I have a question but I'm not sure whether my guess is true.
Given a totally path-disconnected space $X$, since the image of a path connected subset is path connected (simplex in our case), I would like to say that singular simplices are points in $X$ and therefore we have that $B_n(X)=Z_n(X)$, i.e $H_n(X)=\{0\}$ for $n\geq 1$. I've seen in a previous thread (regarding the rational numbers) that this is this the case, but would like to verify it.
Lets see: Since $\Delta^n$ is path-connected, any generator $\Delta^n \to X$ of $C_n(X)$ is a constant map if $X$ is totally path-disconnected.
Therefore $C_n(X)$ is isomorphic to $\mathbb{Z}^{(X)}$ - any singular chain can be thought of as a sum of "weighted" points of $X$.
Now what is $\partial(x)$ for any $x \in X$? Since $x$ is a symbol for the constant map sending all points to $x$, and by the definition of $\partial$ on has, that in degree $n$ $\partial$(x) "=" x, if $n \in 2\mathbb{Z}$, and $\partial(x) = 0$ if $n \in 2\mathbb{Z}+1$. ("=" should indicate that we now consider the constant map sending every point to $x$ starting from a different $\Delta$)
Therefore $Z_{2n}(X) = 0$ and $Z_{2n+1}(X) = C_{2n+1}(X)$. Furthermore we obtain $B_{2n}(X) = 0$ (clear) and $B_{2n+1}(X)=Z_{2n+1}(X)$ (use $\partial$(x) "=" x and $\mathbb{Z}$-linearity of $\partial$).
So I think you're right, in each case one gets $H_k(X) = 0, k >0$.