n??4×0.09
?K= =
0.0592 0.0592 =6.08 K=1.2×10? 3. 1. E1.20×5+0.53×1
IO3-/I-=
6 =1.08V
E1.2×5-1.45×1
IO3-/HIO=4 =1.14V
2. 图中HIO能发生歧化反应 5HIO=HIO3+2I2+2H2O K=
(1.45-1.14)×4
0.0592 =20.946 K=8.83×1020?
五.判断题
1. A:NaI B:NaClO 1). 2NaI+2FeCl3==2FeCl2+2NaCl+I2 2). NaI+AgNO3==AgI↓+NaNO3
3). ClO-+2H?+Cl-==Cl2↑+H2O
4). Cl2+2NaOH(冷)==NaCl+NaClO+H2O 5). 2NaI+NaClO+H2O==I2+NaCl+2NaOH
I2+5NaClO+2NaOH=2NaIO3+5NaCl+H2O 2. 1). A:NaI B:H2SO4(浓) C:I2 D: NaI3 E:Na2S2O3 2). 8NaI+5H2SO4(浓)=4I2+H2S↑+4Na2SO4+4H2O I3-+8Cl2+9H2O=3IO3-+16Cl-+18H? Na2S2O3+2HCl=2NaCl+SO2↑+S↓+H2O
Na2S2O3+4Cl2+5H2O=2H2SO4+2NaCl +6HCl
3. A. NaI B. NaClO (1) 2NaI +2 FeCl3 = 2FeCl2 + 2NaCl + I2 (2) NaI + AgNO3 = AgI↓ + NaNO3 (3) ClO- + 2H?+ Cl- = Cl2↑ + H2O (4) Cl2 + 2NaOH(冷)= NaCl + NaClO + H2O (5) 2 NaI + NaClO + H2O = I2 +NaCl + 2NaOH
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F:Cl2
I2 + 5NaClO + 2NaOH = 2NaIO3 + 5NaCl +H2O
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