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Question

2.68×103 mol of a solution containing an ion An+ required 1.6×103 mol of MnO41 for oxidation of An+ to AO31 ion in acid medium. What is the value of n?

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Solution

An+ is oxidised to AO3
Change in oxidation number =5(in AO3)n(in An+)
=5n.....(i)
2.68×103 mol of An+ ion react with 1.6×103 mol of MnO4 ions
1 mol of An+ ion will react with 1.6×1032.6×103 mol of MnO4 ions
2K+7MnO4+3H2SO4K2SO4+2+2MnSO4+3H2O+5[O]
Number of equivalents of MnO4 used in oxidation of An+ to AO3=0.597×5=2.9853
Thus, from equation (i), 5n=3
n=2.

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