How does equalizing voltage for lead acid battery depend on the temperature? I am using advanced lead-carbon batteries (these ones) in an off-grid solar system. The spec specifies equalizing voltage 14.1 V for 12 V block @ 25 °C (77 °F). I keep the batteries in an outdoor shed, and the temperature where I live tends to hang around 0 °C (32 °F) these days.

So, for lead acid chemistry, would the equalizing voltage go up or down with temperature going down? I feel like it would go up (more voltage needed to do the same job at lower temperatures). How does this reasoning stand? What are the underlying scientific considerations?
Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge.
$$\ce{Pb &lt;=&gt; Pb^2+ + 2 e-}$$
At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$.
$$\ce{Pb^4+ + 2 e+ &lt;=&gt; Pb^2+}$$
For a classical lead–acid battery, the overall cell reaction is approximately
$$\ce{Pb + PbO2 + 2H+ + 2 HSO4- &lt;=&gt; 2 PbSO4 + 2 H2O}$$
As long as $\ce{Pb}$, $\ce{PbO2}$, and $\ce{PbSO4}$ are available at the electrodes, the equilibrium cell voltage depends only on the acid concentration (that's why equilibtrium cell voltage can be estimated based on measured acid density).
The dependence of the equilibrium voltage on concentration is given by the Nernst equation:
$$U_\text{cell}=\left(1.931+0.0592\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
This equation applies to one cell; a battery, however, has six cells in a row. Thus the total voltage is
$$U_\text{battery}=6\left(1.931+\frac{RT}{ZF}\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
where
$R$ is the gas constant,
$T$ is temperature,
$z$ is the number of electrons transferred in the cell reaction, and
$F$ is the Faraday constant.

We are only interested in the dependence of voltage $U$ on temperature $T$. All the other parameters may be taken as constants. Thus, our equation can be simplified to
$$U=\left(11.586+T\cdot k\right)\ \mathrm V$$
where $k$ is a constant.

The exact parameter values for a classical lead–acid battery might differ a bit from a modern battery with lead–carbon electrodes; however, we can calibrate our model using the data given by the manufacturer, i.e. $U=14.1\ \mathrm V$ at $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$
$$14.1\ \mathrm V=\left(11.586+298.15\ \mathrm K\cdot k\right)\ \mathrm V$$
and we find 
$$k=0.008432\ \mathrm{K^{-1}}$$
Therefore, our complete equation is
$$U=\left(11.586+T\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V$$
For a new temperature of $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$, we get an estimate of
$$\begin{align}U&amp;=\left(11.586+273.15\ \mathrm K\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V\\
&amp;\approx13.9\ \mathrm V\end{align}$$