Each half-reaction in the ion-electron method is balanced by adding:

Explanation

Explanation has been added. In the ion-electron method (also known as the half-reaction method), each half-reaction is balanced by adding electrons (e^-) to the reactants or products to balance the charges. In a redox reaction, which involves the transfer of electrons between reactants, the reaction is split into two half-reactions: the oxidation half-reaction and the reduction half-reaction. The oxidation half-reaction involves the loss of electrons, while the reduction half-reaction involves the gain of electrons. To balance these half-reactions, you need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. For example, consider the following redox reaction: Cu(s) + 2Ag^+(aq) → Cu^2+(aq) + 2Ag(s) In this reaction, copper (Cu) is oxidized to copper ions (Cu^2+), losing two electrons. Meanwhile, silver ions (Ag^+) are reduced to solid silver (Ag), gaining two electrons. The two half-reactions are: Oxidation half-reaction: Cu(s) → Cu^2+(aq) + 2e^- Reduction half-reaction: 2Ag^+(aq) + 2e^- → 2Ag(s) To balance these half-reactions, electrons are added to the reactants or products to balance the charges. In this case, two electrons are added to the oxidation half-reaction, and two electrons are added to the reduction half-reaction: Balanced oxidation half-reaction: Cu(s) → Cu^2+(aq) + 2e^- Balanced reduction half-reaction: 2Ag^+(aq) + 2e^- → 2Ag(s) Now, the number of electrons lost in the oxidation half-reaction (2e^-) is equal to the number of electrons gained in the reduction half-reaction (2e^-), ensuring the overall conservation of charge in the redox reaction.