How To Write The Net Ionic Equation For The Reaction Between Copper & Silver Ions

Bring copper and a solution of silver nitrate together, and you initiate a process of electron transfer; this process is described as an oxidation-reduction reaction. The silver serves as an oxidizing agent, causing the copper to lose electrons. The ionic copper displaces the silver from the silver nitrate, producing an aqueous copper nitrate solution. The displaced silver ions in solution are reduced by gaining electrons lost by the copper. During this electron transfer process, solid copper converts to a copper solution, while silver in the solution is precipitated out as a solid metal.

Step 1

Write the oxidation half-reaction. During the process of oxidation, each copper atom (Cu) loses 2 electrons (e-). The copper is in solid, elemental form and this is represented by the symbol (s). The half-reaction is written in symbol form, and an arrow is used to show the direction of the reaction. For example, Cu(s) —> Cu(2+) + 2e(-). Note that the oxidation state (or charged state) is indicated by the integer and sign within the brackets that follow an elemental symbol.

Step 2

Write the reduction half-reaction directly below the oxidation equation, so that the arrows are vertically aligned. Silver is represented by the letters Ag. During the process of reduction, each silver ion (having an oxidation state of +1) binds with one electron released by a copper atom. The silver ions are in solution, and this is indicated by the symbol (aq) which represents the term "aqueous." For example, Ag(+)(aq) + e(-) —> Ag(s).

Step 3

Multiply the reduction half-reaction by 2. This ensures that the electrons lost by the copper during the oxidation reaction are balanced by those gained by the silver ions during the reduction reaction. For example, 2x {Ag(+)(aq) + e(-) —> Ag(s)} = 2Ag(+)(aq) + 2e(-) —> 2Ag(s).

Step 4

Add the oxidation and reduction half-reactions to obtain the net ionic reaction. Cancel any terms that occur on both sides of the reaction arrow. For example, 2Ag(+)(aq) + 2e(-) + Cu(s) —> 2Ag(s) + Cu(2+) + 2e(-). The 2e(-) left and right of the arrow cancel, leaving: 2Ag(+)(aq) + Cu(s) —> 2Ag(s) + Cu(2+) as the net ionic equation.