To determine the number of electrons transferred in the given electrochemical reaction, we first need to write the balanced half-reactions for the oxidation and reduction processes.1. Identify the oxidation and reduction processes:In this reaction, Cu s is oxidized to Cu2+ aq , and Ag+ aq is reduced to Ag s .2. Write the unbalanced half-reactions:Oxidation half-reaction: Cu s Cu2+ aq Reduction half-reaction: Ag+ aq Ag s 3. Balance the atoms other than O and H in each half-reaction:Oxidation half-reaction: Cu s Cu2+ aq Reduction half-reaction: Ag+ aq Ag s 4. Balance the O atoms by adding H2O:There are no O atoms in either half-reaction, so this step is not necessary.5. Balance the H atoms by adding H+:There are no H atoms in either half-reaction, so this step is not necessary.6. Balance the charges by adding electrons e- :Oxidation half-reaction: Cu s Cu2+ aq + 2e-Reduction half-reaction: Ag+ aq + 1e- Ag s 7. Make the number of electrons equal in both half-reactions:The oxidation half-reaction has 2 electrons, and the reduction half-reaction has 1 electron. To make the number of electrons equal, we need to multiply the reduction half-reaction by 2:Oxidation half-reaction: Cu s Cu2+ aq + 2e-Reduction half-reaction: 2Ag+ aq + 2e- 2Ag s 8. Add the half-reactions together and cancel out the electrons:Cu s + 2Ag+ aq Cu2+ aq + 2Ag s The balanced electrochemical reaction is:Cu s + 2Ag+ aq Cu2+ aq + 2Ag s The number of electrons transferred in this reaction is 2.