How is PD = EMF for an open circuit? For an open circuit, $V = E - Ir$. People say since there is no current flowing, $V$ is equal to $E$. However $V = IR$, where $R$ is the external resistance, that doesn't exist either so how can terminal PD exist?
Isn't EMF the work done to move a unit positive charge from the negative to the positive plate? So the only resistance that would affect this is $r$, the internal resistance, which we don't take into consideration because there is no current flowing.
Someone, please enlighten me.
 A: 
However V = IR, where R is the external resistance. That doesn;t exist
  either so how can terminal PD exist?

By KVL, 
$$E = V_R + I\cdot r = V_R + \frac{E}{r + R}\cdot r = V_R + \frac{E}{1 + \frac{R}{r}}$$
Now, see that, as $R \rightarrow \infty$, the right-most term $\frac{E}{1 + \frac{R}{r}} \rightarrow 0$ and thus we conclude that $V_R \rightarrow E$ as $R \rightarrow \infty$
Assuming no external path for charge flow, the source of EMF, e.g., chemical reaction, proceeds only as long as the the electric field due to the separated charge on the plates allows.  At the 'point in time' that the electric field due to the separated charge stops the reaction, the separated charge remains thus producing the open circuit voltage across the source.

"But isn't Vr itself IR, where R tends to infinity"
$$I = \frac{E}{r + R} $$
$$V_R = I \cdot R = \frac{E}{r + R}\cdot R = \frac{E}{\frac{r}{R} + 1}$$
$$\lim_{R \rightarrow \infty}\frac{E}{\frac{r}{R} + 1} = E$$ 
Either way you look at, the voltage $V_R$ exists in the limit as $R \rightarrow \infty$ and equals $E$.

"However, Terminal PD is the work done to take it from A to B in a closed circuit"
Maybe you're over-thinking this.  Consider a cell that is not connected to an external circuit.  The chemical reactions that remove electrons from one plate and deposit electrons on the other have essentially stopped.
But, due to the separated charge, there is an electrostatic field between the terminals of the cell and this gives rise to an electric potential difference (PD) between the terminals.
The (open circuit) terminal PD (multiplied by unit charge) gives the work done on a unit test charge in moving from the positive terminal to the negative terminal.  This is just ordinary electrostatics.
