✔ 最佳答案
1.
Both chlorine water and bromine water can oxidize Fe2+(aq) ion to Fe3+(aq).
Cl2(aq) + 2Fe2+(aq) → 2Cl-(aq) + 2Fe3+(aq) --Standard e.m.f. = +0.59V
Br2(aq) + 2Fe2+(aq) → 2Br-(aq) + 2Fe3+(aq) --Standard e.m.f. = +0.30V
However, iodine solution cannot oxidize Fe2+(aq) (Standard e.m.f. = -0.23 V)
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2.
All of chlorine water, bromine water and iodine solution do not react with HCl.
(The addition of HCl would compress the reactions of halogens with water by shifting the equilibrium positions to the left.)
(X2(aq) + H2O(l) ≒ HX(aq) + HOX(aq))
Reaction of chlorine water with dilute NaOH:
Cold: Cl2(aq) + 2OH-(aq) → Cl-(aq) + OCl-(aq) + H2O(l)
Hot: Cl2(aq) + 6OH-(aq) → 5Cl-(aq) + ClO3-(aq) + 3H2O(l)
Reaction of bromine water with dilute NaOH:
0oC: Br2(aq) + 2OH-(aq) → Br-(aq) + OBr-(aq) + H2O(l)
RT or above: Br2(aq) + 6OH-(aq) → 5Br-(aq) + BrO3-(aq) + 3H2O(l)
Reaction of iodine solution with dilute NaOH:
I2(aq) + 6OH-(aq) → 5I-(aq) + IO3-(aq) + 3H2O(l)
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3.
KCl solution reacts with AgNO3 solution to form a white precipitate of AgCl.
Ag+(aq) + Cl-(aq) → AgCl(s)
AgCl(s) turns grey under sunlight.
2AgCl(s) → 2Ag(s) + Cl2(aq)
AgCl(s) would be dissolved in excess NH3(aq) solution to form a colourless soluton.
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)
KBr solution reacts with AgNO3 solution to form a creamy precipitate of AgBr.
Ag+(aq) + Br-(aq) → AgBr(s)
AgBr(s) slightly turns grey under sunlight.
2AgBr(s) → 2Ag(s) + Br2(aq)
AgBr(s) is slightly dissolved in NH3(aq). It would be dissolved in a large excess of NH3 solution to form a colourless soluton.
AgBr(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Br-(aq)
KI solution reacts with AgNO3 solution to form a yellow precipitate of AgCl.
Ag+(aq) + I-(aq) → AgI(s)
AgCl(s) does not turn grey under sunlight, and would not be dissolved in excess NH3 solution.
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