Question

for each reaction, identify the substance that is the oxidizing agent and the substance that is the reducing agent.
c₂h₄ + h₂ → c₂h₆
fe₂o₃ + co → 2feo + co₂
2ki + cl₂ → 2kcl + i₂
answer bank
oxidizing agent reducing agent

Explanation:

Step1: Recall definitions

An oxidizing agent is reduced (gains electrons) and a reducing agent is oxidized (loses electrons).

Step2: Analyze first reaction ($C_2H_4 + H_2

ightarrow C_2H_6$)
Hydrogen ($H_2$) is added to ethene ($C_2H_4$). $H_2$ is oxidized (loses electrons to form $H^+$ which bonds with $C_2H_4$), so $H_2$ is the reducing agent. $C_2H_4$ is reduced, so $C_2H_4$ is the oxidizing agent.

Step3: Analyze second reaction ($Fe_2O_3 + CO

ightarrow 2FeO+CO_2$)
In $Fe_2O_3$, iron has an oxidation - state of +3. In $FeO$, iron has an oxidation - state of +2. $Fe_2O_3$ is reduced, so $Fe_2O_3$ is the oxidizing agent. In $CO$, carbon has an oxidation - state of +2. In $CO_2$, carbon has an oxidation - state of +4. $CO$ is oxidized, so $CO$ is the reducing agent.

Step4: Analyze third reaction ($2KI + Cl_2

ightarrow 2KCl+I_2$)
In $KI$, iodine has an oxidation - state of - 1. In $I_2$, iodine has an oxidation - state of 0. $KI$ is oxidized, so $KI$ is the reducing agent. In $Cl_2$, chlorine has an oxidation - state of 0. In $KCl$, chlorine has an oxidation - state of - 1. $Cl_2$ is reduced, so $Cl_2$ is the oxidizing agent.

Answer:

• For $C_2H_4 + H_2

ightarrow C_2H_6$: Oxidizing agent: $C_2H_4$, Reducing agent: $H_2$

• For $Fe_2O_3 + CO

ightarrow 2FeO+CO_2$: Oxidizing agent: $Fe_2O_3$, Reducing agent: $CO$

• For $2KI + Cl_2

ightarrow 2KCl+I_2$: Oxidizing agent: $Cl_2$, Reducing agent: $KI$