To calculate the power output of the electrochemical cell, we first need to determine the cell potential E_cell . We can do this using the Nernst equation:E_cell = E_cell - RT/nF * ln Q where E_cell is the standard cell potential, R is the gas constant 8.314 J/molK , T is the temperature in Kelvin 298 K , n is the number of electrons transferred in the redox reaction, F is the Faraday constant 96485 C/mol , and Q is the reaction quotient.First, we need to determine the standard cell potential E_cell . For the given cell, the half-reactions are:Anode oxidation : Zn s Zn aq + 2eCathode reduction : Ag aq + e Ag s The standard reduction potentials E for these half-reactions are:E Zn/Zn = -0.76 VE Ag/Ag = 0.80 VSince the anode reaction is an oxidation, we need to reverse the sign of its standard reduction potential:E Zn/Zn = 0.76 VNow we can calculate the standard cell potential E_cell :E_cell = E cathode - E anode = 0.80 V - 0.76 V = 0.04 VNext, we need to determine the reaction quotient Q . For the given cell, the balanced redox reaction is:Zn s + 2Ag aq Zn aq + 2Ag s The reaction quotient Q is given by:Q = [Zn]/[Ag]^2Given the concentrations of the ions:[Zn] = 0.1 M[Ag] = 0.05 MWe can now calculate Q:Q = 0.1 / 0.05 ^2 = 0.1/0.0025 = 40Now we can use the Nernst equation to calculate the cell potential E_cell :E_cell = E_cell - RT/nF * ln Q E_cell = 0.04 V - 8.314 J/molK * 298 K / 2 * 96485 C/mol * ln 40 E_cell 0.04 V - 0.0129 V = 0.0271 VFinally, we can calculate the power output P of the electrochemical cell using the formula:P = E_cell * Iwhere I is the current 0.5 A :P = 0.0271 V * 0.5 A = 0.01355 WThe power output of the electrochemical cell is approximately 0.01355 watts.