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Grade upto college level Physical Chemistry

2.68 *10-3 moles of a solution containing an ion An+ require 1.61 *10-3 moles of MnO-4 for the oxidation of An+ to AO-3 in acid medium. What is the value of n?

Profile image of Amit Saxena
12 Years agoGrade upto college level
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1 Answer

Profile image of Navjyot Kalra
12 Years ago
Equivalents of A oxidized = Equivalents of A reduced. Since in acidic medium, An+ is oxidized to AO3-, the change in oxidation state from
(+5) to (+n) = 5 – n [∵ O.S. of A in AO-3 = + 5]
∴ Total number of electrons that have been given out during oxidation of 2.68 * 10—3 moles of An+
= 2.68 * 10-3 * (5 -n)
Thus the number of electrons added to reduce 1.61 * 10-3
Moles of Mn-4 to Mn2+, i.e.
(+7) to (+2) = 1.61 * 10-3 * 5
[Number of electrons involved = + 7 – (+2) = 5]
∴ 1.61 * 10-3 * 5 = 2.68 * 10-3 * (5 - n)
5 – n = 1.61 *5/2.68 or n = 5 – 8.05/2.68 = 2