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)
redox reaction?
o yes o no
reducing agent:
oxidizing agent:

cao(s)+co₂(g)→caco₃(s)
redox reaction?
o yes o no
reducing agent:
oxidizing agent:

feo(s)+co(g)→fe(s)+co₂(g)
redox reaction?
o yes o no
reducing agent:
oxidizing agent:

Explanation:

Step1: Calculate oxidation numbers for \(CO(g)+H_2O(g)\to CO_2(g)+H_2(g)\)

• For \(CO\): \(C\) has oxidation number \(+2\), \(O\) has \(- 2\). For \(H_2O\): \(H\) has \(+1\), \(O\) has \(-2\). For \(CO_2\): \(C\) has \(+4\), \(O\) has \(-2\). For \(H_2\): \(H\) has \(0\).
• \(C\) is oxidized (\(+2\to+4\)), \(H\) is reduced (\(+1\to0\)). So it is a redox reaction.
• Reducing agent is \(CO\) (loses electrons, gets oxidized). Oxidizing agent is \(H_2O\) (gains electrons, gets reduced).

Step2: Calculate oxidation numbers for \(CaO(s)+CO_2(g)\to CaCO_3(s)\)

• For \(CaO\): \(Ca = + 2\), \(O=-2\). For \(CO_2\): \(C = + 4\), \(O=-2\). For \(CaCO_3\): \(Ca = + 2\), \(C = + 4\), \(O=-2\).
• No change in oxidation numbers. So it is not a redox reaction.

Step3: Calculate oxidation numbers for \(FeO(s)+CO(g)\to Fe(s)+CO_2(g)\)

• For \(FeO\): \(Fe = + 2\), \(O=-2\). For \(CO\): \(C = + 2\), \(O=-2\). For \(Fe\): \(Fe = 0\). For \(CO_2\): \(C = + 4\), \(O=-2\).
• \(Fe\) is reduced (\(+2\to0\)), \(C\) is oxidized (\(+2\to+4\)). So it is a redox reaction.
• Reducing agent is \(CO\) (loses electrons, gets oxidized). Oxidizing agent is \(FeO\) (gains electrons, gets reduced).

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 \(CaO(s)+CO_2(g)\to CaCO_3(s)\): redox reaction? no
For \(FeO(s)+CO(g)\to Fe(s)+CO_2(g)\): redox reaction? yes; reducing agent: \(CO\); oxidizing agent: \(FeO\)