If [K+] inside a cell is 25 times greater than a [K+] outside the cell, what is the difference in potential between the two sides of the cell wall? (this is a "concentration cell" style of problem)
Ecell = Eo - (0.0592/n)*log(dil/concd)
Ecell = 0 - 0.0592*log(1/25) = ??
To determine the difference in potential between the two sides of the cell wall, we can use the Nernst equation for a concentration cell.
The Nernst equation is given by:
E = (RT/zF) * ln([K+]outside/[K+]inside)
Where:
E is the potential difference (in volts),
R is the ideal gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin,
z is the valence (charge) of the ion (K+ has a valence of 1),
F is the Faraday constant (96,485 C/mol),
[K+]outside is the concentration of K+ outside the cell (in mol/L), and
[K+]inside is the concentration of K+ inside the cell (in mol/L).
In this case, the concentration of K+ inside the cell is 25 times greater than the concentration outside the cell. Let's assume the concentration outside the cell is [K+]outside = x mol/L. Then, the concentration inside the cell would be [K+]inside = 25x mol/L.
Substituting these values into the Nernst equation, we get:
E = (RT/zF) * ln(x/25x)
E = (RT/zF) * ln(1/25)
E = (RT/zF) * ln(0.04)
Now, we need to know the temperature in Kelvin to calculate the potential difference. Let's assume a temperature of 298 K (room temperature).
Substituting the values into the equation:
E = (8.314 J/(mol·K) * 298 K / 1 / 96,485 C/mol) * ln(0.04)
E ≈ 0.0592 V
Therefore, the difference in potential between the two sides of the cell wall is approximately 0.0592 volts.