Question #69245

1 Answer
Jul 7, 2017

Answer:

#LiCl(l) => Li(s) + 1/2Cl_2(g)#

Explanation:

Reduction: #Li^+(l) + e^(-) => Li(s)#

Oxidation: #Cl^(-)(l) => 1/2Cl_2(g) + e^(-)#