This problem amounts to a $2\times 2$ contingency table (actually 2 such tables if you look at pre- and post-intervention). The variable for the columns could be the pass/fail counts, and the variable for the rows could be the treatment condition (control & intervention).
Assuming appropriate conditions have been met, this is a chi-square analysis, and the conventional effect size here is the phi-coefficient: $$\phi = \sqrt{\frac{\chi^2}{N}}$$ (Curiously, this also happens to be the correlation if you code the two variables as ones and zeros.)
The convention with this effect size is to classify with the cut-offs of 0.1, 0.3, and 0.5, for small, moderate, and large, respectively.
Update #1
Based on the information provided in the comments, this would better be described as a 4×2 contingency table. The first variable is the treatment condition (dichotomous), and the second variable is the the pre/post status description. You would have 4 conditions for this variable: achieved at pre and failed to achieve at post, failed to achieve at both pre and post, achieved at both pre and post, and failed to achieve at pre and achieved at post (ordered in "best" for substantiating treatment effect).
In this case, you would use the Cramer’s $V$ as the effect size measure: $$V = \sqrt{\frac{\chi^2}{N \cdot \text{min}(R-1,C-1)}}$$ where $R$ is the number of rows and $C$ is the number of columns in the table. This effect size is compared to the same cut-offs as the $phi$ coefficient.