| Example 3: Separate the redox reaction, Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s). Then balance each half reaction. Let's first separate the two half reactions. In order to do this, we must first look to see what is being oxidized and what is being reduced. We do this by assigning oxidation numbers to everything. The oxidation states for each species in the above reaction are as follows: Cu: 0, Ag+: +1, Cu2+: +2, Ag: 0. From this we see that Cu is being oxidized (went from 0 to +2), and Ag+ is being reduced (went from +1 to 0). Now we need to separate them into half reactions and balance. The half reactions are: Cu(s) → Cu2+(aq) oxidation Ag+(aq) → Ag(s) reduction The rules for balancing half reactions are: - First, balance all atoms except H and O. - Second, balance O atoms by adding as many molecules of H2O as needed. - Third, balance H atoms by adding as many H+ ions as needed. - Lastly, balance the charge by adding as many e- as needed. Let's balance the oxidation half reaction first. Cu(s) → Cu2+(aq) (The number of Cu's are balanced, and there are no oxygens or hydrogens to balance, so all we need to balance now is the number of electrons). The total charge on the left side of the equation is zero, and the total charge on the right side of the equation is +2. Therefore, we need to add two electrons to the right side of the equation. The final half reaction equation looks like this: Cu(s) → Cu2+(aq) + 2e- Now you try balancing the reduction half reaction. Answer: Ag+(aq) + e- → Ag(s) |