Step 1: Use the cell notation:
In the cell $Ni(s)|Ni^{2+}(aq)||Ag^{+}(aq)|Ag$, the left electrode is the anode and the right one is the cathode. The left electrode is nickel.
Step 2: Write the anode process:
At the left side Ni(s) dips in $Ni^{2+}$. The reaction there is $Ni(s) \to Ni^{2+}(aq) + 2e^{-}$.
Step 3: Confirm with the overall reaction:
Add the cathode reaction $2Ag^{+} + 2e^{-} \to 2Ag$ to this. The electrons cancel and give the overall reaction.
Step 4: Eliminate the rest:
Options 2 and 3 are reductions, so they are cathode reactions. Option 4 is oxidation of silver, which runs opposite to the given reaction. Only option 1 remains.
Final Answer:
The anode is where oxidation occurs, so the nickel half reaction in option 1 is anodic.\[ \boxed{\mathrm{Ni(s) \to Ni^{2+}(aq) + 2e^{-}}} \]