What impact does heating or cooling have on the oxidation reduction potential of water?

There are alkalinizing water ionizers on the market that lower the ORP of water. Does heating or cooling that water have an impact on its ORP?

• Another question is, how significantly? What is the context of your question? – Eric Brown Jun 21 '13 at 18:09

First, the temperature effect on ORP. The thermodynamics of oxidation and reduction reactions is governed by the Nernst equation: $$E(ORP) = E_o - \frac{2.3RT}{nF}\times log\frac{[reductant] }{[oxidant]}$$
In this equation, R and F are the gas law and Faraday's constants, respectively, $E_o$ is the standard reduction potential of the system of interest, T is temperature, n is the number of electrons transferred in the reaction and the concentrations of the reductant and oxidant are found in the log term. The coefficient $\frac{2.3RT}{nF}$ in front of the log term can be simplified to $\frac{0.00198\times T}{n}$ and when we perform dimensional analysis on the term we find that it will provide units in Volts. If we are looking for the largest effect that temperature would have on ORP then we want to use the smallest value of n, which would be 1 electron, further simplifying the coefficient to approximately $0.002T$. T has to be in Kelvin so if we take our two extreme values as the freezing (273 K) and boiling (373 K) points of water, we find that the difference in ORP between these two extremes (all else being the same) would be about 200 millivolts.
Yes. I don't have much time to go into details (sorry) but temperature changes the pKa of water which in turn affects the concentrations of [$\ce{H3O+}$] and [$\ce{OH-}$] and that affects both reduction and oxidation potential of water. The effect it negligible for most intentions and purposes over the 100 C of liquid water.