Usually (unit) quaternions represent rotations in 3D. Is that what you mean when you say they "represent an angle"? If you have two (unit) quaternions that each represent a rotation, they're both represented by unit vectors on the hypersphere. There are basically two ways to interpolate between them, a simpler way and a more complicated way that might have preferable properties, depending on what you need this for.
The simpler way is to take convex combinations of the two unit vectors ($\lambda$ times one and $1-\lambda$ times the other, with $\lambda\in[0,1]$) and then normalize them to obtain a unit vector again.
The more complicated way is to find the 4D rotation that rotates the plane they lie in (e.g. by orthogonalizing one of them against the other) and then rotate by some fraction of the angle you need to rotate one into the other.
Both methods yield the same set of interpolation results, but with different parameterizations (the second method yielding a "smoother" and "more natural" parametrization).
Either way, in case your quaternions represent rotations about the same axis through different angles, the interpolated quaternions will also represent rotations about that same axis, through angles in between.
In a sense, there "aren't enough" unit quaternions to describe translations. Let me try to make this more precise, if not completely rigorous.
Conventionally, we can describe rotations using the set of unit quaternions
$H = \{a+bi+cj+dk: a,b,c,d\in\mathbb{R}, a^2+b^2+c^2+d^2=1\}$; each such quaternion can be written in the form $\cos(\frac\alpha 2) + \sin(\frac\alpha 2)(xi + yj + zk)$, representing a rotation by angle $\alpha$ around the axis given by $(x,y,z)$.
Just as the unit circle in $\mathbb{R}^2$ is described by the equation $a^2+b^2=1$, and the unit sphere in $\mathbb{R}^3$ is described by the equation $a^2+b^2+c^2=1$, the above description of $H$ identifies it as the unit hypersphere in $\mathbb{R}^4$. It is thus a compact space, which means that it is "small" in a certain precise sense. As a consequence of this, if you choose any sequence of unit quaternions, then it will have a convergent subsequence - that is, you can throw out enough points from that original sequence and thereby obtain a sequence which converges to a unit quaternion.
Now suppose we choose a way to describe rotations and/or translations by means of unit quaternions. This means that to each quaternion $q$ we associate a rotation or translation $f(q)$. In order to be useful for computation, this association should be a homomorphism: it should satisfy the rule $f(q_1q_2) = f(q_1)\circ f(q_2)$. This ensures that $f$ relates the multiplication of quaternions to the composition of transformations, which is what we usually mean by the quaternions "describing" a set of transformations.
This map $f$ should also be continuous: if $q_1,q_2,\ldots$ is a sequence of unit quaternions which converges to some unit quaternion $q$, then the sequence $f(q_1),f(q_2),\ldots$ should converge to $f(q)$. We impose this continuity condition to avoid strange behaviour, and because in practice just about any reasonable $f$ you can come up with will be continuous.
I claim that no matter how you choose such an $f$, it cannot produce any translation. On the contrary, suppose that we had some $q$ such that $f(q)$ is a translation. Precisely, suppose that $f(q)$ is the translation "add $v$" where $v$ is some nonzero vector in $\mathbb{R}^3$. Since $f$ is a homomorphism, it follows that $f(q^2)$ is the translation "add $v$, then add $v$ again", i.e. "add $2v$". Likewise, for any $n>0$ we have that $f(q^n)$ is the translation "add $nv$".
Consider the sequence $q,q^2,q^3,\ldots$ of unit quaternions. Since $H$ is compact, there are positive integers $n_1< n_2< n_3\cdots$ such that the sequence $q^{n_1}, q^{n_2}, q^{n_3}, \ldots$ converges to a unit quaternion $\bar q$.
Now since $f$ is continuous, it follows that the sequence "add $n_1v$", "add $n_2v$", "add $n_3v$", $\ldots$ converges to $f(\bar q)$, which is some transformation of $\mathbb{R}^3$. But this sequence can't possibly converge to anything, because the sequence of vectors $n_1v,n_2v,\ldots$ shoots off to infinity and so the corresponding sequence of translations does, too. By this contradiction, we conclude that $f(q)$ could not have been a translation.
Best Answer
To rotate a vector $v = ix + jy + kz$ by a quaternion $q$ you compute $v^q = q v q^{-1} $.
So if $q$ and $q'$ are two rotation quaternions, to rotate by $q$ then $q'$ you calculate $(v^q)^{q'} = q' q \,v\, q^{-1} q'^{-1} = q' q \,v\, (q' q)^{-1} = v^{q'q}.$
References
Quaternions and spatial rotation