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?\\(\text{coefficient of }\ce{ag^+} = ?\\)

Explanation:

Step1: Balance electrons in half - reactions

The first half - reaction: $\ce{Ag^{+}(aq) + e^{-} -> Ag(s)}$ involves the gain of 1 electron.
The second half - reaction: $\ce{Zn(s) -> Zn^{2 + }(aq) + 2e^{-}}$ involves the loss of 2 electrons.

To balance the number of electrons transferred, we need to multiply the first half - reaction by 2 so that the number of electrons gained equals the number of electrons lost.

Step2: Multiply the first half - reaction

Multiply the half - reaction $\ce{Ag^{+}(aq) + e^{-} -> Ag(s)}$ by 2. We get: $2\ce{Ag^{+}(aq) + 2e^{-} -> 2Ag(s)}$

Now, we can add this modified half - reaction to the second half - reaction $\ce{Zn(s) -> Zn^{2 + }(aq) + 2e^{-}}$. When we add them, the electrons will cancel out.

The combined reaction is: $2\ce{Ag^{+}(aq) + Zn(s) -> 2Ag(s) + Zn^{2 + }(aq)}$

Answer:

2