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

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