00:01
So now we'll work on problem 102 from chapter 20.
00:10
In this problem, we're told that a voltaic cell is constructed that uses the following half -cell reactions.
00:18
Copper going copper 1 to copper and iodine 2 going to iodine minus.
00:26
The cell is operated at 298 kelvin with concentration of copper is 0 .25 and iodide at 0 .35 molodide at 0 .35 mol.
00:36
Determine e for the cell at these concentrations.
00:42
So we can go ahead and write down the overall reaction here.
00:46
We have i2 plus two copper will produce 2 i minus plus 2 copper plus.
01:04
And our e value here will be 0 .54 minus 0 .52 equal to 0 .02 volts.
01:19
So now we have to take into account the conditions that are specified.
01:24
So e is equal to enoch minus 0 .092 divided by n times the log of q.
01:42
So all of this here is equal to 0 .02 minus 0 .592 divided by 2, since we're transferring two electrons in this reaction.
01:55
Then we get the log, and this is an unusual situation because both of the soluble compounds are on the product side...