Question

decide whether each chemical reaction in the table below is an oxidation - reduction (
edox\) reaction. if the reaction is a redox reaction, write down the formula of the reducing agent and the formula of the oxidizing agent.
co(g)+h₂o(g)→co₂(g)+h₂(g)
fe(s)+cuso₄(aq)→feso₄(aq)+cu(s)
hno₃(aq)+h₂o(l)→h₃o⁺(aq)+no₃⁻(aq)

Explanation:

Step1: Determine oxidation - states

In $CO(g)+H_2O(g)\to CO_2(g)+H_2(g)$:

• In $CO$, $C$ has an oxidation - state of +2 and $O$ has - 2. In $H_2O$, $H$ has +1 and $O$ has - 2. In $CO_2$, $C$ has +4 and $O$ has - 2. In $H_2$, $H$ has 0.
• $C$ in $CO$ is oxidized (from +2 to +4), and $H$ in $H_2O$ is reduced (from +1 to 0). So it is a redox reaction.
• The reducing agent is the substance that gets oxidized, so the reducing agent is $CO$.
• The oxidizing agent is the substance that gets reduced, so the oxidizing agent is $H_2O$.

Step2: Analyze second reaction

In $Fe(s)+CuSO_4(aq)\to FeSO_4(aq)+Cu(s)$:

• $Fe$ has an oxidation - state of 0 in $Fe(s)$, $Cu$ has +2 in $CuSO_4$, $Fe$ has +2 in $FeSO_4$, and $Cu$ has 0 in $Cu(s)$.
• $Fe$ is oxidized (from 0 to +2), and $Cu$ is reduced (from +2 to 0). So it is a redox reaction.
• The reducing agent is $Fe$.
• The oxidizing agent is $CuSO_4$.

Step3: Examine third reaction

In $HNO_3(aq)+H_2O(l)\to H_3O^+(aq)+NO_3^-(aq)$:

• There is no change in oxidation - states of any element. $H$ has +1, $N$ has +5, and $O$ has - 2 throughout the reaction. So it is not a redox reaction.

Answer:

For $CO(g)+H_2O(g)\to CO_2(g)+H_2(g)$:
redox reaction? yes
reducing agent: $CO$
oxidizing agent: $H_2O$
For $Fe(s)+CuSO_4(aq)\to FeSO_4(aq)+Cu(s)$:
redox reaction? yes
reducing agent: $Fe$
oxidizing agent: $CuSO_4$
For $HNO_3(aq)+H_2O(l)\to H_3O^+(aq)+NO_3^-(aq)$:
redox reaction? no
reducing agent: N/A
oxidizing agent: N/A