Answer:
magnesium metal
Explanation:
According to the redox reaction equation, six electrons were transferred hence n=6 and F= Faraday's constant 96500C. ∆G° is given hence E°cell can easily be calculated as follows:
From ∆G°= -nFE°cell
E°cell= -∆G°/nF= -(-411×10^3/96500×6)
E°cell= 0.7098V
But for Al3+(aq)/Al(s) half cell, E°= -1.66V from standard table of reduction potentials.
E°cell= E°cathode- E°anode but Al3+(aq)/Al(s) half cell is the cathode
Hence
E°anode=E°cathode - E°cell
E°anode= -1.66-0.7098= -2.37V
This is the reduction potential of Mg hence the anode material was magnesium metal