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Question

2.68×103 moles of a solution containing an ion An+ require 1.61×103 moles MnO4 for the oxidation of An+ to AO3 in acid medium. What is the value of n?
  1. 2
  2. 4
  3. 3
  4. 1


Solution

The correct option is A 2
Step 1:Write reduction and oxidation half.

Reduction: MnO4+8H++5e Mn2++4H2O...(1)
Oxidation: An++3H2O AO3+6H++(5n)e...(2)

Step 2:To find out value of n.

 in a redox reaction
number of electrons lost = number of electrons gained
 multiply oxidant of equation (1) by (5 - n) and reductant of equation (2) by 5
(5n)MnO4=5An+ i.e, (5 - n) moles of MnO4will oxidise 5 moles of An+
1.61×103 moles of MnO4will oxidise An+=5(5n)×1.61×103moles...(3)
But the moles of  An+ actually oxidised=2.68×103 moles...(4)
On equating equation (3) and (4)
5(5n)×1.61×103=2.68×103
n=5.3352.68=2

 

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