Write the cell reaction for the following cell representation. $$ \mathrm{Zn} / \mathrm{Zn}_{(\mathrm{aq})}^{+2} / / \mathrm{Ag}_{(\mathrm{qq})}^{+} / \mathrm{Ag}_{(\mathrm{s})} $$

First, let's identify the half-cell reactions involved in the given cell representation. At the anode, we have a Zinc half cell with Zn solid and Zn^+2 in aqueous solution. At the cathode, we have a Silver half cell with Ag^+ in aqueous solution and Ag solid.

Now that we have identified the half cells, we can write the respective half-cell reactions. For the Zinc half-cell (anode), the reaction is: $$ \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2 \mathrm{e}^{-} $$ For the Silver half-cell (cathode), the reaction is: $$ \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) $$

Next, we combine the half-cell reactions to create the overall cell reaction. To do this, we need to balance the electrons in both half-cell reactions. The Zinc reaction has 2 electrons, while the Silver reaction has only 1 electron. Therefore, we need to multiply the entire Silver reaction by 2 to balance the electrons: $$ 2 \times (\mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})) $$ This now becomes: $$ 2\mathrm{Ag}^{+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(\mathrm{s}) $$ We can now combine the two balanced half-cell reactions to form the overall cell reaction: $$ \mathrm{Zn}(\mathrm{s}) + 2\mathrm{Ag}^{+}_{(\mathrm{aq})} \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2\mathrm{Ag}(\mathrm{s}) $$

After combining and balancing the half-cell reactions, the final cell reaction is: $$ \mathrm{Zn}(\mathrm{s}) + 2\mathrm{Ag}^{+}_{(\mathrm{aq})} \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2\mathrm{Ag}(\mathrm{s}) $$