(1)Ag ＋ ＋e - =Ag 2NO＋O 2 =2NO 2
(2)Al(OH) 3 , Cu(OH) 2Al(OH) 3 +OH - =AlO 2 — +2H 2 O
(3)4 2 4 O 2
(4) 50 25
(5) Evaporation, concentration, cooling and crystallization
(1) When electrolytically refining Ag, coarse silver is used as the anode, pure silver is used as the cathode, and the electrolyte solution containing Ag + is used as the electrolyte, so the cathode reaction formula is Ag + +e - =Ag, and the filter residue Ag and dilute HNO 3 The reaction generates NO, which is oxidized by O 2 to generate NO 2 . The equation is 2NO＋O 2 =2NO 2 .
(2) NaOH solution reacts with Al 2 (SO 4 ) 3 and CuSO 4 to generate Al(OH) 3 precipitate and Cu(OH) 2 precipitate. If NaOH is excessive, Al(OH) 3 precipitate will dissolve , the ionic equation is Al(OH) 3 +OH - =AlO 2 — +2H 2 O.
(3) Since the valence of Cu decreases before and after the reaction and CuO acts as an oxidizing agent, the only thing that loses electrons is O 2 -, which is oxidized to O 2 and is balanced by the conservation of electrons gained and lost.
(4)4Cu～4CuO～2Al 2 O 3 ～2Al 2 (SO 4 ) 3 ～4CuAlO 2
That is, 2Cu～Al 2 (SO 4 ) 3 ～2CuAlO 2< /p>
n(Cu)＝ =50 mol
So it can be completely converted to generate 50 mol CuAlO 2 ,
At least the volume of Al 2 (SO 4 ) 3 is required. =25.0L.
(5) Since bile alum contains crystal water, in order to prevent the loss of crystal water, methods of heating, concentration, and cooling for crystallization should be adopted.