Oxidation — Reduction and Half Reactions

But if cations are produced at the wire, where did their electrons go The accumulation of silver on the copper wire provides us with the answer. The only apparent way to form metallic silver in this system is from the silver ions in the original solution. So we must conclude that silver cations in solution have accepted the electrons lost by the copper. We say that silver has been reduced, and again we can write an equation to show the change  

The two equations we have written describe what are called half-reactions for the oxidation of copper and the reduction of silver. Neither one can occur on its own because oxidation and reduction must take place in concert with one another. 

As you examine the half-reactions with this in mind, you may notice a small discrepancy. Copper loses two electrons in the oxidation half-reaction, whereas silver gains only one electron in the reduction half-reaction. This tells us that to conserve electrons, two silver ions must be reduced for every copper atom that is oxidized. We can multiply the reduction by two to make this explicit, giving us the following pair of half-reactions  

Here we can more readily see that the silver gains the two electrons that the copper loses. If we add the two half-reactions together, the electrons will cancel and we ll be left with the net ionic equation for the overall redox reaction  

As we discussed in Section 3.3, we could also write this as a molecular equation by including the spectator ions (NO3 in this case)  

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