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Chemistry-

General

Easy

Question

During the electrolytic reduction of alumina, the reaction at cathode is

$2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$
$3 F^{-} \to 3 F + 3 e^{-}$
${Al}^{3 +} + 3 e^{-} \to Al$
$2 H^{+} + 2 e^{-} \to H_{2}$

The correct answer is: $Al to the power of 3 plus end exponent plus 3 straight e to the power of minus not stretchy rightwards arrow Al$

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