Question

2 Al + 8 H2O → 2 H+1 + 3 H2 + 2 Al(OH)4-1 In the equation above, which element was reduced and which element was oxidized?

Answer 1

2 Al + 8 H2O = 2H ^+ + 3H2 + 2 Al(OH)4^-1


the element that was reduced is H
the element which is oxidized is Al

H is reduced since it moves from oxidation state of 1+ to oxidation state O

Al is oxidized since it move from oxidation state 0 to oxidation state of 3+