In a concentration cell Ecell = Eo -0.0592/2*log(dil/concd)
0.015 = 0 - 0.0592/2*log(dil/0.1)
solve for dil
The voltage generated by the zinc concentration cell described by the line notation
Zn(s) || Zn2+(aq,0.100 M) ‖‖ Zn2+(aq,? M) || Zn(s)
is 15.0 mV at 25 °C. Calculate the concentration of the Zn2+(aq) ion at the cathode.
4 answers
After thinking about this problem for an hour I believe that the line notation means that the 0.100 M cell is the smaller concentration and the one on the right is the larger concentration; however, if that is true then the voltage measured should be -0.015 v so perhaps I've misread the problem.
OK. Here's the scoop. In a concentration cell with the line notation as shown, the anode is on the left and is the weaker of the two solutions. The right hand cell is the cathode and is the more concentrated of the two. I just didn't substitute correctly in that equation I gave you. The correct substitution is as follows:
Ecell = Eo -0.0592/2*log(dil/concd)
0.015 = 0 - 0.0592/2*log(0.1/x)
0.015/-0.0296 = -0.507 = log 0.1/x
0.1/x = 0.311
0.1 = 0.311x
x = 0.1/0.311 = 0.322 M which is the concentration of the solution at the cathode.
Ecell = Eo -0.0592/2*log(dil/concd)
0.015 = 0 - 0.0592/2*log(0.1/x)
0.015/-0.0296 = -0.507 = log 0.1/x
0.1/x = 0.311
0.1 = 0.311x
x = 0.1/0.311 = 0.322 M which is the concentration of the solution at the cathode.
DrBob222 posted that last post by anonymous. Sorry about that. Forgot to put my name to it.
2 answers