You are correct that given the standard potentials in the table below, you would expect hydrogen to be reduced first under standard conditions.
\begin{array}{cc}\hline
\text{Equation} & E_0 / \mathrm{V}\\ \hline
\ce{Zn^2+ + 2 e- <=> Zn} & -0.7628\\
\ce{2 H+ + 2 e- <=> H2} & 0\\ \hline \end{array}
The key concept is standard conditions. Standard conditions assume that all relevant ions have a concentration of $1~\mathrm{mol/l}$ — and hydrogen is a relevant ion, thus standard conditions require $\mathrm{pH}\ 0$. To calculate the redox potential at non-standard conditions, the Nernst equation $(1)$ must be used. Thankfully, for the zinc equation it only depends on the concentration of zinc ions. I shall neglect zinc’s concentration for the remainder of this answer, because hydrogen is more interesting.
$$\begin{gather}E = E_0 - \frac{0.059~\mathrm{V}}{z} \lg \frac{[\ce{Red}]}{[\ce{Ox}]}\tag{1}\\[0.6em]
E = 0~\mathrm{V}- \frac{7 \times 0.059~\mathrm{V}}{2} = -0.2065~\mathrm{V}\tag{2}\end{gather}$$
Equation $(2)$ tells us, that under neutral conditions, the potential of hydrogen is already $-0.21~\mathrm{V}$. This is still not enough to explain zinc formation at the cathode, but a big step forwards. More can be achieved by further raising of the $\mathrm{pH}$ value but it won’t get you all the way.
This is where a second issue comes in: overvoltage. Unfortunately, it cannot be expressed easily numerically (at least to the best of my knowledge) but in a nutshell the phenomenon is that it is hard to form gases on certain electrode materials (think activation energy). Due to the help of overvoltage (i.e. the correct choice of cathode), hydrogen liberaion can be almost completely suppressed and zinc formed.