2al + 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?

    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+

Ithinks its false. tell me if im wrong

Compound with sulfer  is bright red