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kym02

  • 11 months ago

An aqueous solution of calcium hydroxide is electrolysed using carbon electrodes. 1) what will be liberated at the anode? 2) It was observed that during the electrolysis, the mass of the anode decreased and a white precipitate was formed. Explain. Thanks :)

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  1. kym02
    • 11 months ago
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    @thomaster i know you're good in chemistry, unlike me.....help? anything's good.

  2. thomaster
    • 11 months ago
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    2 things happen at the anode. First OH- will discharge to give oxygen. second the carbon in the electrodes will react with the oxide giving carbon dioxide. That's an explanation for why the mass of the anode decreases. The white precipitate is the result of a reaction of the carbon dioxide with calcium hydroxide.

  3. thomaster
    • 11 months ago
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    \(\sf Ca(OH)_2 + CO_2 = CaCO_3 + H_2O\) CaCO3 = calcium carbonate, a white powder.

  4. kym02
    • 11 months ago
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    oh but isn't carbon considered an inert electrode?

  5. thomaster
    • 11 months ago
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    Yea but only in certain conditions. Carbon is an inert electrode because it's not an oxidant or reducant in the reaction. That doesn't mean it can't react with the products of the reaction (oxygen in this case).

  6. thomaster
    • 11 months ago
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    Inert electrode isn't the same as an inert element. and carbon is definitely not an inert element under most conditions.

  7. kym02
    • 11 months ago
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    hmm ok thank you very much.....by the way if you don't mind, there's this other question: what happens in The electrolysis of aqueous sodium sulphate using copper electrodes? i know, most of the time, it's carried out with inert electrodes but i saw an example similar to it (the electrolysis of copper sulphate using copper electrodes) and so my answer was something similar to that but, i don't know if its correct since its sodium in this case.

  8. thomaster
    • 11 months ago
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    ehh same question but with sodium sulphate and copper electrodes? Same reaction happens at the anode. copper reacts with oxygen to form copper oxide I guess.

  9. kym02
    • 11 months ago
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    ohh ok thanks for your help!

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