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I was reading the answer of the above question from a book. it said that while using DC current, electrolysis occurs. Hence it affects the conductance of the solution. But if any gaseous product is created then polarization happens. Then why is polarization not furthermore discussed. So I request you answer what is it and how it affects while trying to get the conductance ?

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closed as off-topic by airhuff, aventurin, Tyberius, M.A.R., Pritt Balagopal Mar 27 '18 at 17:02

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    $\begingroup$ I cannot answer your question. However, I speak as a moderator. Here on SE we operate on a basis of respect for each other. By repeatedly editing your post, especially without any addition to the content, you are bumping your post to the top of the front page, which seems to be what you want. However, this also pushes everybody else’s questions lower, which is disrespectful. I will warn you that there is no purpose in trying to bargain; we have the ability to lock questions to prevent them from being edited. $\endgroup$ – orthocresol Mar 26 '18 at 14:48
  • $\begingroup$ I hence apologize to all for my behavior. $\endgroup$ – user187604 Mar 28 '18 at 4:10
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When a current is passed through an electrolyte, a chemical reaction takes place. This alters the composition of the solution over time and you won't actually be measuring the conductivity(or conductance) of the initial solution.

Since in AC current, equal current flows in both directions over a given amount of time(larger enough than time period), reaction takes place in both the directions. On an average, no chemical change in composition would be observed.(Of course, that doesn't mean no reaction has taken place.) Therefore the conductivity(or conductance) you measure will be the conductivity you wanted to measure.

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  • $\begingroup$ Please see my question body and say what polarization is and how does it affect the conductance? $\endgroup$ – user187604 Mar 24 '18 at 8:26
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    $\begingroup$ @user187604 I think the full question in your book maybe helpful to answer your question. $\endgroup$ – Govind Balaji S Mar 24 '18 at 15:29

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