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In electrochemistry, the rate of electrolysis depends on rate of charge entering the cell. Second most important point is that one can either control potential or current but not both during electrolysis i.e. you cannot have both values set. If you are fixing potential at 2.5 V, the value of current is not in your hand. I don't think Ohm's law remains valid ...


For a general electrolytic reaction, $$\ce{M^{n+} + $n$ e^- -> M}$$ where $n$ moles of electrons are consumed in depositing 1 mole of substance ($n$ is also known as $n$-factor). So, for $x$ moles of substance, $nx$ moles of electrons would be consumed. Therefore, $$Q = nxF,$$ dividing by $nF$ on both sides, we have $$\frac{Q}{nF} = x,$$ which is ...


The misconception you have in the question is that the concept of tabulated electrode potentials do not apply to molten salts. The values in general chemistry textbooks have been calculated in water under specific concentrations. No need to extrapolate electrode potentials to molten states.


$\ce{OH-}$ can't be reduced. It may only be oxidized by giving an electron at the anode, while $\ce{H+}$ is reduced instead of $\ce{OH-}$. Secondly, the reducing property of $\ce{H2O}$ is greater than that of $\ce{H+}$ ion. Moreover, it is greater in number than other ions like $\ce{H+}$ ions etc. So it will be preferentially discharged although it is ...

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