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Chemistry Paper 12 Nov 09 Qs (Electrolysis)

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Hi Saturation. Me again :D

First of all, the answer is D not B.

So, basically, at anode, chlorine gas is formed. The gas bleaches the indicator, so it becomes colourless (test for chlorine gas).

At the cathode, it a bit confusing. Hydrogen gas is formed, but that has nothing to do with the colour change. Hydrogen forms and leaves. What important is what's left behind. We had Na+ and Cl- from NaCl and H+ an OH- from water. So H+ and Cl- leave. This means Na+ and OH- remain, making NaOH, A BASE :D Thats why it turns blue.

Hope that helped.
 
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Whoops! Sorry!
Oh in that case it makes sense!:), but i never knew that chlorine could bleach a Universal Indicator solution!
Guess i know now!
Thank you! :) :) :)
 
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Please read this question very well if posibly now. Chlorine gas formed from chloride ion is released at the anode because of its concentration. The correct colour of the indicator is RED. Hydrogen ion formed hydrogen gas and is also released at the cathode. Then, after some time, Na+ and OH- will form an alkaline solution which turns the indicator blue/purple. The answer is B.
 
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