Is co-cohomology the same as homology?

Not in general.

Suppose that all of the $C_n = \oplus_{i = 0}^{\infty} \mathbb{k}$, with differentials $0$, and let $G = \mathbb{k}$. Then $Hom(C_n, \mathbb{k}) = \Pi_{i = 0}^{\infty} \mathbb{k}$. Taking the $\mathbb{k}$ dual again gives something containing as a subspace $\Pi_{i = 0}^{\infty} \mathbb{k}$, with zero as the differentials. The cohomology groups are different, as $(\Pi_{i = 0}^{\infty} \mathbb{k})^* \not = \oplus_{i = 0}^{\infty} \mathbb{k}$

If you have torsion in your chain groups: for example $\mathbb{Z}/2$ as some component of some $C_n$, this data will disappear when you double dual. (You can concoct an example again with zero differentials.)

I guess one situation in which you can say that they will be the same is when you are working in the category of finite dimensional vector spaces.

In the more common algebraic topology situation (for example cellular homology), then $C_*$ is a bounded complex of finitely generated free $R$ modules, where $R$ is a PID. In this case one can say something by repeated applications of the universal coefficients theorem and computational facts about exts - the formula you'll get for ${}_iH$ (the ith cocohomology) will be some messy combination of direct sums and compositions of Exts and Homs, involving the groups $H^{i}, H^{i-1}, H_i, H_{i-2}$: https://en.wikipedia.org/wiki/Universal_coefficient_theorem

(Maybe somebody with more sophistication than me can interpret this in terms of composition of the left derived functors of $Hom(\_,G)$?)

Maybe you can play around with that to get some conditions under which the cocohomology agrees.

If $G = \mathbb R$ (or any field), and $C$ is the simplical chain complex associated to a countably infinite discrete space (e.g., the integers), then $H_0 = Z_0$ will be an infinite direct sum of copies of the reals, but $Z^0$ will be the dual space of that, and the double dual will not be isomorphic to the original (because of the inifinite dimensions), hence I believe that the $0$th co-cohomology will not be the same as the $0$th homology.