Manganese (II) oxidation by air

I am studying the reaction of dissolved dioxygen with manganese(II) hydroxide for the Winkler method:



```
                   2Mn(OH)2(s)+O2(aq)⟶2MnO2(s)+2H2O(l)

```

The data I have been given is Mn(OH)2's pKs=12.7, as well as the standard potential for MnO2(s)/Mn(OH)2(s), which is 1.23V, and the one of O2/H2O, which is also 1.23V; nothing more. Using Gibbs free energy for each half reaction, I have managed to calculate the quilibrium constant for the reaction at 298K, which I found to be 10^-25.4, making it dramatically unfavorable to products.


Does anyone find the same thing for this constant, or have I made a mistake? If it is the right constant, how come the reaction takes place?


I am aware that it is not the commonly used reaction when it comes to the Winkler method, but its the only one I can propose when it comes to having the dioxygen react with manganese hydroxide owing to what few data I have. Also sorry for my shaky English, I'm french.


Thanks in advance for your answers!


