New answers tagged electrolysis
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The voltage won't vary as it depends on the materials used as active electrodes and their respective standard reduction potentials.
The current will vary with plate area due to varying resistance of the cathode plate.
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Seems your cell is operating as,
Anode: $\ce{H2 = 2H + 2e'}$
Cathode: $\ce{2H(+) + 2e' = 2H}$, and then $\ce{2H + 1/2O2(g) = H2O(l)}$ -Heat.
E = 1.4 V gives, 1.4 = [2.303RT/F] log R, where R = ratio H activity between cathode and anode R = 10^-23.65 if your anode assumes as activity H = 1.0, at the cathode H activity is 10^-23.65 the removal H by the ...
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Yes, it is true that you cannot use the standard electrode potential to determine which cation would be reduced and vise versa during an electrolysis of mixed molten salts, in your case, $\ce{AlBr3}$ and $\ce{MgI2}$. That's because most tabulated values of standard electrode potentials are determined for aqueous solutions. Therefore, they can be used only as ...
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There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and ...
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