In all cases there is diffusion of ions in both directions if
the permeability of the membrane allows. But where there is a
concentration difference, there will be more net movement from
the area of high concentration.
Diffusion of K+ and Cl- ions results in
a net movement from A to B, until equilibrium is reached at 55mmol
on each side.
Back to Q.1
In net terms, K+ moves down the concentration gradient (from A to B).
So, in terms of electrical charge, A becomes less positive and B more positive because the Cl- cannot diffuse.
This gives an electrical gradient, with the relative negativity of A attracting K+ ions to A from B.
Eventually, equilibrium reached as balance of concentration and
potential gradients.
Back to Q.2
No net movement of Cl-, as same concentration in A and B.
K+ will diffuse more from B to A, down its concentration gradient - so A becomes more positive and B relatively negatively charged.
This gives an electrical gradient attracting K+ to B from A.
Eventually, equilibrium reached as balance of concentration and potential gradients for K+.
Area A is electrically positive to area B.
Back to Q.3
Same as in 3, but there will also be a slower net movement of
Na+, so that eventually the electrical and concentration
gradients will disappear as both Na+ and K+
reach the same concentration (50mmol) in areas A and B.
Back to Q.4
A Na+ pump is needed to counteract the slow "leak"
of Na+ ions.
Back to Q.5
Just the opposite of 3.
Na+ ions diffusing from A to B and eventual balance
between electrical and concentration gradients, with area B electrically
positive relative to area A.
Back to Q.6