answer to the question is IODINE

              Cr2O4-  OXIDATION STATE  OF Cr=+3

               MnO4-  OXIDATION STATE  OF Mn=+7

            NO3-  OXIDATION STATE  OF N=+5

 

NOW THE ABOVE ELEMENTS CAN GET REDUCED (accept electrons to reach their normal configuration)

AS THEY ARE AT MAXIMUM OXIDATION STATE

  

           OXIDATION STATE  OF I2=0 ( I2 is iodine)

     I2 CAN GAIN 2 ELECTRONS TO FORM 2I-   , .PRESENCE  OF I- IN THE   SOLUTION INTERFERES WITH THE H[+] CONCENTRATION THEREBY BRINGING A CHANGE IN pH