[Sayood_93]

Question: The reduction of hydrogen peroxide to water happens in an acidic solution. Which concentration of H2O2 must have been in the solution if the pH = 4.3, and the potential difference against the hydrogen half cell was measured at +1.50 V. Use E0 = +1.78V.

[Babcock_Hall]

According to forum rules, you must show your attempt before anyone can help you.

[Sayood_93]

Thanks for the quick reply!

[H3O+] = 10^-4.3
 
E = E0 + 0.059/2 log (red/oxd)

-0.28/0.0295 = log [H2O2] + log [H3O+]

[H202] = 0.128 mol/L


Sorry I'm writing this from my phone, therefore I wrote it skipping a few steps.

[Sayood_93]

The equation I should have mentioned:

H202 + 2e + 2 H3O+ --> 4H2O

[Borek]

E = E0 + 0.059/2 log (red/oxd)

And not

[tex]E = E_0 + \frac {0.059} 2 \log \frac {ox}{red}[/tex]

?

[Sayood_93]

I don't know, red/oxd or oxd/red?

[Borek]

Then check. It is a matter of a correct combination of sign and red/ox or ox/red ratio. For obvious reasons not every combination is a correct one.

[Sayood_93]

Ahh so it's
1) + with log ox/red
Or
2) - with log red/ox

Coming to a concentration os H2O2 1.23 x 10 ^ 18 mol/L

?

[Borek]

Ahh so it's
1) + with log ox/red
Or
2) - with log red/ox

Yes.

Coming to a concentration os H2O2 1.23 x 10 ^ 18 mol/L

Not what I got.

Not that I like the result I got.

[Sayood_93]

What did you get?

1.5 = 1.78 - ((0.059/2) log (red/oxd))

-0.28 = -  0.0295 log([H3O+]^2 [H2O2])

9.4915254 = log [H3O+]^2 + log [H2O2]

So concentration of H3O+ 10^4.3 = 5.011 x 10^-5

Squaring this gives you 2.51x10^-9.

Log of this is - 8.6

So 9.49 + 8.6 = log [H2O2]

10^18.09 = [H2O2]

= 1.23 x 10^18

[Borek]

I got the same result you got using wrong Nernst equation. I didn't like it, as the initial equation was wrong. Surprisingly, on the further analysis, it seems to be a correct result, as you initially made two mistakes, canceling each other. (Unless I am completely wrong, I just did here locally things that make me doubt my sanity ).

What is the oxidized and what is the reduced form?

[Sayood_93]

I'm not really following you, mu first nernst equation waas wrong.

The second one is correct.

H3O+ & H202 were reduced.

Oxd was H20 but its a constant so it can be 'left out'.

So which of my answers is correct, the second or the first. You seem to be talking in riddles lol.

[Borek]

Could be we are talking apples and oranges.

H3O+ & H202 were reduced.

Do you mean they are the reduced form of the hydrogen peroxide, or do you mean they are reduced in the reaction?

I will repeat the same question I asked in my previous form: what is the oxidized and what is the reduced form here?

[Sayood_93]

H2O is the reduced form of H2O2.

H2O2 is the oxidized form of H2O.

[Borek]

H2O2 is the oxidized form of H2O.

OK. So in the equation of the form you used in your calculations

[tex]E = E_0 - \frac {0.059} 2 \log \frac {red}{ox}[/tex]

should it be in the nominator, or denominator?