Electrical physics question

[PROTON77]

I have got this question stating that: Immersing a pair of gold electrodes into a NaCl salt solution in the presence of a fixed voltage causes the capacitance and current to:
either increase, decrease, remain unchanged or increase briefly and then return to its original value..
I think that by the equation C= Q/V that capacitance should increase since there is more charge present due to the Na+ and Cl - but i don't know if it will only stay increased for a brief period.
Also since capacitance increases, i would assume that current should increase as it is directly proportional to capacitance.

 

[davenn]

causes the capacitance and current to:
either increase, decrease, remain unchanged or increase briefly and then return to its original value..

I think that by the equation C= Q/V that capacitance should increase since there is more charge present due to the Na+ and Cl - but i don't know if it will only stay increased for a brief period.
Also since capacitance increases, i would assume that current should increase as it is directly proportional to capacitance.

hi there
welcome to PF

increase or decrease relative to what ??

Dave

 

[PROTON77]

What i mean by that is if it was in distilled water or freshwater and then placed into salt water.

 

[meBigGuy]

distilled water is an insulator. Saltwater is a good conductor.

Not sure what happens to the capacitance. Gets swamped by the conduction, but there must be something there.

 

[DrZoidberg]

Due to the formation of an electric double layer the capacitance will greatly increase.
http://en.wikipedia.org/wiki/Double-layer_capacitance

If the voltage is high enough to cause electrolysis a constant current will flow. Otherwise a current will only flow for a very short time until the capacitance is charged.

 

[PROTON77]

I think what i meant to say was:
over time what happens to capacitance and the current, do they increase, decrease or increase briefly and then return to the original value or just remain unchanged. So what the question actually meant was from time 0 to time x what will happen to these two variables if voltage remained the same in the salt solution.

 

[meBigGuy]

Due to the formation of an electric double layer the capacitance will greatly increase.
http://en.wikipedia.org/wiki/Double-layer_capacitance

If the voltage is high enough to cause electrolysis a constant current will flow. Otherwise a current will only flow for a very short time until the capacitance is charged.

How can a double layer capacitor form in a conductive solution without a separator?

I would have though that saltwater would conduct at any voltage.

What voltage is required to cause electrolysis with 2 gold terminals and saltwater?

I think I am missing something.

 

[DrZoidberg]

The separator is only there to prevent the two electrodes from touching. It is not needed to form a double layer.
Saltwater can in principle conduct at any voltage but depending on the electrode material that current can be very low. I don't have the exact numbers but here is some information that might help you to figure it out.
http://www1.lsbu.ac.uk/water/electrolysis.html
http://en.wikipedia.org/wiki/Exchange_current_density
http://en.wikipedia.org/wiki/Butler–Volmer_equation

But perhaps you don't need to know exact numbers. The point is that you have two effects - a constant current and a short lasting charging current due to the capacitance. The constant current could potentially be extremely small or it could be large depending on voltage and electrode material.

 

[PROTON77]

I understand that the capacitance will increase because the dielectric constant increases but will it go back to the original value since the voltage remains constant.

 

[tech99]

I have got this question stating that: Immersing a pair of gold electrodes into a NaCl salt solution in the presence of a fixed voltage causes the capacitance and current to:
either increase, decrease, remain unchanged or increase briefly and then return to its original value..
I think that by the equation C= Q/V that capacitance should increase since there is more charge present due to the Na+ and Cl - but i don't know if it will only stay increased for a brief period.
Also since capacitance increases, i would assume that current should increase as it is directly proportional to capacitance.

I should have though that the capacitance between the plates will increase when you immerse them in water. The permittivity of water is about 81. The NaCl ionises in solution, so there will also be a good conductor between the plates. The voltage will now have to charge the additional capacitance and also drive a continuous current through the solution.

 

[PROTON77]

Ok that makes sense i thought that the capacitance would increase and that it would conduct a current, however will the current increase when placed in the solution or stay the same as it was before it was placed in the salt solution since the voltage is constant.

 

[tech99]

Ok that makes sense i thought that the capacitance would increase and that it would conduct a current, however will the current increase when placed in the solution or stay the same as it was before it was placed in the salt solution since the voltage is constant.

Imagine a capacitor in parallel with a resistor. When you connect a voltage to the pair, a current will flow for a moment whilst the capacitor charges. But the resistor will continue to draw a current continuously.

 

[meBigGuy]

WOW --- I'm totally ignorant of what is happening here.

If I connect 0.1V to two gold electrodes immersed in saltwater, what current will result?

I'm assuming 1cm area, 0.1 mol/liter gives about 10mS which means 0.1X.01 = 1ma. Or, am I completely out to lunch and it does it not conduct until the voltage reaches a certain level?

I'm missing some basic knowledge here.

 

[davenn]

WOW --- I'm totally ignorant of what is happening here.
If I connect 0.1V to two gold electrodes immersed in saltwater, what current will result?
I'm assuming 1cm area, 0.1 mol/liter gives about 10mS which means 0.1X.01 = 1ma. Or, am I completely out to lunch and it does it not conduct until the voltage reaches a certain level?
I'm missing some basic knowledge here.

don't feel too bad mate ... I am in the same boat

D