Kernel in categorical sense The question that i have is to understand better this concept. This notion of kernel is really general, and i want to know the real difference to the algebric kernel, for exemple, in categories like Grp and Ring, all morphisms have kernels in the categorical sense? 
 A: The notion of kernel is defined in a category with zero object(you can find more different definitions at nlab article).
Definition 1.(zero object) Let $A$ be a category, $0\in A$ be its object. The object $0$ is called the zero object of $A$ iff $0$ is both an initial object and a terminal object of $A$. In this case $A$ is called a category with zero object. If zero object exists, then it is unique up to isomorphism. For any two objects $a,b\in A$ we can define the zero morphism $0_{a,b}\colon a\to b$, which is a unique morphism from $a$ to $b$, that factors through $0$.
Definition 2.(equalizer) Let $A$ be a category, $a,b\in A$ be its objects, $f,g\colon a\to b$ be its arrows. Let $c\in A$ be an object of $A$ and $h\colon c\to a$ be an arrow. Then the pair $(c,h)$ is called the equalizer of the pair $(f,g)$(denoted by $Eq(f,g)$) iff the following two properties hold:
1). $f\circ h=g\circ h$;
2).for any object $d\in A$ and for any arrow $k\colon d\to a$, such that $f\circ k=g\circ k$, there exists a unique $x\colon d\to c$, such that $h\circ x=k$.
Definition 3.(kernel) Let $A$ be a category with zero object $0$, and let $f\colon a\to b$ be its arrow. The kernel of $f$(denoted by $Ker(f)$), if it exists, is the equalizer of the pair $(f,0_{a,b})$. So we have a formula:
$$
Ker(f)=Eq(f,0_{a,b}).
$$
The advantage of such categorical definition is that in this definition kernel is a limit(because equalizer is a limit(it is a limit of a diagram $\downarrow\downarrow$)), and hence it can be determined by its universal property. So if a category with zero object is complete(like $\mathbf{Grp}$), or at least finitely complete, then all its morphisms have kernels.
Edit. The category of groups $\mathbf{Grp}$ is obviously a category with zero object(trivial group), but $\mathbf{Ring}$ is not(see Martin's comment for details).
A: I assume you are referring to the notion of kernel pair. That is, in a category $\mathsf C$, the pullback of a arrow $f \colon A \to B$ along itself : the square
$$ \require{AMScd}
\begin{CD}
\ker f @>>> A \\
@VVV @VVfV \\
A @>>f> B
\end{CD}
$$
is cartesian.
If $\mathsf C$ is the category of sets, then $\ker f = \{ (a,a') \in A \times A : f(a) = f(a') \}$ is an equivalence relation on $A$. Consider then $\mathsf C$ to be $\mathsf{Grp}$, $\mathsf{CRing}$, $R\text{-}\mathsf{Mod}$ etc. : the forgetful functor $U \colon \mathsf C \to \mathsf{Set}$ commutes with (small) limits and so $A \mathop / \ker f$ is the same sets with both definition of $\ker$.
(The "etc." need to be specified, but I'm not qualified enough to do it : don't hesitate to read this section of the kernel page on nLab.)
