So the answer is: neutron number; 5; None;
(2) To prepare KIO3 by electrolysis, the anode undergoes oxidation reaction, the electrode reaction formula is I-+3H2O-6e-=IO3-+6H+, and the cathode reaction formula is 2H2O+2e-=H2↑+2OH-, and the cathode PH increases.
So the answer is: I-+3H2O-6e-= IO3-+6h+; Increase;
(3)①IO3- is oxidizing, and reacts with reducing I- to generate I2 under acidic conditions, and the ionic equation of the reaction is IO3-+5I-+6H+=3I2+3H2O.
So the answer is: io3-+5i-+6h+= 3i2+3H2O;
② According to the reaction IO3-+5I-+6H+=3I2+3H2O, I2+2S2O32- (colorless) →2I-+S4O62- (colorless), the reaction relations IO3-~ 3 I2 ~ 6s2o 32- were obtained.
n(na 2 S2 o 3)= 0.0 10mol/l×0.0 12l = 0.000 12mol,
Then io3-~ 3i2 ~ 6s2o32-
1? six
n(IO3-)? 0.00012mol
N(IO3-)= 0.00002 mol,
m(KIO 3)= 0.00002mol×2 14g/mol = 0.00428g = 4.28mg,
Therefore, in 1kg salt, m (kio3) =10× 4.28mg = 42.8mg,
Where m (I) = 42.8 mg×127214 = 25.4 mg > 20 mg,
So it can be judged that the salt is qualified.
So the answer is: 42.8; Qualified.