How many electrons involved in the reduction of one mole of Al2o3 to aluminium metal
For every mole Al2O3 you have 2 mols Al. There are 6.02E23 atoms in a mol Al and each atom has 3e involved.
2Al^3+ ==> 2Al + 6e
To determine the number of electrons involved in the reduction of one mole of Al2O3 to aluminum metal, we need to look at the balanced chemical equation for the reaction.
The balanced equation for the reduction of Al2O3 to aluminum metal is:
2 Al2O3 + 3 C → 4 Al + 3 CO2
From this equation, we can see that 2 moles of Al2O3 react to produce 4 moles of aluminum.
Since each mole of Al has a charge of +3, we can deduce that each aluminum ion (Al3+) gains 3 electrons during the reduction.
Therefore, for the reduction of one mole of Al2O3, which produces 4 moles of aluminum, the total number of electrons involved would be:
4 moles of Al × 3 electrons/mole = 12 electrons
Hence, 12 electrons are involved in the reduction of one mole of Al2O3 to aluminum metal.
To determine the number of electrons involved in the reduction of one mole of Al2O3 to aluminum metal, we need to examine the balanced chemical equation for the reaction.
The balanced equation for the reduction of Al2O3 to aluminum is:
2 Al2O3 + 3 C → 4 Al + 3 CO2
From the equation, we can see that two moles of Al2O3 are reduced to four moles of aluminum (4 Al). Since each mole of aluminum requires three electrons to be reduced from Al3+ to Al, we can calculate the total number of electrons involved.
Number of moles of Al = 4 moles
Number of moles of electrons per mole of Al = 3 moles
Total number of electrons involved = (Number of moles of Al) x (Number of moles of electrons per mole of Al)
Total number of electrons involved = 4 moles x 3 moles = 12 moles
Therefore, in the reduction of one mole of Al2O3 to aluminum metal, 12 moles of electrons are involved.