Question

A potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.15 x 10-20 J of work is required to eject a positive sodium ion (Na ) from the interior of the cell, what is the magnitude of the potential difference between the inner and outer surfaces of the cell?

Answer 1

Concepts and reason

The concepts required to solve this problem is the expression of work done in terms of potential difference.

First, write the expression of work done and then rearrange the expression of work done to obtain an expression for potential difference. The expression for work done shows the relation that the potential difference and the charge are directly proportional to it.

Fundamentals

Write the expression for work done in terms of potential difference.

$W = VQ$

Here, $W$ is the work done, $V$ is the potential difference and $Q$ is the charge.

The charge on the sodium ion $N{a^ + }$ will be equal to $1.6 \times {10^{ - 19}}\;{\rm{C}}$ because a sodium atom is neutral in nature and $N{a^ + }$ will be formed with the loss of one electron causing a positive charge equal to the charge of the electron.

Write the expression for work done in terms of potential difference.

$W = VQ$

Here, $W$ is the work done, $V$ is the potential difference and $Q$ is the charge.

Rearrange for $V$.

$V = \frac{W}{Q}$

Write the expression of the potential difference obtained.

$V = \frac{W}{Q}$

Here, $W$ is the work done, $V$ is the potential difference and $Q$ is the charge.

Substitute $1.15 \times {10^{ - 20}}\;{\rm{J}}$ for $W$and $1.6 \times {10^{ - 19}}\;{\rm{C}}$ for $Q$.

$\begin{array}{c}\\V = \frac{{1.15 \times {{10}^{ - 20}}\;{\rm{J}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C}}}}\\\\ = 0.0718\;{\rm{V}}\\\end{array}$

Ans:

The magnitude of the potential difference between the inner and outer surfaces of the cell is $0.0718\;{\rm{V}}$.