Oxidation of alcohols by Ceric Ammonium Nitrate
In this reaction, the ion $\ce{Ce^{4+}}$ is a powerful oxidant. It needs one electron to be transformed into the more stable ion $\ce{Ce^{3+}}$. This electron may come from the central function $\ce{-CHOH-}$ of any secondary alcohol $\ce{R-CHOH-R'}$. The bond between $\ce{O}$ and $\ce{H}$ is able to loose $1$ electron and $1$ $\ce{H+}$ ion. Same thing happens with the bond $\ce{C-H}$ belonging to the same group $\ce{-CHOH-}$. As a consequence, two electrons are emitted with two $\ce{H+}$ ions, and the remaining organic molecule becomes a ketone $\ce{R-CO-R'}$. The two electrons react with two ions $\ce{Ce^{4+}}$. The two half-equations are : $$\ce{Ce^{4+} + e^- -&gt; Ce^{3+}}$$  $$\ce{R-CHOH-R'-&gt;  R-CO-R' + 2 H+ + 2 e-}$$ so that the final equation is : $$\ce{2 Ce^{4+} + R-CHOH-R' -&gt; 2 Ce^{3+} + R-CO-R' + 2 H^+}$$