To calculate the cell potential for the given electrochemical reaction, we first need to identify the oxidation and reduction half-reactions. In this case, Cu s is being oxidized to Cu2+ aq , and Ag+ aq is being reduced to Ag s .The given standard reduction potentials are:Cu2+ aq + 2 e- Cu s E = 0.34 VAg+ aq + e- Ag s E = 0.80 VSince Cu s is being oxidized, we need to reverse the first half-reaction and change the sign of its standard reduction potential:Cu s Cu2+ aq + 2 e- E = -0.34 VNow, we can add the two half-reactions together:Cu s Cu2+ aq + 2 e- E = -0.34 V2 Ag+ aq + e- Ag s E = 2 x 0.80 VThe overall reaction is:Cu s + 2 Ag+ aq Cu2+ aq + 2 Ag s To calculate the cell potential, we add the standard potentials of the two half-reactions:E cell = E oxidation + E reduction E cell = -0.34 V + 2 x 0.80 V E cell = -0.34 V + 1.60 VE cell = 1.26 VSo, the cell potential for this electrochemical reaction is 1.26 V.