Need to calculate the resistance accross membrane in water

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SUMMARY

The discussion focuses on calculating the resistance across an ionic membrane in a homemade acrylic cell containing 0.1M NaCl. A DC voltage of 9.6V was applied, resulting in a measured voltage of 4.45V across the membrane with a current of 0.93mA. The calculated resistance is 4.45V divided by 0.93mA, yielding approximately 4785 Ohms. Concerns were raised about the accuracy of the voltage measurement and the placement of voltmeter leads, indicating potential issues with the experimental setup.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with ionic membranes and their properties
  • Basic knowledge of electrochemistry
  • Experience with using a multimeter for voltage and current measurements
NEXT STEPS
  • Research the principles of ionic conductivity in membranes
  • Learn about proper voltmeter lead placement for accurate measurements
  • Explore the effects of different salt concentrations on membrane resistance
  • Investigate the relationship between voltage drop and current in electrochemical cells
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Chemists, electrochemists, and researchers involved in membrane technology and electrochemical measurements will benefit from this discussion.

M. Hammad Khan
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Hello everybody,
I am chemist. and I prepare ionic membrane and placed in a home made acrylic cell containing 0.1M NaCl separated into two portions by ionic membrane. I applied DC voltage of 9.6V (using 9V battery). Using multi-meter the measured DC voltage in water was 4.45V. The distance between the electrodes is 20cm. The measured amperes were 0.93mA. Can I calculate the resistance from voltage drop? and how? Please help or guide if I m doing wrong.
 
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It's not clear to me what you are doing, but a diagram might help. My best guess is that the resistance is 4.45V divided by 0.93 mA which is 4785 Ohms. But without an understanding of what you are doing, that caculcation might not be the right one, or the experimental method could be fundamentally flawed.

The thing that seems strange is the battery is loaded down to 4.45 V with less than 1 mA of current. Not sure why this is.
 
M. Hammad Khan said:
Using multi-meter the measured DC voltage in water was 4.45V.
Voltage is always between two points.
"voltage in water" is just three words that convey no information.

Where were the two voltmeter leads physically placed ?
 

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