Amembrane at 298 k has an interior concentration of 1mm of na* and an exterior concentration of 20 mm. assume that equilibrium has been reached and the free energy difference is zero. calculate the voltage of the interior relative to the exterior. a. -31 mv b.- 60 mv c.-76 mv d. -94 mv

C. -76 mV

Explanation:

The potential of a cell can be calculated by the Nernst equation:

E = E° - (RT/nF)*lnQ

Where E° is the standard potential, R is the gas constant, T is the temperature, F is the Faraday constant, n is the number of electrons at the redox reaction, and Q is the reaction coefficient (products/reactants).

For T = 298 K, the equation can be modified for:

E = E° - (0.0592/n)*logQ

And

E° = -ΔG°/nF

Where ΔG° is free energy, so E° = 0V.

Because the voltage will be calculated of the interior relative to the exterior, the interior is the reactant, and the exterior is the product. So, Q = [Na⁺]exterior/[Na⁺]interior. 1 electron has been replaced in this reaction, so n = 1.

E = -0.0592*log(20/1)

E = -0.076 V

E = -76 mV

i believe the answer is true.