Current in a solution of NaCl ions.

  • Thread starter Goldenwind
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  • #1
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[SOLVED] Current in a solution of NaCl ions.

Homework Statement


Current passes through a solution of sodium chloride. In 1.00 second, 2.68*10^16 Na+ ions arrive at the negative electrode and 3.92*10^16 Cl- ions arrive at the positive electrode.

A) What is the current passing between the electrodes?
B) What is the direction of the current? (Toward, or away from the negative electrode)


I haven't encountered a problem like this before in my Physics course, and I haven't taken Chem in a while. For part A, could I be given a hint on what equation to use?

For part B, how would I figure this out on my own (Without being told the answer)?
In theory, it'd be going towards the negative electrode, as we consider electricity to go from positive, to negative.

From what I understand, we basically have saltwater or saltliquid of somesort... which is a mix of Na+ and Cl-... when we pass current through this, the Cl- will shoot towards the positive electrode, and the Na+ towards the negative. But how do I use these facts to calculate a current?
 

Answers and Replies

  • #2
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Any comments at all? Theories, or anything?
I haven't a clue on this one.

One formula (Which everyone probably knows) is I = Q/t, which might help.
Where I = Current, Q = Charge, t = Time. We're given time.

Is it possible to figure out charge of this current from the number of ions?

Edit: Omg, another formula I just remembered is q = ne!!!

We have n. How do we find the charge on these ions? My Chem is rusty >.<
 
  • #3
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They're Na+ and Cl- ions, deficient and in excess respectively of one electron.

[tex]e = 1.602 176 487 \cdot 10^{-19}C[/tex]
 
  • #4
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They're Na+ and Cl- ions, deficient and in excess respectively of one electron.

[tex]e = 1.602 176 487 \cdot 10^{-19}C[/tex]

Genious! :D

Thank-you, thank-you, thank-you! :D
 
  • #5
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For part a) Find out the current due to +ive ions and -ive ions. Add them to get total current...
For part b) You already know the answer >>> +ive to -ive
 
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