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there was this statement in my chemistry textbook....

However, for measuring the resistance of an ionic solution we face two problems. Firstly, passing direct current (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or another solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power<

I don't understand how using AC over Dc would help, I mean electric current would pass ( although less than it would have otherwise) but still it could change the composition of a solution by electrolyzing it. So why AC?

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    $\begingroup$ Aside of electrolyte changes, electrode processes caused by DC significantly affect the potential difference applied to the solution itself and therefor the passing current. $\endgroup$
    – Poutnik
    Jan 20 at 13:27
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    $\begingroup$ The voltage applied is very small and the current passed is hardly enough to do more than rearrange the ions at the surface of the electrodes. The electrode double layers act more like capacitors, and the conductivity (inverse of resistance) is controlled by the movement of ions thru the electrolyte. Whatever electrode process happens during one half cycle (almost nothing) is completely reversed during the next half cycle. If you beef up the voltage (and current) and slow down the frequency, you could get some change in composition (like gas evolution). $\endgroup$ Jan 20 at 14:57

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  • Since AC can be capacitively coupled to the solution, metallic electrodes don't need to contact the solution -- they can be outside an insulating container, e.g., glass or PTFE, avoiding introducing metal ions.
  • At moderate AC frequencies, ions won't migrate an appreciable distance in one direction before returning to that position on the opposite half of the cycle. Steam vaporizers running at 50 or 60 Hz, using electrodes immersed in salt water, do not produce hydrogen and chlorine, just hot water and steam, for example.
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  • $\begingroup$ Thank you, didn't think about capacitative tendencies in the solution 🙏 $\endgroup$ Jan 21 at 6:00
  • $\begingroup$ @PriyanshuChoubey If this answer helped you, make sure to accept the answer by clicking the checkmark. $\endgroup$ Jan 21 at 6:45

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