I really like Ivan's answer above, but wanted to add one other perspective.
Static discharge is a significant possibility, and that might change the chemistry (e.g., redox events). If that doesn't happen, I'd argue that very little will happen because these fields are very common on the nanoscale.
Let's imagine we have ~1MV/m applied to the vessel. That works out to $1\times 10^6\; \mathrm{V/m}$. So about 1V per µm. Even if you have a much smaller reaction flask (10 cm maybe), that's 1V per 100 nm.
Consider the molecular scale. If I have a monocation separated 1.0 nm from a molecule, the molecule experiences ~1V/nm in vacuum. That would be $1\times 10^9\; \mathrm{V/m}$. Granted, in any reasonable environment, the dielectric constant of the solvent will decrease the field substantially, but even in water, the field would be ~$1\times 10^7\; \mathrm{V/m}$.
There's an atomic unit of electric field: $5.142 \times 10^{11}\; \mathrm{V/m}$. That corresponds to the interaction energy between a proton and an electron in a hydrogen atom.
Your fields are much lower than common ion-molecule interaction fields or proton-electron fields.