The E^0_ cell of Al _ (s) | Al ³⁺(1 M ) || Pb ²⁺(1 M ) | Pb _ (s) cell is 1.5 V if E^0_ Pb is -0.14 V then E^0_ Al will be

Step-by-step solution

The standard cell potential is given by the difference between the standard reduction potentials of the cathode and the anode. E⁰_ cell = E⁰_ cathode - E⁰_ anode From the given cell representation, aluminium (Al) undergoes oxidation and acts as the anode, while lead (Pb) undergoes reduction and acts as the cathode. E⁰_ cell = E⁰_ Pb ²⁺/ Pb - E⁰_ Al ³⁺/ Al Substituting the given values: 1.5 = -0.14 - E⁰_ Al ³⁺/ Al E⁰_ Al ³⁺/ Al = -0.14 - 1.5 = -1.64 V Answer: -1.64 V