3(gained 10 electrons)

 2(+5)=+10                                                   2(0)=0  
2H2O+10MnO2+2ClO3¯ → 3Cl2+ 10MnO4¯ +4H+
+4                                                                                                                                                                     +7  

10(lost 3 electrons)        
 

2nd: Calculate the change in oxidation number for the reduced and oxidized element.

 

MnO2
           +4          2(-2)   = 0		 ClO3¯
       +5     3(-2)      = -1		 Cl2
        2(0)    = 0		 MnO4¯
           +7        4(-2)     = -1

 

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