Oxidation of Cr^3+ to Cr^6+ by CIO^- in alkaline solution (CIO^- is converted to
ID: 1056668 • Letter: O
Question
Oxidation of Cr^3+ to Cr^6+ by CIO^- in alkaline solution (CIO^- is converted to Cl). The confirming test for Fe^3+ A solution that may contain any of the group III cations. Treatment of the solution with CIO^- alkaline medium yields a yellow solution and a colored precipitate. The acidified solution unaffected by treatment with NH_4OH. The colored precipitate dissolves in nitric acid; addition of excess NH_4OH to this acidic solution produces only a blue solution. Which group III ions are present? Which are absent? Which are in doubt? How would you remove all doubt?Explanation / Answer
GROUP III CATIONS :- The four group III cations are Cr+3, Al+3, Fe+3, and Ni+2.
The first step in analysis involves separating the ions into two subgroups by treating the solution with NaOH and NaOCl.
The hypochlorite ion oxidizes Cr(III) to a higher,more stable oxidation state (namely Cr(VI)) which is soluble:
2Cr+3(aq) + 3OCl-1(aq) + 10 OH-1(aq) ----> 2 CrO4-2(aq) + 3 Cl-1(aq) + 5 H2O(l)
In addition, Al+3 forms a soluble hydroxo-complex ion in the presence of excess hydroxide:
Al+3(aq) + 4 OH-1(aq) Al(OH)4-1(aq)
In contrast, Ni+2 and Fe+3 do not readily form hydroxo-complexes and are not oxidized by hypochlorite.
They forminsoluble hydroxides under these conditions: Ni+2(aq) + 2 OH-1(aq) Ni(OH)2(green solid)
Fe+3(aq) + 3 OH-1(aq) Fe(OH)3(red solid)
To separate aluminum and chromium, the solution containing CrO4-2 and Al(OH)4- is acidified to destroy the hydroxo-complex:
Al(OH)4-1(aq) + 4 H+ Al+3(aq) + 4 H2O(l) (aq)
Treatment with aqueous ammonia gives a gelatinous white precipitate of aluminum hydroxide. The concentration of hydroxide in ammonia is too low to form the hydroxo-complex:
Al+3(aq) + 3 NH3(aq) + 3 H2O(l) 3 NH4+(aq) + Al(OH)3(white solid)
The chromate ion remains in solution. It can be tested and confirmed by precipitation as yellow BaCrO4:
Ba+2(aq) + CrO4-2(aq) BaCrO4(yellow solid)
The BaCrO4 precipitate dissolves in acid.
The solution is then treated with NH4OH to produce a deep blue color due to
the presence of a peroxo-compound, probably CrO5:
2 BaCrO4(s) + 4H+ + 2NH4OH(aq) 2 Ba+2(aq) + 3H2O(l) + 2NH3 + 2CrO5(blue, aq)
so present cation is Cr3+ and absent Al3+ but Fe3+ and Ni2+ not react so no Fe3+ and Ni2+
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