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