To calculate the number of electrons transferred in the electrochemical reaction between silver nitrate and a copper electrode, we need to first write the balanced half-reactions and the overall redox reaction.The half-reactions are:1. Oxidation half-reaction Cu -> Cu + 2e :Cu s Cu aq + 2e2. Reduction half-reaction Ag + e -> Ag :2Ag aq + 2e 2Ag s Now, we can combine the half-reactions to get the overall redox reaction:Cu s + 2Ag aq Cu aq + 2Ag s In this reaction, one copper atom loses two electrons oxidation and two silver ions each gain one electron reduction . Therefore, the total number of electrons transferred in this electrochemical reaction is 2.