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.

Question

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.

Answer 1

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.

Answer 2

To calculate the molarity of Ag+ in the unknown solution, we need to use the Nernst equation, which relates the cell potential to the concentrations of the species involved.

The Nernst equation is given by:

Ecell = E°cell - (RT/nF) * ln(Q)

In this case, since we have a concentration cell, the cell potential (Ecell) is given, and the reaction quotient (Q) can be calculated from the concentrations of Ag+ in both half-cells.

Given:
Ecell = 226 mV = 0.226 V
T = 25°C = 298 K
R (gas constant) = 8.314 J/(mol·K)
n (number of moles of electrons transferred) = 1
F (Faraday constant) = 96,485 C/mol

First, we need to calculate the standard cell potential (E°cell). Since both half-cells contain silver, the standard reduction potential for Ag+ to Ag is zero. Therefore, E°cell = 0.

Now we can substitute the given values into the Nernst equation:

0.226 V = 0 - (8.314 J/(mol·K) * 298 K / (1 mol * 96,485 C/mol)) * ln(Q)

Simplifying and solving for ln(Q):

ln(Q) = -0.226 V / (8.314 J/(mol·K) * 298 K / (1 mol * 96,485 C/mol))

Now we can convert the natural logarithm to exponential form:

Q = exp(-0.226 V / (8.314 J/(mol·K) * 298 K / (1 mol * 96,485 C/mol)))

Now we need to find the concentration of Ag+ in the unknown solution. Let's call this concentration x M.

In the first half-cell, the concentration of Ag+ is 1.00 M, and in the second half-cell, it is x M. Therefore, the reaction quotient Q can be written as:

Q = [Ag+]second half-cell / [Ag+]first half-cell = x M / 1.00 M = x

Now we can substitute x into the expression for Q:

x = exp(-0.226 V / (8.314 J/(mol·K) * 298 K / (1 mol * 96,485 C/mol)))

Using a scientific calculator, we can calculate the value of x:

x ≈ 5.59 * 10^(-3) M

Therefore, to 3 significant figures, the molarity of Ag+ in the unknown solution is approximately 5.59 * 10^(-3) M.