What would be the voltage if Fe|Fe+2 || Pb+2 Pb
volts

Net Cell Voltage = 0.28v

Explanation:

The Rxn process here is

Fe°(s) + Pb⁺²(aq) => Fe⁺²(aq) + Pb(s)

∴ Half-Rxns are:

Fe°(s) => Fe⁺²(aq) + 2e⁻      E°(Oxdn) = -0.41v*

Pb⁺²(aq) + 2e⁻ => Pb°(s)      E°(Redn) = -0.13v*

Net Cell Voltage = E°(Redn) - E°(Oxdn) = (-0.13v) - (-0.41v) = 0.28v

*From Table of Standard Cell Potentials at 25°C, 1Atm Pressure