Answer:
0 current
Explanation:
Given that:
The concentration of Na inside the cell is 100 mmol/liter
The concentration of Na outside the cell is 100 mmol/liter
where;
z(valence of the ionic species for Na⁺ ) = +1, and the membrane voltage is 58 millivolt
Therefore;
[tex]E_{ion} = \dfrac{58}{z} \times log _{10} \bigg (\dfrac{[ion]_{out}}{[ion]_{in}}\bigg)[/tex]
[tex]E_{ion} = \dfrac{58}{1} \times log _{10} \bigg (\dfrac{100}{100}\bigg)[/tex]
[tex]E_{ion} =58 \times log _{10} \bigg (1\bigg)[/tex]
[tex]E_{ion} = 58 \times 0[/tex]
[tex]\mathbf{E_{ion} =0}[/tex]