Crossed Cannizzaro products
The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.

$$\begin{aligned}
&amp;\ \ \ \ \  \ce{O} &amp;&amp; \ \ \ \ \ \ce{O-} \\
&amp;\ \ \ \ \  \parallel &amp;&amp; \ \ \ \ \ \mid \\
&amp;\ce{R-C-H +OH- -&gt;}  &amp;&amp; \ce{R-C-H} \\
&amp; \ &amp;&amp; \ \ \ \ \ \mid \\
&amp; \ &amp;&amp; \ \ \ \ \ \ce{OH} 
\end{aligned}$$

Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:


Sterics: The Ph group is larger than hydrogen, so nucleophiles will have more trouble getting to the aldehyde in benzaldehyde
Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.


If formaldehyde reacts faster than benzaldehyde with hydroxide anion, then more of formaldehyde's tetrahedral intermediate will form, which means there will be more benzaldehyde left for the second step - hydride transfer.