Question

In the balanced reaction: 2 I^- + Cl_2 → 2Cl^– + I_2, ________ is the reducing agent, and ________ is the oxidizing agent.

Options:
a. I^–, Cl^–
b. Cl^–, I^–
c. I^–, I_2
d. Cl_2, Cl^–

Answer 1

In the reaction 2 I^- + Cl_2 → 2Cl^- + I_2, the reducing agent is I^- as it loses electrons, and the oxidizing agent is Cl_2 as it gains electrons. Option c, I^-, Cl_2 is correct.

Step-by-step explanation:

In the balanced reaction 2 I^- + Cl_2 → 2Cl^- + I_2, the reducing agent is I^-, and the oxidizing agent is Cl_2. This can be determined by examining the changes in oxidation states for the species involved in the reaction. The iodide ion, I^-, loses electrons as it is oxidized to I_2, making it the reducing agent because it provides electrons to another substance. On the other hand, Cl_2 gains electrons and is reduced to Cl^-, which makes it the oxidizing agent because it accepts electrons from another substance.

Therefore, the correct option is c. I^-, Cl_2. It's important to note that in redox reactions, the substance that gets oxidized (loses electrons) is the reducing agent, and the substance that gets reduced (gains electrons) is the oxidizing agent.