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The overall reactions with the enthalpies of water electrolysis at room temperature can be represented by the equation below:

$$\ce{2H2O + \pu{237.2 kJ mol-1} + \pu{48.6 kJ mol-1} → 2 H2 + O2}$$

Where $\pu{237.23 kJ mol-1}$ is the maximum electrical work can perform and should be calculated by:

$$ \begin{align} \Delta H^\circ &= \pu{0 kJ mol-1} - (\pu{-285.83 kJ mol-1}) \\ &= \pu{285.83 kJ mol-1} \end{align} $$

$$ \begin{align} \Delta S^\circ &= 0.5 \times\pu{205 J mol-1 K-1} + \pu{130.6 J mol-1 K-1} - \pu{70 J mol-1 K-1} \\ &= \pu{163.1 J mol-1 K-1} \end{align} $$

$$ \begin{align} \Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ \\ &= \pu{285.83 kJ mol-1} - \pu{298 K}\times\pu{163 J mol-1 K-1} \\ &= \pu{237.23 kJ mol-1} \end{align} $$

where $\pu{48.6 kJ mol-1}$ is the net value of heat absorption, and can be calculated as the following:

$$ \begin{align} T\Delta S^\circ &= \pu{298 K}\times\pu{163.1 J mol-1 K-1} \\ &= \pu{48.6 kJ mol-1} \end{align} $$

This is the overall heat absorption during water electrolysis. I still don't know the heat release/absorption for the anode and cathode electrochemical reaction separately.

For the anode electrochemical reaction:

$$\ce{2 H2O → O2 + 4 H+ + 4 e-}$$

I believe this is a heat absorption process.

For the cathode electrochemical reaction:

$$\ce{2 H+ + 2 e- → H2}$$

I believe this is a heat release process.

How to calculate the value of heat absorption at anode, and heat release at cathode separately?

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