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`        please please answer this question...pleaseIn the given equationS2 O82-  + 2e-  ------> 2SO42-Mn2+ +4H2O -------> MnO4- + 8H+ + 5e-how many moles of S2O82- ions are required to oxidize 1mole of Mn2+and thanks in advance`
one year ago

Sahani Kumar
99 Points
```							(Mn2+) + (H2O) ------->(MnO4-) + (8H+ )+ (5e-) so n=5(S2O82-) + (2e-) --------->2SO4^2-      so n=2These two reaction show that 1 mole of Mn2+ yields ( 5e-)but 1 mole S2O8^2- needs only 2e-So now let's how many moles of S2O8^2- will accept 5e-    2e-  -------accepted by--------1 mole                      5e- --------will be accepted by--------?moleSo ?mole =( 1mole * 5e- )/2e- =5/2 =2.5mole 5e- will be accepted by 5/2 mole  =2.5 mole So answer is that 2.5 mole of S2O8^2- will accept 5e- and therefore will oxidise 1 mole of Mn2+Regards Askiitians members If you like this answer please encourage us by approving this answer.
```
one year ago
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