Cerium electronic configuration change from 0 to +2 oxidation state
The lanthanides are most stable in their 3+ oxidation state. For example, lanthanum is perfectly happy changing from

$$\mathrm{La}^0 = [\ce{Xe}]\mathrm{(5d)^1(6s)^2}$$ to 

$$\mathrm{La}^{3+} = [\ce{Xe}]\mathrm{(5d)^0(6s)^0}$$

Then, as you go along the lanthanide series, you're adding more 4f electrons. For example:

$$\mathrm{Ce}^{3+} = [\ce{Xe}]\mathrm{(4f)^1(5d)^0(6s)^0}$$
$$\mathrm{Pr}^{3+} = [\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$
$$\mathrm{Nd}^{3+} = [\ce{Xe}]\mathrm{(4f)^3(5d)^0(6s)^0}$$

etc.

This electronic configuration is extremely stable for all lanthanides, and there are very few exceptions to it.

Ce(II) is extremely unstable, and if there's one way to stabilise it, that would be adoption of the Pr(III) electronic structure. This way Ce has no 5d or 6s electrons, which makes lanthanides happy (apologies for the non-scientific way of framing this).