[purpledwarak]

I am trying to understand the impact of the concentration of H₂O₂ on the oxidation potential of an aqueous solution under acidic conditions.

The half-cell reaction for H₂O₂ under acidic conditions (pH = -0.4 or H+=2.511 M) in an aqueous solution is

H₂O₂ + 2e^- + 2H⁺ = 2H₂O

The standard oxidation potential (E_o) for the above reaction is 1.7 V.

The leaching processes I am looking at requires an oxidation potential(E) of 0.7 volts.

I am using the Nernst equation for non-standard conditions to arrive at E=0.7 volts by varying the concentration of H₂O₂.

[tex] E= E_o-\left[\frac {0.0591}{n}×log_{10}\left(\frac {1}{[H_2O_2][H^+]^2}\right)\right] [/tex]

Substituting values Eo =1.7, E = 0.7, n=2, required concentration [H₂O₂] =X, and  H+ = 2.511

[tex] 0.7= 1.7-\left[\frac {0.0591}{2}×log_{10}\left(\frac {1}{X ×[2.511]^2}\right)\right][/tex]

X=7.13×10^{-33} M and this is a really low value of H₂O₂ concentration.

I would like to know where I am going wrong while applying the Nernst equation under non-standard conditions?

[mjc123]

You have got the sign wrong (or alternatively, you have got the log expression upside down, which amounts to the same thing). This is something I have difficulty remembering, so I always think in terms of Le Chatelier's principle. What will be the effect of increasing [H₂O₂]? It will push the equilibrium to the right; that is, it will make the oxidation (of something else by H₂O₂) more favourable, so it will increase the electrode potential.

[purpledwarak]

mjc123, thank you for the reply.

If the H₂O₂ concentration (i.e. X) decreases to a value below 1 in the Nernst equation then the ratio within the log₁₀ expression is greater than one. Therefore, E ≤ E_o.

So I think the above expression is correct and by decreasing the  H₂O₂ concentration E value reduces. Alternately, according to the Le Chatelier's principle a lower H₂O₂ concentration will drive the reaction to the left.

Hope this reasoning is Ok. My problem is that I am getting a very low value for H₂O₂ concentration to reduce E to 0.7 V. Still trying to figure out where I am going wrong.

[mjc123]

On reflection you're right, the minus sign and the upside down log expression combine to make it right. I don't think your answer is right - I get 2 x 10^-35M - but it's in the right regime, i.e. very small. Think about it; the factor 2.303RT/F = 0.059V occurs very often in electrochemistry, a number you should memorise. If n = 2, then E changes by 30 mV per decade of concentration, so you'll need about 33 orders of magnitude to get a 1V change. 1V is a very big change for an electrode potential. Do you have to use H₂O₂? Does the pH have to be -0.4? Varying the H⁺ concentration will have twice the effect of varying [H₂O₂].

[purpledwarak]

Yes, the process I am looking at (oxidative leaching of a metal) requires H₂O₂ and a pH in the range of -0.5 to 0.

From an electrochemical perspective, should I include the other half reaction where the metal gets oxidized and combine it with this half reaction of hydrogen peroxide gets reduced? This will reduce the E_o.

This is because the electrons required for the reduction of H₂O₂ should be generated by the oxidation of the metal.

[mjc123]

Yes, the process I am looking at (oxidative leaching of a metal) requires H2O2 and a pH in the range of -0.5 to 0.

Then, as we've seen, you're pretty much constrained to an electrode potential in the region of 1.7 V. (Not exactly - at concentrations like 2M, the activity coefficient will not be close to 1.) What is wrong with this? Why does it need to be 0.7V?

should I include the other half reaction where the metal gets oxidized and combine it with this half reaction of hydrogen peroxide gets reduced?

Good question. You said "The leaching processes I am looking at requires an oxidation potential(E) of 0.7 volts". Did you mean the electrode potential of the oxidiser? Or the cell potential, i.e. the oxidiser electrode potential needs to be 0.7V higher than that of the metal? Is that made clear? It would be helpful if you could quote the actual question or task specification. Does it need to be at least 0.7V? or no more than 0.7V? or exactly 0.7V? What is the metal, and what is its electrode potential? Is there anything else present that might get unwantedly oxidised at 1.7V?