To gain 1 mole Mg from Mg^{2+} and 1 mole A | vedclass

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Question

(A) The reduction reactions are as follows: $Mg^{2+} + 2e^- \rightarrow Mg$ To produce $1 \ mole$ of $Mg$,$2 \ moles$ of electrons are required. Charg

Vedclass · Accepted Answer

$193000 \ C$ and $289500 \ C$

Vedclass · Answer

(A) The reduction reactions are as follows: $Mg^{2+} + 2e^- ightarrow Mg$ To produce $1 \ mole$ of $Mg$,$2 \ moles$ of electrons are required. Charge required $= 2 	imes 96500 \ C = 193000 \ C$. $Al^{3+} + 3e^- ightarrow Al$ To produce $1 \ mole$ of $Al$,$3 \ moles$ of electrons are required. Charge required $= 3 	imes 96500 \ C = 289500 \ C$. Therefore,the required charges are $193000 \ C$ and $289500 \ C$.

To gain $1 \ mole$ $Mg$ from $Mg^{2+}$ and $1 \ mole$ $Al$ from $Al^{3+}$,how many Coulombs of electricity are required respectively?

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