Ek chemistry 1001#881

[inaccensa]

H202 can behave as both oxidizing and reducing agent as shown by the two half reactions below labeled R1 ad R2, which is true?

R1: O2+2H++2e- --> H2O2 E=.68V
R2: 2H+ +H2O2+2e- -->2H2O E=1.78V-->(higher reduction potential)

--> R1 demonstrates that H202 behaves as a reducing agent basic solutions
-->R2 "
-->R1 demonstrates that H2O2 behaves as an oxidizing agent in basic solution
-->R2 ''

I understand the following

In R1, oxygen's oxidation number decreases from zero to -1, hence it underwent reduction
In R2, oxygen's oxidation number decreases from -1 to -2, its again undergoing reduction

So wouldnt reaction2 illustrate that H2O2 is an oxidizing agent in a basic solution

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[PingPongPro]

H202 can behave as both oxidizing and reducing agent as shown by the two half reactions below labeled R1 ad R2, which is true?

R1: O2+2H++2e- --> H2O2 E=.68V
R2: 2H+ +H2O2+2e- -->2H2O E=1.78V-->(higher reduction potential)

--> R1 demonstrates that H202 behaves as a reducing agent basic solutions
-->R2 "
-->R1 demonstrates that H2O2 behaves as an oxidizing agent in basic solution
-->R2 ''

I understand the following

In R1, oxygen's oxidation number decreases from zero to -1, hence it underwent reduction
In R2, oxygen's oxidation number decreases from -1 to -2, its again undergoing reduction

So wouldnt reaction2 illustrate that H2O2 is an oxidizing agent in a basic solution

I don't know what the answer is, but I think it might have to do with H+ not being a basic solution.

 

[inaccensa]

Someone please

 

[vayntraubinator]

If you flip R1 backwards, H2O2 is oxidized to oxygen (Oxygen here -1==> 0), and is therefore a reducing agent.

Since the H+ is on the right, my guess is that this reaction is more favorable in basic solutions because the emf is negative.

In R2, H2O2 is reduced to H2O (Oxygen here -1 ==> -2), so it is an oxidizing agent. I think since the H+ is on the left side, this reaction is not favorable in basic solution (more favorable in acidic). This leaves me only choices A and C.

Notice, that if R1 is kept in the original direction, O2 is the oxidizing agent.

R1 and R2 both in their original directions are being reduced (therefore, they've got reduction potentials).

As we said before, though, in R1 H2O2 behaves as a reducing agent.

I'd go with A, for your safest bet don't have that book tho so let us know!!

 

[vayntraubinator]

This problem will stick if you relate it to peroxisomes in biology.

Their job is to remove H2O2 (poisonous) by reducing it to water.

 

[inaccensa]

got it. Thanks for breaking it down, the way you did. EK explanation was incomprehensible.. It makes a lot of sense. You are right A is the right answer.

If you flip R1 backwards, H2O2 is oxidized to oxygen (Oxygen here -1==> 0), and is therefore a reducing agent.

Since the H+ is on the right, my guess is that this reaction is more favorable in basic solutions because the emf is negative.

In R2, H2O2 is reduced to H2O (Oxygen here -1 ==> -2), so it is an oxidizing agent. I think since the H+ is on the left side, this reaction is not favorable in basic solution (more favorable in acidic). This leaves me only choices A and C.

Notice, that if R1 is kept in the original direction, O2 is the oxidizing agent.

R1 and R2 both in their original directions are being reduced (therefore, they've got reduction potentials).

As we said before, though, in R1 H2O2 behaves as a reducing agent.

I'd go with A, for your safest bet don't have that book tho so let us know!!