How to write the last cell reaction for the following cell?
The species do cancel out. However, the concentrations of the species do not. It is a bit of a matter of preference as to how you express this. The way I would write it is:

$$\ce{Zn^{2+}_{(aq)} ($c = 0.01\text{ M}$) + Zn_{(s)} &lt;=&gt; Zn^{2+}_{(aq)} ($c = 0.1\text{ M}$) + Zn_{(s)}}$$

for which $E = E^0 - \frac{RT}{2F}\,\ln\!\left(\frac{0.1}{0.01}\right) = -0.030 \text{ V}$ at 298 K, since $E^0 = 0 \text{ V}$. A negative value of $E$ agrees with the reverse reaction above being spontaneous, which is of course what you would expect since $0.1 &gt; 0.01$.