posted by Shelly on .
"will .250 M CuSO4 power a battery the same as .100 M CuSO4? or will it be weaker?"
"power" is a rather loose term here. Power depends on current flow, and the stronger electrolye will allow for better current flow, but this is a very general statement.
I don't know what you mean by weaker or stronger. However, for the Cu half cell, the Nernst equation is
E = Eo - (0.0592/2)*log(Cu/Cu^+2)
Eo for Cu = 0.337 written as a reduction.
As the solution becomes more dilute the denominator is smaller and the fraction is larger. The log of a larger number is greater and that times a negative sign gives that part of the equation a more negative value. That added to a positive number (0.337 for Cu) makes the number smaller. So the voltage of the cell will be smaller (everything else being equal) and assuming the Cu/Cu^+2 is the negative electrode of the cell. The cell will last longer, however, for the higher concn.