To calculate the overpotential of a lead electrode, we need to use the Tafel equation, which relates the overpotential to the current density i and the exchange current density i0 : = 2.303 * R * T / n * F * log10 i / i0 where R is the gas constant 8.314 J/molK , T is the temperature 298 K , n is the number of electrons transferred in the redox reaction 2 for Pb2+ / Pb , and F is the Faraday constant 96485 C/mol .First, we need to calculate the equilibrium potential E using the Nernst equation:E = E0 + R * T / n * F * log10 [Pb2+] where E0 is the standard potential of the Pb2+ / Pb electrode -0.13 V and [Pb2+] is the concentration of Pb2+ ions 0.001 M .E = -0.13 + 8.314 * 298 / 2 * 96485 * log10 0.001 E -0.13 + 0.043 * log10 0.001 E -0.13 - 0.086E -0.216 VNow, we need to find the current density i at this equilibrium potential. Since the exchange current density i0 is given as 0.015 A/m2, we can use the Tafel equation to find the overpotential : = 2.303 * 8.314 * 298 / 2 * 96485 * log10 i / 0.015 We can rearrange the equation to solve for i:i = 0.015 * 10^ * 2 * 96485 / 2.303 * 8.314 * 298 Since the overpotential is the difference between the equilibrium potential E and the standard potential E0 : = E - E0 = -0.216 - -0.13 = -0.086 VNow, we can plug this value back into the equation for i:i = 0.015 * 10^ -0.086 * 2 * 96485 / 2.303 * 8.314 * 298 i 0.015 * 10^-0.086 * 23.15 i 0.015 * 10^-1.99 i 0.015 * 0.0102i 0.000153 A/m2Finally, we can plug this value of i back into the Tafel equation to find the overpotential : = 2.303 * 8.314 * 298 / 2 * 96485 * log10 0.000153 / 0.015 0.043 * log10 0.0102 0.043 * -2 -0.086 VSo, the overpotential of the lead electrode in contact with a solution containing [Pb2+] = 0.001 M at 298 K is -0.086 V.