How does electrical energy develop by concentrating ions on one side of a biomembrane? In biology there is a topic called oxidative phosphorylation. 
In the first step of this process substrates such as isocitrate is oxidised and the electrons are transferred to coenzymes NAD+ or FAD to form NADH or FADH2. These high-energy electrons are then transferred through a series of electron carriers of electron transport chain. The energy released is used to translocate protons from the matrix to intermembrane space establishing 
a proton electrochemical gradient across the inner mitochondrial membrane.
In the next step the protons move down the electrochemical gradient, through an ATP-synthesising complex. The energy stored in the gradient is used to synthesis ATP.
How is electrical energy developed by concentrating ions on one side of a biomembrane?
Thank you.
 A: This may be slightly more technical then you are after but may be of use to others.
There are two things going on here.


*

*As mentioned by others you have the separation of charge by a dielectric and hence have a capacitor. If you start of in an electrically neutral situation and move $n$ protons from outside the membrane to the inside a voltage of: $$V=\frac{en}{C}$$ will develop across the membrane (use of $Q=CV$)
2.The other thing is to do with entropy. Imagine you had a box of $N$ particles. If you where to look in this box, it would be very odd to find all the particles in one side - the entropy of such a situation is low. When moving ions across the membrane you are creating a situation like this, something which is going to cost an amount of 'free energy' per molecule:
$$G=k_BT \ln\left( \frac{C_{in}}{C_{out}}\right)$$
So since I have mentioned it let me explain what 'free energy' is. Free energy is energy with the ability to do work. We can quantify this into something called the proton motive force. This is the free energy per charge to take one proton from outside of the cell to the inside. It takes the form:
$$PMF=V+\frac{G}{e}$$
$$=\frac{en}{C}+\frac{k_BT}{e} \ln\left( \frac{C_{in}}{C_{out}} \right)$$
So if I were to take one proton across the membrane I would lose an amount of free energy equal to $e \times PMF$. And hence lose this amount of useful energy.
As you can see the proton motive force consists of a part due to the electrostatic attractions between separated molecules (in the form of a capacitor) and part due to the entropy change associated with moving ions. 
A: Let's say we already have an assembly of 3 static charges, with initial energy $E_0$.

In order to place an additional charge, we need to push the charge in, thus applying work (Energy) equivalent to
$W_4 = kq_4(q_1/r_{14} + q_2/r_{24} + q_3/r_{34})$
And now, the whole assembly of 4 charges has an additional $W_4$ of energy:
$E_{tot} = E_0 + W_4$
This is the same process as translocating protons in step 1.
In step 2, to provide energy for the production of ATP, We need to release energy from the assembly of charges.
In the example, say we release $q_4$, removing the charge $q_4$ from the 4-charge assembly will release Energy equal to $E_{released}=W_4$, and also lowering the energy in the assembly.
