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Aluminum redox potential has nothing to do with the voltage of the cell you have build with it. Let me explain why, because it is not obvious. Electrolytic cells made with aluminum anodes always yield rather low voltages. The reason is that usual pieces of aluminum are always covered by a thin, continuous and colorless layer of aluminum oxide $\ce{Al2O3}$. ...


6

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's ...


6

You may heat mixture of $\ce{MnO2}$ with charcoal to higher than $\pu{850 °C},$ in the absence of air. Manganese will be produced in the solid state by reduction of the oxide. But it is sensitive to air oxidation and easily reoxidized. So better use an excess of charcoal. In the industry, rough $\ce{MnO2}$ out of the mine is first reduced to $\ce{MnO}$ by $\...


3

The reaction $$\ce{Al2O3 + 3 Mg -> 3 MgO + 2 Al}$$ is slightly exothermic, with a $\Delta H = - 130 kJ/mol$. It is also exoergic at all reasonable temperatures, as $\Delta S $ is very small ($- 1.5 J/mol$). So the reaction is in principle feasible. The trouble is that $\ce{Al2O3}$ and $\ce{Mg}$ are solid at room temperature. And the reactions between ...


2

Research notes in this ACS article Kinetics of Hydrogen Reduction of Manganese Dioxide that even heating $\ce{MnO2}$ in an atmosphere of hydrogen gas only results in $\ce{MnO}$. The source further notes that very high temperatures (some 1,600 K) are required for any appreciable formation of the metal. As such, a more facile approach, simply dissolve the ...


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