Question 17 of 31advanced🔧 ApplyLong Answer5 marks

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Correct Answer

The statement is justified based on the definitions of oxidising agent and reducing agent. An oxidising agent increases the oxidation number of an element in a given substance, whereas a reducing agent lowers the oxidation number of an element in a given substance. If the reducing agent is in excess, it causes a greater decrease in oxidation number, forming a lower oxidation state compound. If the oxidising agent is in excess, it causes a greater increase in oxidation number, forming a higher oxidation state compound.

Illustration 1: Reaction of phosphorus with chlorine. When reducing agent P4P_4 is in excess: P4+3Cl2→2PCl3P_4 + 3Cl_2 \rightarrow 2PCl_3 (Oxidation state of P is +3). When oxidising agent Cl2Cl_2 is in excess: P4+10Cl2→4PCl5P_4 + 10Cl_2 \rightarrow 4PCl_5 (Oxidation state of P is +5).

Illustration 2: Reaction of sodium with oxygen. When reducing agent NaNa is in excess: 4Na+O2→2Na2O4Na + O_2 \rightarrow 2Na_2O (Oxidation state of O is -2). When oxidising agent O2O_2 is in excess: 2Na+O2→Na2O22Na + O_2 \rightarrow Na_2O_2 (Oxidation state of O is -1).

Illustration 3: Reaction of carbon with oxygen. When reducing agent CC is in excess: 2C+O2→2CO2C + O_2 \rightarrow 2CO (Oxidation state of C is +2). When oxidising agent O2O_2 is in excess: C+O2→CO2C + O_2 \rightarrow CO_2 (Oxidation state of C is +4).

Exercise: EXERCISES | Q: 7.11 | (Page Page 18)
For More Understanding

Explanation

The answer applies the textbook definitions of oxidation (increase in oxidation number) and reduction (decrease in oxidation number) to explain how the excess of a particular reagent dictates the final oxidation state of the product. The three standard illustrations (P4/Cl2, Na/O2, and C/O2) demonstrate that an excess of the reducing agent limits the oxidation process, yielding a lower oxidation state, while an excess of the oxidising agent drives the oxidation further, yielding a higher oxidation state.

Solution Steps

  1. Step 1: State the definitions of oxidising and reducing agents in terms of oxidation number changes.

  2. Step 2: Explain how excess reducing agent leads to a lower oxidation state and excess oxidising agent leads to a higher oxidation state.

  3. Step 3: Provide the first illustration (Phosphorus and Chlorine) showing variable oxidation states of P (+3 and +5).

  4. Step 4: Provide the second illustration (Sodium and Oxygen) showing variable oxidation states of O (-2 and -1).

  5. Step 5: Provide the third illustration (Carbon and Oxygen) showing variable oxidation states of C (+2 and +4).

PreviousNext