\text{C}$ be $z$. Then $z+2\times(-2)=0$, so $z = + 4$. The oxidation state of $\text{C}$ increases from + 2 to + 4, so $\text{C}$ is oxidized.

Step4: Analyze the third reaction

For the reaction $\text{H}_2\text{S}(aq)+2\text{NaOH}(aq)
ightarrow\text{Na}_2\text{S}(aq)+2\text{H}_2\text{O}(l)$. In $\text{H}_2\text{S}$, let the oxidation state of $\text{S}$ be $w$. Since hydrogen has an oxidation state of + 1, $2\times(+1)+w = 0$, so $w=-2$. In $\text{Na}_2\text{S}$, let the oxidation state of $\text{S}$ be $v$. Since sodium has an oxidation state of + 1, $2\times(+1)+v = 0$, so $v=-2$. The oxidation state of $\text{S}$ does not change, so $\text{S}$ is neither oxidized nor reduced.

Step5: Analyze the fourth reaction

For the reaction $2\text{CrO}_{4}^{2 - }(aq)+2\text{H}_3\text{O}^+(aq)
ightarrow\text{Cr}_2\text{O}_{7}^{2 - }(aq)+3\text{H}_2\text{O}(l)$. In $\text{CrO}_{4}^{2 - }$, let the oxidation state of $\text{Cr}$ be $m$. Since oxygen has an oxidation state of - 2, $m + 4\times(-2)=-2$, so $m = + 6$. In $\text{Cr}_2\text{O}_{7}^{2 - }$, let the oxidation state of $\text{Cr}$ be $n$. Then $2n+7\times(-2)=-2$, so $n = + 6$. The oxidation state of $\text{Cr}$ does not change, so $\text{Cr}$ is neither oxidized nor reduced.

Answer:

1. reduced
2. oxidized
3. neither oxidized nor reduced
4. neither oxidized nor reduced