Question

A potential difference of 100 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to eject a positive sodium ion (Na ) from the interior of the cell

Answer 1

1.602 × 10⁻²⁰ J

Step-by-step explanation:

Since the sodium ion Na has a positive charge, the work done in moving it from the interior of the cell to the exterior of the cell is W = qΔV where q = charge on sodium ion = +1 and ΔV = potential difference between inner surface of the cell and outer surface of the cell. Since the inner surface of the cell is negative relative to the outer surface, the sodium ion moves from a lower potential to a higher potential. So, the potential difference is ΔV = + 100 mV = 0.1 mV = 10⁻¹ V

Also, q = +1 which is one unit of charge. So, q = +e where e = electron charge = +1.602 × 10⁻¹⁹ C

So, W = qΔV (Note that the work done is positive since we have to input work into the cell to move a positive charge from lower to higher potential)

So, substituting the values of the variables into the equation, we have

W = qΔV

W = +1.602 × 10⁻¹⁹ C × 10⁻¹ V

W = +1.602 × 10⁻²⁰ J