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Reduction by solvated electron

  1. Oct 15, 2011 #1
    When sodium metal is added to ammonia, some of the sodium dissolves. Each of the dissolving sodium atoms loses an electron and becomes a cation. Both the cation and the free electron are solvated by ammonia molecules.

    Na(s) + (x+y) NH3(l) → Na(NH3)x+(aq) + e-(NH3)y(aq) (blue)

    At this stage the solution is blue because of the presence of solvated electron.
    After adding a catalyst Fe III which is rusty iron bar, the reaction proceed this way

    2 NH3(l) + 2 Na (s) (Fe2O3(rust) catalyst) --> 2 NaNH2 + H2

    after a while the solution turn from blue to milky.
    why is the color change this way? Is it because the consumption of the solvated electron? And why the above reaction write Na (s) isn't it dissolved to be (aq)? And I want to know how the Fe III catalyze the above reaction.
    then excess amount of NH4Cl is added and the solution turned brown.
    I know the addition of NH4Cl will produce more NH3 because
    2NH3<--->NH2^- + NH4^+
    Then why it is brown I don't know what is the reason for the brown.

    I also want to know the reduction of aromatic compound like naphthalene in liquid ammonia.
    first Na(s) is added to liquid NH3, then after it turned blue naphthalene solid (orange) is added.
    the solution turn to light green and then colorless.
    I propose the following explanation please tell me weather I am right.
    first the Na(s) loss a electron in NH3(l) it is solvated like the above part. After the addition of naphthalene, it is reduced by the mechanism described for Birch reduction. In the course of the reduction NH2^- is produced and react with Na(s) to give NaNH2 and H2
  2. jcsd
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