Step 1: choose the right tool.
The standard value $\mathrm{-0.76\,V}$ is true only when the zinc ion concentration is exactly 1 M. Here it is far smaller, 0.01 M, so we must adjust the potential using the Nernst equation, which corrects the standard value for the real concentration at 298 K.
Step 2: write the Nernst form for this half-cell.
For the reduction $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$, two electrons are involved, so $n=2$, and the equation reads:
\[ E = E^\circ - \frac{0.059}{2}\log\frac{1}{[\mathrm{Zn^{2+}}]} \]
Step 2 continued: put the numbers in.
With $[\mathrm{Zn^{2+}}]=0.01$, the term $\dfrac{1}{0.01}=100$, and $\log 100 = 2$. So:
\[ E = -0.76 - \frac{0.059}{2}(2) = -0.76 - 0.059 \]
Step 3: get the value.
\[ E = -0.819\ \text{V} \]
The potential becomes more negative than the standard value, which makes sense because lowering the ion concentration makes it harder to deposit zinc.
\[ \boxed{E = -0.819\ \text{V}} \]