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Both methods yield $\pu{4.45E-3}$ mol $\ce{O2}$. But to get $\pu{1 mol}$ $\ce{O2}$, you need $\pu{4 mol}$ electrons, as the half-reaction at the anode is : $$\ce{2 H2O -> 4H+ + O2 + 4 e-}$$So to get $\pu{4.45E-3 mol}$ $\ce{O2}$, the needed amount of electrons is $4$ times the amount of $\ce{O2}$, or $4$· $\pu{4.45E-3}$ = $\pu{1.78E-2}$ electrons. Now $2$ ...


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There is oxidized oxygen in water $$\ce{2 H2O -> 4 H+(aq) + O2(g) + 4 e-}.$$ $\ce{HSO4-}$ in sulphuric acid solution can get oxidized as well at high current density and therefore high enough anodic potential: $$\ce{HSO4-(aq) -> S2O8^2-(aq) + 2 H+(aq) + 2 e- }$$ This is the industrial way to produce potassium peroxodisulphate.


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Rate of electrolysis is only dependent on the current (Faraday's law) passing through the solution. Remember that in an electrochemical experiment you cannot independently vary voltage or current. One factor has to be fixed. These two modes are called constant current electrolysis or constant potential electrolysis. Once you have fixed voltage, you can ask ...


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The equation noted corresponds well to that of the first half-cell: on the other hand, the potential indicated is not the good one but just that of the standard potential. The Nernst relation is written $\ce{E=E^0 +0.06*log \frac{[Co^{3 +}]}{[Co^{2 +}]}}$: the potential therefore depends on the concentrations You can therefore calculate the potential of this ...


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In the "Index of Electrolytes" on page 3 of the document at the source of the problem (Ref. 1) note that the third column is titled "Maximum conductance and point of inflection at 25°C [$\pu{μmhos/cm}$/% by wt.]". However, the entries are given in units of conductivity ($\pu{μmhos/cm}$ or $\pu{μS/cm}$)! The first number for each entry in ...


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The basic reason for using graphite as an anode is in Hall–Héroult process in the electrolytic reduction of alumina to aluminum metal is because graphite being an allotrope of carbon and an inert electrode reacts with oxygen to give out carbon dioxide which thus prevents the liberation of oxygen as a final product at the anode. Had there been any other metal ...


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The answer is quite simple, if you look at the equation carefully. Conductance and conductivity are related by cell constant, $\displaystyle\frac{\text{Area}}{\text{Distance}},$ where the area is the area of the electrodes and the distance is the distance between the electrodes. $$G = \sigma\frac{A}{l}$$ In conductivity measurements, the cell constant was ...


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