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
2 cro₄²⁻(aq)+2 h₃o⁺(aq)→cr₂o₇²⁻(aq)+3 h₂o(l)
2 cl⁻(aq)+2 h₂o(l)→2 oh⁻(aq)+h₂(g)+cl₂(g)
4 hf(g)+sio₂(s)→sif₄(g)+2 h₂o(g)
4 ki(aq)+2 cucl₂(aq)→2 cui(s)+i₂(aq)+4 kcl(aq)

Explanation:

Step1: Determine oxidation - state rules

Oxidation state of oxygen is - 2 (except in peroxides), hydrogen is + 1 (except in metal hydrides), and in a neutral compound, the sum of oxidation states is 0, and in an ion, it is equal to the charge of the ion.

Step2: Analyze first reaction ($2CrO_{4}^{2 - }(aq)+2H_{3}O^{+}(aq)\to Cr_{2}O_{7}^{2 - }(aq)+3H_{2}O(l)$)

For $CrO_{4}^{2 - }$, let the oxidation state of $Cr$ be $x$. Then $x + 4\times(-2)=-2$, so $x = + 6$. For $Cr_{2}O_{7}^{2 - }$, let the oxidation state of $Cr$ be $y$. Then $2y+7\times(-2)=-2$, so $y = + 6$. The highlighted $Cr$ atom has an oxidation state of + 6 on both sides, so it is neither oxidized nor reduced.

Step3: Analyze second reaction ($2Cl^{-}(aq)+2H_{2}O(l)\to2OH^{-}(aq)+H_{2}(g)+Cl_{2}(g)$)

The oxidation state of $Cl$ in $Cl^{-}$ is - 1, and in $Cl_{2}$ is 0. The $Cl$ atom is oxidized as its oxidation state increases from - 1 to 0.

Step4: Analyze third reaction ($4HF(g)+SiO_{2}(s)\to SiF_{4}(g)+2H_{2}O(g)$)

In $SiO_{2}$, the oxidation state of $Si$ is + 4 ($x+2\times(-2) = 0$, so $x = + 4$), and in $SiF_{4}$, the oxidation state of $Si$ is also + 4 ($x + 4\times(-1)=0$, so $x = + 4$). The highlighted $Si$ atom is neither oxidized nor reduced.

Step5: Analyze fourth reaction ($4KI(aq)+2CuCl_{2}(aq)\to2CuI(s)+I_{2}(aq)+4KCl(aq)$)

In $CuCl_{2}$, the oxidation state of $Cu$ is + 2. In $CuI$, the oxidation state of $Cu$ is + 1. The $Cu$ atom is reduced as its oxidation state decreases from + 2 to + 1.

Answer:

reaction	highlighted atom is being...
$2Cl^{-}(aq)+2H_{2}O(l)\to2OH^{-}(aq)+H_{2}(g)+Cl_{2}(g)$	oxidized
$4HF(g)+SiO_{2}(s)\to SiF_{4}(g)+2H_{2}O(g)$	neither oxidized nor reduced
$4KI(aq)+2CuCl_{2}(aq)\to2CuI(s)+I_{2}(aq)+4KCl(aq)$	reduced