For a given half cell, Al ³⁺+3 e ⁻ Al on increasing of aluminium ion, the electrode potential will

Step-by-step solution

Al ³⁺+3 e ⁻ Al ( s ) aligned & E _ Red = E _ Red ^ o - 0.0591 3 [ Al ( ~s )] [ Al ³⁺ ] & E _ Red = E _ Red ^ o - 0.0591 4 1 [ Al ³⁺ ] [ Since active mass of solid =1] & E _ Red = E _ Red ^ o + 0.0591 3 [ Al ³⁺ ] aligned So E_ Red [ Al ³⁺ ] conc ^ n of Al ³⁺