How to deal with the unit "Coulomb per equivalent"? In a question that I came across they defined $1\,\pu{F}$ as 96500 coulomb per equivalent. Now as far as my knowledge is concerned $1\,\pu{F} = 96500\,\pu{C/mol}$. The question involved asks to calculate the current that needs to be passed (or may be a function of that) wherein I mistook the $n$ factor and the answer was just half of what it was supposed to be. However that taking $1\,\pu{F}$ as "coulomb per equivalent" would lead me to a antoher unit (equivalent per mole) in order to get the desired unit in the question.

In this context I would like to:

a)  Get an intuitive explanation into the number of equivalents (just like moles is Avogadro number of substances).

b)  Know what the unit "equivalents/mole" means (is it the same thing as $n$ factor?) How does the $n$ factor treat number of moles and equivalents so differently?
For an electrochemical reaction, you count the atoms / ions by mol, and use the coulomb as a counting unit of charge. For a more intuitive explaination of the $n$ factor in the Faraday equation, try this analogy:

The summer olympics include swimming in a pool with lanes $50\,\pu{m}$ long.  Among the typical competitions are runs about $50$, and $100\,\pu{m}$.  Now, for the sake of the analogy, think the swimmers were representatives of ions.  Each of the swimmers could represent


a $\ce{Cl-}$ to yield $\ce{Cl}$, a change by one unit of charge.  So they swimm only one lane in one competition, $n = 1$; the single discharge happens when the swimmers clap the wall with their hand.
a $\ce{Cu^{2+}}$ wanting to yield $\ce{Cu}$. As they start with two charges now, two discharges in this single race are required ($n = 2$). Their first discharge is when they contact and turn at the opposite wall ($50\,\pu{m}$), their second and final discharge when they clap the other wall with their hand ($100\,\pu{m}$).