Question

the two half - reactions are given below.
$ce{ag^{+}(aq) + e^{-} -> ag(s)}$
$ce{zn(s) -> zn^{2+}(aq) + 2e^{-}}$
what is the coefficient for the silver ion in the combined balanced net ionic equation?
coefficient of $ce{ag^{+}}$ = ?

Explanation:

Step1: Balance electrons in half-reactions

The first half - reaction: $\ce{Ag^{+}(aq) + e^{-} -> Ag(s)}$ has 1 electron gained.
The second half - reaction: $\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{-}}$ has 2 electrons lost.
To balance the electrons, we multiply the first half - reaction by 2 so that the number of electrons gained equals the number of electrons lost.
After multiplying the first half - reaction by 2, we get: $2\ce{Ag^{+}(aq) + 2e^{-} -> 2Ag(s)}$

Step2: Combine the two half - reactions

Now we add the modified first half - reaction and the second half - reaction:
$2\ce{Ag^{+}(aq) + 2e^{-}}+\ce{Zn(s)} -> 2\ce{Ag(s)}+\ce{Zn^{2+}(aq) + 2e^{-}}$
The electrons ($\ce{2e^{-}}$) cancel out on both sides, and the combined balanced net ionic equation is $2\ce{Ag^{+}(aq)}+\ce{Zn(s)} -> 2\ce{Ag(s)}+\ce{Zn^{2+}(aq)}$

Answer:

2