2.68 × 10 − 3 moles of a solution containing an ion A n+ require 1 .61 × 10 − 1 moles of MnO 4 - ion for the oxidation of A n+ to AO 3 - in acidic medium. What is the value of n?

Step-by-step solution

A n + → A + 5 O 3 − + 5 − n e − × 5 , Oxidation half reaction MnO 4 - + 8 H + + 5 e - → Mn 2 + + 4 H 2 O × ( 5 - n ) , Reduction half reaction Applying reaction stoichiometry, 5 moles of A n+ = 5-n moles of MnO 4 - 5 5-n = 2.68 × 1 0 - 3 1.61 × 10 - 3 = 1.66 = 5 3 5 - n = 3 ∴ n = 2