Why does the internal resistance of a rechargeable battery depend upon whether it's charged or discharged? The following text is from Concepts of Physics by Dr. H.C.Verma, chapter 32, "Electric Current in Conductors", page 199, 19:

The internal resistance of an accumulator battery of emf $6\ \mathrm V$ is $10\ \Omega$ when it is fully discharged. As the battery gets charged up, its internal resistance decreases to $1\ \Omega$. [...]

From the Wikipedia article on rechargeable battery, I came to know that "accumulator battery" is an alternative name for it. But I don't understand why the internal resistance of the battery must vary depending upon the fact whether it's charged or discharged? I guess it is something to do with the electrochemical reactions inside the battery. 
Further, what is the nature of variation of internal resistance with the energy stored in the battery - is it linear or something else? Also, how can a discharged battery have a non-zero emf as stated above ("The internal resistance of an accumulator battery of emf $6\ \mathrm V$ is $10\ \Omega$ when it is fully discharged.")?
 A: Towards the end of its discharge, that is when the cell has almost run out of energy, the acid becomes weaker, that is $\text{SO}_4^{--}$ and $\text H^+$ ions are lost from the electrolyte (because of the chemical reactions occurring at the plates). Owing to the loss of these current-carrying ions, the electrolyte's resistivity increases, hence so does the internal resistance of the cell, as this is mainly the resistance of the electrolyte.
As for any relationship between internal resistance and energy stored, I suspect that the internal resistance only starts to rise appreciably when almost all the cell's energy has been used. Car batteries have to be almost 'flat' (no energy left) before they won't turn the starter-motor, and an appreciable internal resistance would prevent them from doing so. This paragraph, though, is merely speculative.
"The internal resistance of an accumulator battery of emf 6 V is 10 Ω when it is fully discharged." I think the writer is talking about a battery of nominal emf 6 V, that is 6 V when charged. 
A: Internal Resistance depends on the ability of the battery to supply current. A discharged battery cannot simply supply current. When high current/low resistance is attached across it, voltage drops across the terminals. 
When battery is fully charged, it can supply high current while maintaining voltage across it, hence low internal resistance.
A: In an electrolytic cell, the conduction of electric current is mainly due to the presence of the ions in the electrolyte. When the battery is fully charged, the concentration of ions is maximum. As the energy is utilized from the battery, the concentration of the ions reduces. As concentration of ions decreases, the electrical conductivity of the battery reduces. Or the internal resistance increases. This answers why the internal resistance of the battery depend upon whether the battery is charged or discharged. 
Generally, the nature of variation of internal resistance with the energy contained in the battery depends upon the type of the battery and also on other factors like temperature. The following graph shows the variation of internal resistance with the electromotive force (emf) of a lead acid battery:

Image description: SoC - State of Charge; Typical internal resistance readings of a lead acid wheelchair battery. The battery was discharged from full charge to $10.50\ \text V$. The readings were taken at open circuit voltage (OCV).
Image source: How does Internal Resistance affect Performance? - Battery University
It can be clearly seen that the internal resistance increases with decrease in the emf. There will be a minimum potential difference below which the battery could not supply a significant amount of electric current due to its high internal resistance. This explains why a fully discharged battery (in the question) has a non-zero internal resistance.
