There is much, much material behind the situation you describe. Here is one aspect of it.
The scheme $E$ described bt $f(x,y)=0$ is a subscheme of the affine plane over $R$, namely $E=V(I)\subset \mathbb A^2_R$. Its ring of regular functions is $A=R[X,Y]/(f)$, so that you may write $E=Spec(A)$.
The interesting point is that you have a morphism $f: E\to Spec(R)=\mathbb A^5_{\mathbb Q}$.
So in reality, you are studying a family of affine curves, one for each $s\in \mathbb A^5_{\mathbb Q}$.
And believe me: $\mathbb A^5_{\mathbb Q}$, five-dimensional space over the rationals, is really, really big and complicated!
For example, whenever you choose five rational numbers $q_1,...,q_5\in \mathbb Q$, you get a so-called "rational point" point $q=\lt \alpha_1-q_1,..., \alpha_5-q_5\gt\in Spec (R)=\mathbb A^5_{\mathbb Q}$ whose fiber with respect to $f$ is the affine curve $E_q \subset \mathbb A^2_{\mathbb Q}$
given by the equation $f(x,y)=y^2+q_1 xy+q_3 y-x^3-q_2 x^2-q_4 x-q_5.$
At the other extreme, if you take the generic point $\eta=(0)\in Spec (R)=\mathbb A^5_{\mathbb Q}$ its fiber will be the generic curve $E_\eta\subset \mathbb A^2_{ \mathbb{Q}(\alpha_1,\ldots,\alpha_5)}$ given ( a little confusingly!) by the original equation $f(x,y)=y^2+\alpha_1 xy+\alpha_3 y-x^3-\alpha_2 x^2-\alpha_4 x-\alpha_5.$ And, as mentioned before, there are many, many other points in $\mathbb A^5_{\mathbb Q}$, for example $\lt\alpha_2 ^3-\alpha_2 +1, \alpha_4^{2011}-17 \gt \;\; \in Spec(R) \quad $ (and it will be worse next year...)
This illustrates Grothendieck's philosophy that you should not study a scheme like $E$ per se, but rather as a family of schemes: here the family is the set of fibers of $f$, parametrized by $\mathbb A^5_{\mathbb Q}$.
Edit
Nadori has now completely changed his question in an an edit and his new question is : what is the field of functions of $E$ ?.
The affine scheme $E$ corresponds to the ring $A=R[X,Y]/(f)$.
Since the polynomial $f$ contains the isolated term $-\alpha_5$, the ring $A$ is isomorphic to $\mathbb Q[\alpha_1, \alpha_2,\alpha_3, \alpha_4; X,Y] $ , so that $E\simeq \mathbb A^6_{\mathbb Q}$ and the required function field is $\mathbb Q(\alpha_1, \alpha_2,\alpha_3, \alpha_4; X,Y)$, a purely transcendental extension of $\mathbb Q$.