Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Question #69245

1 Answer

Jul 7, 2017

$L i C l (l) \Rightarrow L i (s) + \frac{1}{2} C l_{2} (g)$ $LiCl(l) => Li(s) + 1/2Cl_2(g)$

Explanation:

Reduction: $L i^{+} (l) + e^{-} \Rightarrow L i (s)$ $Li^+(l) + e^(-) => Li(s)$

Oxidation: $C l^{-} (l) \Rightarrow \frac{1}{2} C l_{2} (g) + e^{-}$ $Cl^(-)(l) => 1/2Cl_2(g) + e^(-)$

Related questions

See all questions in Redox Reactions

Impact of this question

2415 views around the world

You can reuse this answer
Creative Commons License