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 c(s)+o₂(g)→co₂(g) na₂co₃(s)+h₃po₄(aq)→na₂hpo₄(aq)+co₂(g)+h₂o(l) 2h₂(g)+o₂(g)→2h₂o(g) 4hf(g)+sio₂(s)→sif₄(g)+2h₂o(g)

Explanation:

Step1: Determine oxidation - state rules

Oxidation is an increase in oxidation state, and reduction is a decrease in oxidation state. For elements in their elemental form, the oxidation state is 0. For oxygen in compounds (except in peroxides), the oxidation state is - 2.

Step2: Analyze first reaction $C(s)+O_2(g)\to CO_2(g)$

The oxidation state of $C$ in $C(s)$ is 0. In $CO_2$, let the oxidation state of $C$ be $x$. Since $O$ has an oxidation state of - 2 and there are 2 $O$ atoms, $x + 2\times(-2)=0$, so $x = + 4$. The oxidation state of $C$ increases from 0 to + 4, so $C$ is oxidized.

Step3: Analyze second reaction $Na_2CO_3(s)+H_3PO_4(aq)\to Na_2HPO_4(aq)+CO_2(g)+H_2O(l)$

In $Na_2CO_3$, let the oxidation state of $C$ be $x$. $Na$ has an oxidation state of + 1 and $O$ has - 2. So, $2\times(+1)+x + 3\times(-2)=0$, $x=+4$. In $CO_2$, using the same rule for oxygen (- 2), for $CO_2$, $x+2\times(-2) = 0$, $x = + 4$. The oxidation state of $C$ does not change, so $C$ is neither oxidized nor reduced.

Step4: Analyze third reaction $2H_2(g)+O_2(g)\to 2H_2O(g)$

The oxidation state of $O$ in $O_2$ is 0. In $H_2O$, $H$ has an oxidation state of + 1. Let the oxidation state of $O$ be $x$. Then $2\times(+1)+x=0$, $x=-2$. The oxidation state of $O$ decreases from 0 to - 2, so $O$ is reduced.

Step5: Analyze fourth reaction $4HF(g)+SiO_2(s)\to SiF_4(g)+2H_2O(g)$

In $SiO_2$, let the oxidation state of $Si$ be $x$. $O$ has an oxidation state of - 2, so $x + 2\times(-2)=0$, $x = + 4$. In $SiF_4$, $F$ has an oxidation state of - 1. Let the oxidation state of $Si$ be $x$. Then $x+4\times(-1)=0$, $x = + 4$. The oxidation state of $Si$ does not change, so $Si$ is neither oxidized nor reduced.

Answer:

reaction	oxidized	reduced	neither oxidized nor reduced
$Na_2CO_3(s)+H_3PO_4(aq)\to Na_2HPO_4(aq)+CO_2(g)+H_2O(l)$			C
$2H_2(g)+O_2(g)\to 2H_2O(g)$		O
$4HF(g)+SiO_2(s)\to SiF_4(g)+2H_2O(g)$			Si