Question

(1) Solution A and B are separated by a membrane that is permeable to Mg2 and impermeable to Cl-. Solution A contains 0.1 mM MgCl2, solution B contains 100 mM MgCl2. Mg2 will be at electrochemical equilibrium when solution A is around ____________ mV. 180 mV - 180 mV 90 mV - 90 mV 60 mV - 60 mV

Answer 1

The potential difference (mV) at which Mg2+ will be at electrochemical equilibrium between solution A and B is around 180 mV.

Step-by-step explanation:

In this case, the difference in concentration of MgCl2 between solution A and solution B creates an electrochemical gradient. The membrane allows Mg2+ ions to pass through, but does not allow Cl- ions to pass through. As a result, Mg2+ ions will move from the higher concentration in solution B to the lower concentration in solution A until the electrochemical equilibrium is reached.

Since Mg2+ ions are positively charged, the movement of ions will create an electrical potential difference. The magnitude of this potential difference can be calculated using the Nernst equation:

E = (RT/nF) ln([Mg2+ in solution A]/[Mg2+ in solution B])

Given the concentrations of 0.1 mM in solution A and 100 mM in solution B, the potential difference (E) will be around 180 mV.

Answer 2

90 mV

Step-by-step explanation:

Faraday constant, F = 96487

Gas constant, R = 8.314

Number of mole of electron transferred, Z = 2

MgCl in solution A, = 0.1 mm

MgCl in solution A, = 100 mm

Using the relation :

E=(RT/ZF) In(Mg out/Mgin)

E = (8.314*298) / (2*96485) * In(100/0.1)

E = (2477.572 / 192970) * 6.9077552

E = 0.0128391 * 6.9077552

E = 0.0886897 V

E = 88.68 mV (approximately 90 mV)