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How do I know how much energy I need to obtain 20 g alumunium from alumina with 5 V?

I calculated the quantity of electricity is 2.22 F.

Any help appreciated!

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Required energy $E$ can be found as a product between voltage $U,$ current $I$ and time $t$:

$$E = UIt\label{eqn:1}\tag{1}$$

Unknown multiplier $It$ can be found using united Faraday law:

$$m = \frac{MIt}{zF}\quad\implies\quad It = \frac{mzF}{M}\label{eqn:2}\tag{2}$$

where $m$ is the mass of the substance liberated at the electrode; $M$ is the molar mass; $z$ is the number of electrons transferred per ion; $F$ is Faraday constant. Considering aluminium is produced via Hall–Héroult process or similar, at the cathode the following reaction takes place:

$$\ce{Al^3+ + 3 e− → Al}$$

and $z = 3.$ Plugging \eqref{eqn:2} into \eqref{eqn:1} allows to finally find the energy:

$$ \begin{align} E &= \frac{mzUF}{M}\\ &= \frac{\pu{20 g} × 3 × \pu{5 V} × \pu{96485.33 C mol-1}}{\pu{26.98 g mol-1}}\\ &= \pu{1.07 MJ}\tag{3} \end{align} $$

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  • $\begingroup$ With much respect, can you assist me to understand chemistry? Can i have your email or whatever platform, so i can ask simple chemistry question to you? Basically, i dont have privilege to ask many questions in this forum. @andselisk $\endgroup$ – Lifeforbetter Oct 29 '19 at 6:03
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    $\begingroup$ @Lifeforbetter I prefer not to share my personal information and I don't do private teaching. If you have a question, the better way would be to ask community and post a new question. Also, please note this is not really a forum, but a Q&A site with somewhat restricted back-and-forth communication capabilities. In the meantime, I'd suggest to look at Resources for learning Chemistry, or enroll in free chemistry courses online. $\endgroup$ – andselisk Oct 29 '19 at 6:25

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