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Shriver & Atkins: Inorganic Chemistry 4e

Chapter 16

The starch-iodine clock reaction
 This particular version is, in fact, the Landolt Reaction. The first, very slow step, is the rate determining step in which iodate is reduced to iodide:

HSO3-(aq) + H2O (l) → HSO4-(aq) + 2H+(aq) + 2e-

IO3-(aq) + 6H+ (aq) → I-(aq) + 3H2O(l)

The iodide ion then undergoes a fast conproportionation reaction with excess iodate ion:

2I-(aq) → I2 (aq) + 2e-

2IO3- (aq) + 12H+(aq) + 10e-→ I2 (aq) + 6H2O(l)

The iodine is then reduced to iodide ion again in an ‘immeasurably fast’ reaction with hydrogen sulfite ion:

I2 (aq) + 2e-→ 2I-(aq)

HSO3-(aq) + H2O(l) → HSO4-(aq) + 2H+(aq) + 2e-

It is only when all the hydrogen sulfite ion is completely consumed that the iodine concentration suddenly starts to build up and the characteristic deep blue-black starch-iodine complex forms (this consists of the starch molecules wrapping themselves helically around a chain of iodine molecules).

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Comparative solubility of the silver halides
 Addition of chloride ion to silver ion produces a precipitate of white silver chloride.

Ag+(aq) + Cl-(aq) → AgCl (s)

Addition of iodide ion to silver ion produces a precipitate of yellow silver iodide.

Ag+(aq) + I-(aq) → AgI (s)

The more soluble silver chloride reacts with ammonia to give the soluble colorless diamminosilver(I) complex ion

AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)

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Hypochlorite ion as an oxidizing agent I
Hypochlorite ion reacts with the chloride ion in a conproportionation reaction to give chlorine gas:

2ClO-(aq) + 4H+(aq) + 2e-→ Cl2(g) + 2H2O(l)

2Cl-(aq) → Cl2(g) + 2e-

to give a combined reaction of:

ClO-(aq) + Cl(aq) + 2H+(aq) → Cl2(g) + H2O(l)

Damp colorless starch-iodide paper turns purple-black (the reaction between iodine and starch) upon exposure to the chlorine gas:

2I-(aq) → I2(s) + 2e-

2Cl2(g) + 2e-→ 2Cl-(aq)

Blue litmus paper turns a pale pink as a result of the combination of acidic and oxidizing properties of chlorine gas.

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Hypochlorite ion as an oxidizing agent II
Initially, lead(II) ion reacts with hypochlorite ion to give a white precipitate of lead(II) hypochlorite:

Pb2+(aq) + 2ClO-(aq) → Pb(ClO)2(s)

The lead(II) hypochlorite is then oxidized to lead(IV) oxide. Reaction is slow at room temperature but much faster when the tube is placed in boiling water.

Pb2+(aq) + 2H2O(l) → PbO2(s) + 4H+(aq) + 2e-

ClO-(aq) + 2H+(aq) + 2e-→ Cl-(aq) + H2O(l)

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