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Solution: Step 1: Determine the overall cell potential (Eocell) For the given reaction : Pb2+ (ag) + Culs) - Pb(s) + Cuz+ ( 2q) Eocell = Edcathode) - Edanode) . Cathode ( reduction): Pb2+(aq) + 2e- - Pb(s), Eo = -0.13 V . Anode (oxidation): Culs) - Cuz+ (aq) + 20-, EQ = - 0.34 V ( reverse of the given half-reaction) Eocell = (-0.13 V) - (-0.34 V) = 0.21 V Step 2: Calculate the equilibrium constant (Kc) The relationship between Excell and Ke is given by the Nernst equation: Eocell = (RTInF ) In Kc Where: . R = 8.314 JAmol.K ) ( gas constant ) .T = 298 K ( temperature in Kelvin) .A = 2 ( number of electrons transferred ) "F = 96485 C/mol ( Faraday's constant ) Rearranging the equation to solve for Kc: Kc = e"(n FEocell/RT ) Kc = e ((2 )(96485 C/mol)(0.21 V)//8.314 J/(mol. K)(298 KC ~ 8.9 x 107