Let $[n]:=\{1,....,n\}$ and define the $2^n$-dimensional $\mathbb{R}$-algebra $C_n$ with basis $e_I$, $I \subset [n]$, such that $e_\emptyset = 1, e_ie_j = -e_je_i$ for $i \not =j, e_j^2 = 1 $ and $e_I={e_i}_1....{e_i}_k$ for $I = \{i_1,.....,i_k\}$ such that $i_1< ....< i_k \in [n]$
(i) Then how would I show that the $\mathbb{R}$-algebra $C_n$ is semi-simple?
(ii) Also how would I show the that the center of $C_n$ is $\mathbb{R}.1$ if $n$ is even and $\mathbb{R}.1 \oplus \mathbb{R}.e_1e_2.....e_n$ when $n$ is odd thefore implying that the center is of the following cases:
$\mathbb{R}$ if $n\equiv 0\mod 2$
$\mathbb{R \times R}$ if $n \equiv 1\mod 4$
$\mathbb{C}$ if $n \equiv 3 \mod 4$
You're essentially asking for piecemeal results of the classification of real Clifford algebras. The last line in your first paragraph is an out-of-context reference to the standard basis of $2^n$ elements which I'll label as $b_i$'s.
There is a pretty good presentation that would cover this in Jacobson's Basic algebra II section 4.8, if you can get your hands on it.
I think I've seen a direct analysis of the centers before, but it does not seem too straightforward. The easiest way to draw this conclusion is to study the structure theorem and find out that these algebras are all of these types:
A full matrix ring over $\Bbb R$ or $\Bbb H$
A full matrix ring over $\Bbb C$
$R^2$ where $R$ is of type 1 above.
The first case has center $\Bbb R$, the second $\Bbb C$, and the third $\Bbb R\times \Bbb R$. Their occurrence is controlled by $n$.