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Current in a solution

by chainsaw
Tags: current, ionic, solution
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chainsaw
#1
Apr10-09, 07:53 PM
P: 2
I don't even know what equation to use!

In an ionic solution, 5.0*10^15 positive ions with +2e charge pass to the right each second while 6.0*10^15 negative ions with charge -e pass to the left.

What is the magnitude of current in the solution?


(5.0*10^15)*(2*1.6*10^-19) - (6.0*10^15)*(1.6*10^-19)=8.3*10^-4
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rl.bhat
#2
Apr10-09, 07:57 PM
HW Helper
P: 4,433
Check the calculation.
chainsaw
#3
Apr10-09, 08:03 PM
P: 2
my calculation was wrong, when I recalculated it i got 6.4E-4, and that was still wrong
What equation should i use?

rl.bhat
#4
Apr10-09, 08:24 PM
HW Helper
P: 4,433
Current in a solution

There must be two different solutions in the electrolyte. So when +2e charge from one solution moving toward right,-2e charge is moving towards left. This adds to -e charge from the second solution. So net flow of charge is -3e towards left.


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