Upon using aluminium Electrodes for a concentrated solution of NaCl, i conducted electrolysis on the solution for nearly 4-5 hours. There was definitely a gas possibly hydrogen being evolved at the cathode. The anode was also dissolving at a fast rate. After the electrolysis i evaporated the remainder of the water and was left with a silvery sludge. Could someone clear me up on the identity of the sludge in question and the anode reaction?
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1$\begingroup$ You already asked that and I already told you that it is Al(OH)3. $\endgroup$– Ivan NeretinCommented Feb 16, 2021 at 9:19
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1$\begingroup$ you were not satisfied with the answer given chemistry.stackexchange.com/questions/146322/…? $\endgroup$– NicolasCommented Feb 16, 2021 at 9:20
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$\begingroup$ Donald Trump or my cat just would not take "no" as an answer. Well, apparently some people just would not take Al(OH)3 as an answer. $\endgroup$– Ivan NeretinCommented Feb 16, 2021 at 10:17
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1 Answer
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At the anode, two reactions may happen simultaneously : One due to the electrolysis, and another one due to the reaction of Al on water. The reaction due to the electrolysis is probably $$\ce{Al(s) -> Al^{3+}(aq) + 3 e-}$$ followed by the hydrolysis reactions : $$\ce{Al^{3+} + H2O -> [Al(OH)]^{2+}(aq) + H^+(aq)}$$ $$\ce{[Al(OH)]^{2+} + H2O -> [Al(OH)2]^+ (aq) + H+ (aq)}$$ $$\ce{[Al(OH)2]^+ + H2O -> Al(OH)3 (s) + H+(aq)}$$ On the other side, the simple reaction of $\ce{Al}$ with water without any electricity is $$\ce{2Al(s) + 6 H2O -> 2 Al(OH)3 (s) + 3 H2(g)}$$ followed by the reaction happening at the anode $$\ce{H2 -> 2 H+ + 2 e-}$$ As a consequence, the final result in both cases is the anodic production of a slurry of gelatinous $\ce{Al(OH)3}$(s), and $\ce{H+}$(aq) in solution.
