In a category with a $0$ object any two kernels of a morphism $f$ are isomorphic "in an evident sense" This is from Weibel.  A kernel of a morphism $f : B \to C$ is a morphism $i : A \to B$ such that $fi = 0 = (A \to 0 \to C)$ and for any other $e: A' \to B$ such that $fe = 0$, then $e$ factors uniquely through $i$: there exists a unique $e' : A' \to A$ such that $e = i e'$.  A kernel is monic.

and any two kernels of $f$ are isomorphic in an evident sense;

What do they mean by isomorphic?
 A: If $i : A \to B$ and $i^\prime : A^\prime \to B$ are both kernels of a morphism $f : B \to C$, then there is a unique isomorphism $g: A \to A^\prime$ such that $i = i^\prime \circ g$. This is what is meant by "two kernels are isomorphic in an evident sense."
A: As usual with universal properties: Suppose $i \colon A \to B$ and $j \colon D \to B$ are kernels of $f \colon B \to C$. Then as $fj = 0 \colon D \to B$ and $i$ is a kernel, there is an $e \colon D \to A$ such that $j = ie$. On the other hand, $fi = 0$, so since $j$ is a kernel, $i = je'$ for some $e' \colon A \to D$. Now 
$$ i = je' = iee' \colon A \to B $$
As $i$ is monic, $\def\i{\mathrm{id}}\i_A = ee'$. Analogously, $j = je'e$ gives $e'e = \i_D$. So $i$ and $j$ are isomorphic.
A: 
Proposition 1. Let $Y$ and $X$ denote terminal objects of a category $\mathbf{C}$. Then the unique morphism $Y \leftarrow X$ is an
  isomorphism.

(This is easy to prove directly, but I think its best viewed as a corollary of the following observation: call an object $X$ endomorphically rigid iff for all $f : X \leftarrow X$ we have $f = \mathrm{id}_X$. Now verify that given arrows $f : Y \leftarrow X$ and $g : X \leftarrow Y$, it holds that if $Y$ and $X$ are endomorphically rigid, then $f$ and $g$ are inverses.)
Okay, lets conceptualize a bit. Given a morphism $f : Y \leftarrow X$ in a category with a distinguished choice of $0$ morphisms, let us define the category of "prekernels" of $f$ as follows. Objects are morphisms $i : X \leftarrow I$ satisfying $0 = f \circ i$. Arrows are commutative triangles (please comment if my meaning isn't clear.) We can then define that a kernel of $f$ is a terminal object in the category of prekernels of $f$. Now apply Proposition 1, taking $\mathbf{C}$ to be the category of prekernels of $f$. This is a good way of understanding what Weibel means.
A: A kernel is a limiting cone (I let you say over which diagram), hence is unique up to unique isomorphism (of cones).
