Posted by **Student** on Tuesday, February 26, 2008 at 11:02am.

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.

- Chemistry -
**DrBob222**, Tuesday, February 26, 2008 at 5:28pm
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.

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