Chemistry Question: Decomposition of Water by Electrolysis

[Xylitol]

I understand that if we run an electrical current through water, it will decompose into its component parts (oxygen and hydrogen) according to the following chemical equation:

2 H2O -> 2 H2 + O2

I also know that this is an endothermic reaction (since the combustion of H2 is definitely exothermic). Because it is an endothermic reaction, theoretically, the temperature of the water that has a current running through it should go down.

However, I was thinking that even though the chemical reaction noted above would cause the temperature of the solution to drop, wouldn't the current itself send enough energy into the system to cause the overall temperature of water to increase? Essentially, my question is:

Will the temperature of water that has an electrical current running through it INCREASE even though the decomposition of water is an ENDOTHERMIC reaction?

Thanks!

 

[sao123]

I understand that if we run an electrical current through water, it will decompose into its component parts (oxygen and hydrogen) according to the following chemical equation:

2 H2O -> 2 H2 + O2

I also know that this is an endothermic reaction (since the combustion of H2 is definitely exothermic). Because it is an endothermic reaction, theoretically, the temperature of the water that has a current running through it should go down.

However, I was thinking that even though the chemical reaction noted above would cause the temperature of the solution to drop, wouldn't the current itself send enough energy into the system to cause the overall temperature of water to increase? Essentially, my question is:

Will the temperature of water that has an electrical current running through it INCREASE even though the decomposition of water is an ENDOTHERMIC reaction?

Thanks!

this depends simply on the amount of energy put into the water from the electricity.

There is a certain minimal amount of energy which must be added to the system before the reaction will begin to take place. anything continually added above that will be considered excess, and will add to the total heat of the system.

So now, you should be able to determine the amount of energy which must be added to begin to break the H bonds, and compare it to what you are actually adding from the current. Then you'll have your answer.

 

[Xylitol]

this depends simply on the amount of energy put into the water from the electricity.

There is a certain minimal amount of energy which must be added to the system before the reaction will begin to take place. anything continually added above that will be considered excess, and will add to the total heat of the system.

So now, you should be able to determine the amount of energy which must be added to begin to break the H bonds, and compare it to what you are actually adding from the current. Then you'll have your answer.

Hmm that sounds like a logical response. I have 9V batteries on me that I can use to test how 9 volts (and whatever respective current is sent through my local tap water) affects the temperature of water. I was just lazy and didn't want to bust them out if I didn't have to. Guess it's time to set up a good 'ol experiment right now.

 

[Red Squirrel]

Yeah the current will actually heat the water. 120v with two very close together nails will bring the water to a pretty hot temperature. It actually looks like it's boiling at that point but that's just the reaction.

I don't think a 9 volt will do much though, as it wont be able to provide enough current. Best bet is 12v from a computer psu. 120v if you are crazy. (do use a GFCI outlet if you're going to do that)