Can someone direct me on the question below? I am not sure how to work this problem. Thank you.

Chambers A and B are separated by a membrane. Chamber A contains 15.7 M KNO3 and chamber B contains 10.98 M KNO3. If K+ is in equilibrium across the membrane, what is the electrical (potential) difference (Resting membrane potential)? Which side is electrically positive?

To determine the electrical potential difference (resting membrane potential) and which side is electrically positive in this scenario, you need to use the Nernst equation. The Nernst equation is as follows:

E = (RT/nF) * ln([oxidized]/[reduced])

In this equation, E represents the electrical potential difference or resting membrane potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is Faraday's constant (96,485 C/mol), and [oxidized]/[reduced] represents the ratio of concentrations of the oxidized and reduced species.

In this case, the species of interest is K+, and the reaction that occurs across the membrane is K+ oxidized on one side, and K+ reduced on the other side. Since K+ is in equilibrium, the concentrations of oxidized and reduced species will be the same.

Let's denote the concentration of KNO3 in chamber A as [KNO3]A and the concentration of KNO3 in chamber B as [KNO3]B.

Now, using the Nernst equation, we can determine the electrical potential difference (E) between chambers A and B. Since [KNO3]A and [KNO3]B are the same, we can substitute these values into the equation as follows:

E = (RT/nF) * ln([KNO3]A/[KNO3]B)

To determine which side is electrically positive, you need to compare the concentration of KNO3 in chamber A with that in chamber B. Whichever side has a higher concentration of KNO3 will be electrically positive.

Remember to convert temperature to Kelvin and plug in the appropriate values for R, n, and F.