As other answers have already pointed out, the notion of "standard basis" only applies to the very particular $K$-vector spaces $K^n$, whose elements are $n$-tuples of elements of$~K$ (scalars). I've emphasised "are", because in any $n$-dimension $K$-vector space you can represent vectors by $n$-tuples of scalars (after having chosen an ordered basis; the scalars are the coordinates of these vectors in this basis); however in general a vector and its $n$-tuple of coordinates remain two different things.
You know what the standard basis of $K^n$ is (and it is actually an ordered basis: more than just a set of vectors, it is a list where each basis vector has its own place). The one point I would like to add is mention the property that makes this particular basis stand out among other bases, either of $K^n$ or of other spaces.
Property. Any $v\in K^n$ is equal to the $n$-tuple of coordinates of $v$ with respect to the standard basis of$~K^n$.
Clearly for such a property to hold, it is necessary that such vectors $v$ be $n$-tuples of scalars in the first place, in other words that the vector space in question be $K^n$. Moreover, the coordinates $(c_1,\ldots,c_n)$ of $v$ with respect to an ordered basis $(\mathbf e_1,\ldots,\mathbf e_n)$ are by definition the scalars such that $v=c_1\mathbf e_1+\cdots+c_n\mathbf e_n$ holds; if (with $v\in K^n$) one requires that moreover $v=(c_1,\ldots,c_n)$ as stated in the property, this requires that
(c_1,\ldots,c_n)=c_1\mathbf e_1+\cdots+c_n\mathbf e_n.
This should hold for all possible values of $c_1,\ldots,c_n\in K$. In particular one can take some $c_i=1$ and all other $c_j$ zero, and this leads to the conclusion that $\mathbf e_i\in K^n$ has to be $(0,\ldots,0,1,0,\ldots,0)$ with the nonzero component at position$~i$; this should be so for every$~i$. Once these cases are checked, all others follow by linearity. So the stated property holds for the standard basis, and also characterises the standard basis of$~K^n$.