posted by Student
One half-cell of a concentration cell consists of a silver wire dipping into a 1.00 M solution of Ag+. The second half-cell consists of a silver electrode in a solution containing an unknown concentration of Ag+. The measured cell potential was determined as 226 mV at 25°C. Calculate the molarity of Ag+ in the unknown solution to 3 significant figures.
E(cell) = Eo - (0.0592/n)log(dil/concd)
Does the problem give any information to tell you which way the electrons are flowing? If not, all I know to do is to work it both ways; i.e., we don't know which solution is the more concentrated. IF we assume the concn of the unknown is less (often the known is made to be 1 M), then we substitute
0.226 for Ecell
0 for Eo
n = 1
dil we will call the x value
concd we will call 1 M.
Solve for x, the concn of the more dilute solution.
NOW, you can reverse that if you wish and recalculate x (making dil 1 and concd in the denominator x) and solve for x and compare the two values. One will be realistic and the other will not. Post your work if you get stuck. I worked the problem and my answer was 1.52 x 10^-4 M (the other value was 6750 M if I reversed the numbers). Clearly, 6750 M is not very realistic. Check my thinking. Check my arithmetic.