Given that the electric potential is 7.0×10^−2 V higher outside the cell than inside the cell, and that the cell membrane is 0.10 thick, calculate the work that must be done (in joules) to move one sodium ion from inside the cell to outside.

work= difference in potential * charge

There are more complicated ways to do this .

To calculate the work done to move one sodium ion from inside the cell to outside, we need to know the electric field and the distance traveled by the ion.

The electric field can be determined using the given information that the electric potential is higher outside the cell than inside. The change in electric potential (ΔV) can be calculated by subtracting the potential inside the cell from the potential outside. In this case, ΔV = 7.0×10^−2 V.

The electric field (E) can be obtained by dividing the change in electric potential by the distance traveled. The distance traveled is the thickness of the cell membrane (d) which is given as 0.10 m.

So, E = ΔV / d = (7.0×10^−2 V) / (0.10 m).

Now, we have the electric field (E) required to calculate the work done.

The work done (W) is given by the formula: W = q * ΔV, where q is the charge of the sodium ion.

To calculate the charge of a sodium ion, we need to know the elementary charge, denoted as e. The elementary charge is the charge of an electron and is approximately equal to 1.6×10^−19 C.

The charge of a sodium ion (q) is the same as the elementary charge, but with a positive sign, since sodium ions have a positive charge. Therefore, q = +1.6×10^−19 C.

Now, we can substitute the values into the work formula:

W = q * ΔV = (+1.6×10^−19 C) * (7.0×10^−2 V).

Calculating the product of the charge and the change in electric potential will give us the work done in joules.

To calculate the work done to move one sodium ion from inside the cell to outside, we need to find the change in electric potential energy.

The change in electric potential energy (ΔPE) is given by the equation:

ΔPE = q * ΔV

Where:
ΔPE is the change in electric potential energy,
q is the charge of the sodium ion,
ΔV is the change in electric potential.

We are given that the electric potential is 7.0×10^−2 V higher outside the cell than inside, so ΔV = 7.0×10^−2 V.

The charge of a sodium ion is equal to the elementary charge, which is 1.6 × 10^−19 C.

ΔPE = (1.6 × 10^−19 C) * (7.0×10^−2 V)
ΔPE = 1.12 × 10^−20 J

Therefore, the work done to move one sodium ion from inside the cell to outside is approximately 1.12 × 10^−20 Joules.