Q.A cell is built from a nickel electrode dipped in $1\ \text{M}\ \text{NiSO}_4$ (anode) and a silver electrode dipped in $1\ \text{M}\ \text{AgNO}_3$ (cathode). Write its cell notation and the overall balanced cell reaction.

[!TLDR] Nickel (anode, oxidized) is written first, silver (cathode, reduced) last; balance electrons ($n=2$) between the two half-reactions for the overall equation. [!ANSWER] $\text{Ni}(s)\ |\ \text{Ni}^{2+}(1\ \text{M})\ ||\ \text{Ag}^{+}(1\ \text{M})\ |\ \text{Ag}(s)$; overall reaction $\text{Ni}(s) + 2\text{Ag}^{+}(aq) \to \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$.

Since nickel is given as the anode, it is oxidized: $\text{Ni}(s) \to \text{Ni}^{2+}(aq) + 2e^{-}$. Since silver is given as the cathode, it is reduced: $\text{Ag}^{+}(aq) + e^{-} \to \text{Ag}(s)$. Following the anode-first, cathode-last convention, the cell notation is $\text{Ni}(s)\ |\ \text{Ni}^{2+}(1\ \text{M})\ ||\ \text{Ag}^{+}(1\ \text{M})\ |\ \text{Ag}(s)$. To combine the half-reactions into the overall cell reaction, the electrons lost must equal the electrons gained: nickel loses $2e^{-}$ per atom, but silver gains only $1e^{-}$ per ion, so the silver half-reaction must be doubled: $2\text{Ag}^{+}(aq) + 2e^{-} \to 2\text{Ag}(s)$. Adding the two half-reactions and cancelling the $2e^{-}$ on each side gives the overall balanced equation, $\text{Ni}(s) + 2\text{Ag}^{+}(aq) \to \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$. [!ANSWER] Cell notation: $\text{Ni}(s)\ |\ \text{Ni}^{2+}(1\ \text{M})\ ||\ \text{Ag}^{+}(1\ \text{M})\ |\ \text{Ag}(s)$; overall reaction: $\text{Ni}(s) + 2\text{Ag}^{+}(aq) \to \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$.

Write each half-reaction from the given anode/cathode roles, scale so electrons lost = electrons gained, then add and cancel electrons.

Forgetting to double the silver half-reaction before adding, which leaves electrons uncancelled in the 'overall' equation.