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What's the kinetic validity of the following way of balancing the above redox reaction?

I've done some searching of the literature; I've never come across oxide anions being mentioned in redox reaction mechanisms. Also, I read something about how two-electron reduction is impossible in a single step - something about spin pairing.

I'm fairly sure the following method is a good thermodynamic method of thinking about the reaction, but what about its kinetic validity?

I'm thinking that the below method isn't kinetically valid, but neither was it meant to be kinetically valid, and neither are the traditional methods of balancing redox reactions kinetically valid. Traditional methods call for balancing "O" and "H." But we aren't dealing with 0 oxidation state oxygens and hydrogens, are we?

Here is the thought process that my professor would take in solving this problem:

1) Skeleton reaction. Note that sulfur is being reduced from a +6 oxidation state to a +4 oxidation state. This suggests that the solid zinc loses two electrons.

$\ce{Zn + HSO4^- ->SO_2 +Zn^2+}$

2) Note that two negative two oxidation state oxygens are "disappearing" from left to right. In other words, what happened to the oxygens in the hydrogen sulfate ion?

$\ce{Zn + HSO4^- ->SO_2 +Zn^2+}$

$\ce{4~ negative ~two~ oxidation~ state ~oxygens -> 2 ~negative ~2~ oxidation~ state~ oxygens }$

3) The special thing about these oxygens is that they're negative two oxidation state oxygens. This suggests that these oxygens are the oxide anion. The oxide anion is highly basic in solution. Therefore, the oxide anions must have reacted with some system component, which caused them to "disappear." This would be the strongest acid in the system.

$\ce{HSO4^- ->SO_2 + 2O^2- (unstable)}$

$\ce{2O^2- + 4H3O^+ ->6H2O}$

4) Also note that a hydrogen proton disappears from left to right in the skeleton equation. Because the proton is highly acidic, it is attacked by the strongest base in the solution - water (since we're in an acidic solution).

$\ce{HSO4^- ->SO_2 + 2O^2- + H^+}$

$\ce{H^+ + H2O -> H3O^+}$

5) Final balanced equation:

$\ce{Zn + HSO4^- + 4H3O+ + H2O ->SO_2 +Zn^2+ +6H2O + H3O+}$

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  • $\begingroup$ whats left then you've got your answer but i too think that kinetic validation should be considered $\endgroup$ Jun 28, 2014 at 9:27
  • $\begingroup$ So it's not kinetically valid? $\endgroup$
    – Dissenter
    Jun 28, 2014 at 18:29
  • $\begingroup$ cant say any thing about that i am also confused at this situation $\endgroup$ Jun 28, 2014 at 18:34
  • $\begingroup$ I highly doubt it's kinetically valid and even if it were so I doubt one could prove it. Studying kinetics and verifying mechanisms is a time-consuming task. $\endgroup$
    – Dissenter
    Jun 28, 2014 at 18:36
  • $\begingroup$ i too think the same it is really a time consuming process $\endgroup$ Jun 28, 2014 at 18:37

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When you balance the equation for a redox reaction, the process of balancing it or the balanced equation itself say nothing about kinetics. That said, redox reactions, as the name implies, deal with electron transfer. Unless you are dealing with unusual molecules (bulky ligands for example) or unusual conditions (solids, glasses, very low temperature, etc.) electron transfer reactions are close to diffusion controlled, so kinetics is usually not an issue and is often glossed over in such reactions.

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  • $\begingroup$ Are you saying that electrons diffuse around? $\endgroup$
    – Dissenter
    Jun 29, 2014 at 1:04
  • $\begingroup$ The term "diffusion controlled" describes a process that happens so quickly (electron transfer in this case) that it is only limited by the rate at which the electron donor can diffuse through the solvent and encounter the electron acceptor. The rate constant for such a process is usually quite high. $\endgroup$
    – ron
    Jun 29, 2014 at 3:43
  • $\begingroup$ Isn't electron transfer followed by the transfer of other elements? Is that the whole idea behind redox? Electrons move ... but so can/will other elements? $\endgroup$
    – Dissenter
    Jun 29, 2014 at 3:45
  • $\begingroup$ Sure, but the electron transfer is the key step. Once this happens the die is cast. $\endgroup$
    – ron
    Jun 29, 2014 at 3:54

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