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write the half reactions for 2Li + H2 -> 2LiH
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\[Li \rightarrow Li ^{+} + e ^{-}\] and \[H _{2} \rightarrow 2H ^{+} + 2e ^{-}\]
Although this doesn't quite make sense as the electrons of 2 half equations should be on the opposite side of the arrow
Well one is oxidation and the other is reduction, but what happened to the 2 for the Li
Oh sorry, the charge of hydrogen will be 1- as the compound is a metal hydride so \[H _{2} + 2e^{-} \rightarrow 2H ^{-}\]
Then you just double the half equation for lithium so the number of electrons are the same and will cancel off when you combine the 2 half equations
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ah yes, overlooked that as well so 2Li + H2 + 2e- --> 2Li+ + 2H- +2e- so cancel 2Li + H2 --> 2Li+ + 2H-
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