# Question #01c4d

May 5, 2015

Start by assigning oxidation numbers to the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 1}}{A {g}^{+}} + \stackrel{\textcolor{b l u e}{0}}{F e} \to \stackrel{\textcolor{b l u e}{+ 2}}{F {e}^{2 +}} + \stackrel{\textcolor{b l u e}{0}}{A g}$

Notice that silver is being reduced from an oxidation state of +1 on the reactants' side, to an oxidation state of zero on the products' side.

At the same time, iron is being oxidized from an oxidation state of zero to an oxidation state of +2.

Write the reduction and oxidation half-reactions

$\stackrel{\textcolor{b l u e}{+ 1}}{A {g}^{+}} + 1 {e}^{-} \to \stackrel{\textcolor{b l u e}{0}}{A g}$ $\to$ reduction half-reaction;

$\stackrel{\textcolor{b l u e}{0}}{F e} \to \stackrel{\textcolor{b l u e}{+ 2}}{F {e}^{2 +}} + 2 {e}^{-}$ $\to$ oxidation half-reaction.

In a redox reaction, the total number of electrons gained during reduction must be equal to the total number of electrons lost during oxidation.

Multiply the reduction half-reaction by 2 to get an equal number of electrons transferred

$\left\{\begin{matrix}\stackrel{\textcolor{b l u e}{+ 1}}{2 A {g}^{+}} + 2 {e}^{-} \to \stackrel{\textcolor{b l u e}{0}}{2 A g} \\ \stackrel{\textcolor{b l u e}{0}}{F e} \to \stackrel{\textcolor{b l u e}{+ 2}}{F {e}^{2 +}} + 2 {e}^{-}\end{matrix}\right.$

Add the two half-reactions to get

$2 A {g}^{+} + \cancel{2 {e}^{-}} + F e \to 2 A g + \cancel{2 {e}^{-}} + F {e}^{2 +}$

The balanced equation will be

$2 A {g}^{+} + F e \to 2 A g + F {e}^{2 +}$