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 Cu aq , and Ag aq is being reduced to Ag s .The given standard reduction potentials are:Cu aq + 2e 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 Cu aq + 2e E = -0.34 VNow, we can add the two half-reactions together:Cu s Cu aq + 2e E = -0.34 V2 Ag aq + e Ag s E = 2 +0.80 V Cu s + 2 Ag aq Cu aq + 2 Ag s The overall cell potential is the sum of the standard potentials of the two half-reactions:E cell = E oxidation + E reduction E cell = -0.34 V + 2 +0.80 V E cell = -0.34 V + 1.60 VE cell = 1.26 VSo, the cell potential for the given electrochemical reaction is 1.26 V.