Half reaction for 2Li + H2–> 2LiH?

For the life of me I don’t get it. I have 5 attempts and i’ve used 3 and they’re all wrong.

Oxidation:

Li —-> Li+ + e- …(1)

Reduction:

H2 + 2e- ——> 2H- …(2)

Multiplying (1) by 2 and adding in (2) gives the required equation….

2Li + H2 ———> 2LiH

Source(s): Hope this help =)