To determine the correct half-reaction when hydrogen reacts with chlorine to form hydrogen chloride ([tex]\( H_2 + Cl_2 \rightarrow 2 HCl \)[/tex]), let's carefully examine the half-reactions. In this reaction, hydrogen ([tex]\( H_2 \)[/tex]) is oxidized, and chlorine ([tex]\( Cl_2 \)[/tex]) is reduced.
1. Oxidation and Reduction Basics:
- Oxidation involves the loss of electrons.
- Reduction involves the gain of electrons.
2. Examining Each Option:
Option A: [tex]\( H_2 \rightarrow 2 H^- + 2 e^- \)[/tex]; oxidation
- This indicates hydrogen (H_2) loses electrons and forms hydride ions ([tex]\( H^- \)[/tex]).
- This is incorrect because hydrogen does not form hydride ions in this reaction.
Option B: [tex]\( H_2 + 2 e^- \rightarrow 2 H^+ \)[/tex]; reduction
- This suggests hydrogen (H_2) gains electrons to form protons (H+).
- This is incorrect because the equation represents reduction where [tex]\( H_2 \)[/tex] should be losing electrons, not gaining them.
Option C: [tex]\( H_2 + 2 e^- \rightarrow 2 H \)[/tex]; reduction
- This also indicates hydrogen (H_2) gaining electrons.
- This is incorrect for the same reason as Option B.
Option D: [tex]\( H_2 \rightarrow 2 H^+ + 2 e^- \)[/tex]; oxidation
- This shows hydrogen (H_2) loses electrons and forms protons (H+).
- This correctly represents oxidation as hydrogen loses electrons in this process.
3. Conclusion:
The correct half-reaction that correctly describes the change occurring when hydrogen reacts with chlorine is:
[tex]\(\boxed{4}\)[/tex] [tex]\( H_2 \rightarrow 2 H^+ + 2 e^- \)[/tex]; oxidation
