What is the oxidizing agent in the following oxidation- reduction equation?

H+ + I– + Cr2O72– —> Cr3+ + I2 + H2 O

Answer:
i think the 1st 2 answer is WRONG...oxidizing agent itself go through reduction...so it would be acidified potassium dichromate(VI),Cr2O72-.u can c that oxidation number of Cr frm +6 reduce to +3...i can sure i am rite..trust me...haha
phosphorous
the oxidizing agent loses electrons. the oxidizing agent would be the Iodine
An oxidising agent is one which oxidises other substances and itself gets reduced.Now reduction involves decrease of oxidation number and oxidation number.Now the oxidation number of Cr in Cr2O72-is 6 and that in Cr3+ is +3.so its oxidationnumber has decreased .so it has been reduced and hence is the oxidizing agent.I- is the reducing agent and it suffers oxidation to iodine.

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