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In the electrolysis of alumina using cryolite, the reaction that takes place at cathode is

(1) $ 4H_2O+4e^{-} \longrightarrow 2H_2+4\: OH^{-}$

(2) $4AI^{3+}+12e^{-} \longrightarrow 4AI$

(3) $6F_2+2AI_2O_3 \longrightarrow 4AIF_3 + 3O_2$

(4) $12F \longrightarrow 6F_2+12e^{-}$
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1 Answer

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(2) $4AI^{3+}+12e^{-} \longrightarrow4AI$
answered Nov 7, 2013 by pady_1

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