To determine how many electrons are gained in the given half-reaction \( O_2 + \text{electrons} \rightarrow 2 O^{2-} \), let's break down the process step-by-step:
1. Identify the Initial and Final States:
- The initial reactant is \( O_2 \), which is a diatomic oxygen molecule.
- The final products are two oxide ions, \( 2 O^{2-} \).
2. Analyze the Oxidation States:
- In \( O_2 \), each oxygen has an oxidation state of 0 (the default state for elements in their diatomic form).
- In \( O^{2-} \), each oxygen has an oxidation state of -2.
3. Calculate the Change in Oxidation State for One Oxygen Atom:
- The oxidation state changes from 0 (in \( O_2 \)) to -2 (in \( O^{2-} \)).
- Therefore, each oxygen atom gains 2 electrons to reduce its oxidation state from 0 to -2.
4. Account for Both Oxygen Atoms:
- In \( O_2 \), there are two oxygen atoms.
- Each oxygen atom gains 2 electrons, so for two oxygen atoms, the total number of electrons gained is:
[tex]\[
2 \text{ (oxygen atoms)} \times 2 \text{ (electrons per atom)} = 4 \text{ electrons}
\][/tex]
Therefore, the total number of electrons gained in the half-reaction is 4.
The correct choice is C. 4
