Ice cube in a thermally isolated system ... Will any of it melt? If an ice cube at $0$ °C is in a thermally isolated system on its own, will any of it melt?  
The chemistry teacher says it will reach a state of equilibrium of half ice and half water due to variation in kinetic energy of the particles but I don't understand where the latent heat energy would come from, unless half of the ice cube ends up as several degrees below zero to supply the energy to melt the other half, which goes against thermal equilibrium principles.
 A: To go from water(solid) to water (liquid) you need to increase the internal energy of the water (break bonds).
As there is no heat input to the water or work done on the water, the internal energy of the water will not change and thus the water will stay in the solid state.
A: Your teacher seems to be mislead, it would follow that if the system is in isolation at 0 degrees, and the water was already in a solid state, then there would have to be some energy input to provide latent heat for the phase change. While it is true that the temperatures of the individual particles will follow the Maxwell-Boltzman Distribution, the total energy of the system is still not enough to trigger a phase change. 
A: Your teacher is wrong. 
The ice will stay at the same ice-water fraction it was when you sealed the container.
If the system wasn't in thermal equilibrium when you closed it, then you could have a phase change since the energy would already be there. But in that case you're not talking about thermodynamics since thermo only deals with systems in equilibrium.
A: I agree with your assessment.
I'm also not sure what your teacher is trying to get at here.
If it's pure ice, and the system is thermally isolated, there's no reason for it to gain additional energy required to melt some of the ice.  Like you also said, if the temperature of the ice dropped to provide that energy, it would violate the equilibrium of the system; so the system would never settle in that state.
If the ice were in thermal equilibrium with its surroundings at $0°C$, it would not make a difference; because there would be no net heat transfer.  It would still be the same as if it were isolated.  I could see some merit in what your teacher said.  That said, it's still a bit misleading.  
At the phase transition temperature, the mixture would be quasi-stable in a solid-liquid mixture.  For such a mixture, the descriptions we've been using for equilibrium and quasi-equilibrium mechanics cannot be applied to the same level of accuracy.  Basically, you get into the realm of statistics and microscopic effects.  Because you aren't in equilibrium at all times; the path you take to get the system to $0°C$ will also matter.
See this question for some more insights on what happens with the mixture at $0°C$. 
Thanks to John Bollinger in the comments for letting me see that my second part of the answer wasn't really relevant (and also makes it less clear to me what the teacher was talking about).
