# Chemistry

Hydrogen peroxide can act as either an oxidizing agent or a reducing agent, depending on the species present in solution. Write balanced half-reaction equations for each of the following:

(a) H2O2(aq) acting as an oxidizing agent in an acidic solution. Write the balanced half-reaction equation. Phases are optional.

(b) H2O2(aq) acting as a reducing agent in an acidic solution. Write the balanced half-reaction equation. Phases are optional.

(c) A disproportionation reaction is one in which a single species oxidizes and reduces itself. Write the complete balanced equation for the disproportionation reaction of H2O2(aq). Phases are optional.

I tried doing the problems but they all seems wrong
a)4H^+--->H2O2
4H + 2e^- --->H2O2
b) O2--->H2O2
O2^2- + 4e^- --->H2O2
c) I am not too sure how to do it.

Could someone please help me? I need to turn in my hw before 11 pm but I don't know how to figure this one out. Thanks

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1. KCr2O7 with H2O2, the peroxide must act as a reducing agent

http://www.ibchem.com/faq/?p=24

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bobpursley
2. a. oxidizing agent
H2O2 + 2e + 2H^+ ==> 2H2O

b. reducing agent
H2O2 ==> O2 + 2e + 2H^+

c. disproportionation. One O goes from -1 to zero; the other O goes from -1 to -2. -1 to zero is loss of electrons; -1 to -2 is gain of electrons. Therefore, one O atom is oxidized and the other one is reduced. Another example of disproportionation is
Hg2Cl2 ==> Hg + HgCl2
One Hg atom is oxidized and the other one is reduced.
2H2O2 ==> 2H2O + O2

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3. Thank you, DrBob222!

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4. KCr2O7+H2O2 → ?

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5. HASldkhlakdjf;ksjdh

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6. Thanks Dr. Bob always saving the day for noobs like me

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