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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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Given the whitening observed at the cathode, one potential explanation is that calcium salts were precipitating there, forming a layer blocking passage of part of the current. I usually write the cathodic hydrogen-evolution reaction in neutral solution as: $$\ce{2 H2O + 2 e- -> H2 + 2OH-}$$ Thus, the pH in the vicinity of the cathode will be ...

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First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $\ce{Zn^{2+}/Zn}$ at $\pu{-0.76 V}$. Then, going to the right, you find $\ce{V^{III}/V^{II}}$ at $\pu{-0.26 V}$. Now you find $\ce{Pb^{2+}/Pb}$, and so on. Now try to learn how to use this scale. The ...

2

The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to E_\ce{Pb} = \pu{-0....

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For equally charged ions a smaller radius implies a higher charge density, and therefore stronger Coulombic interactions with other charges in solution, including stronger ion-dipole interactions with water molecules, resulting in a more stable hydration sphere, and stronger interactions with counterions (here chloride). Such interactions reduce the mobility ...

1

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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Let's see if we can tease out the main points here. The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook. Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight ...

1

I suspect the problem is not mathematical. It is in the meaning of the wave function. Well. I will now show you how I explain it qualitatively in my high school classes. I state that the electron is like a violin string, but a string with three dimensions that is vibrating in the fourth dimension. As I know this last words are not understandable, I start ...

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Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric ...

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Theoretically, we could do any reaction. Practically, some are only possible in one direction; others are under special conditions. For the case of copper: the standard potential of the pair $\ce {Cu^{2 +} / Cu}$ being positive, this means that naturally the reaction is done in the direction $\ce {Cu^{2+} -> Cu}$. But we can still force the reaction ...

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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 standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $H ^ +$ and not with $HO ^ -$

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During an titration, one passes from a state where the titrated species is in excess to a state where the titrating reagent is in excess. Let us now take the case of the dosage of a solution of ions $\ce{Fe^2+}$ by the permanganate ions $\ce{MnO4-}$: we follow this dosage by immersing a platinum electrode and a reference electrode. At the start, the solution ...

2

Why does each atom wait to turn into gas until they reach a particular electrode? There are no oxygen atoms or hydrogen atoms formed in the solution. There is also no "electricity" going through the solution. The processes happen at the electrodes because the cathode is able to provide electrons (to reduce hydrogen ions to dihydrogen) and the ...

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In electrolysis, why does oxygen only appear on the anode? Easy way to remember this is that oxidation occurs at the anode (both start with vowels), and oxidation implies a loss of hydrogen. If we lose hydrogen from water, we are left with oxygen. If you wish to understand it crudely, the concept is that anode the surface in the solution where the electrons ...

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Even pure water is somewhat ionized. If water is more conductive than pure brake fluid, the test is valid. Why don't you try this as an experiment? To sample(s) of brake fluid fresh from a new, unopened, can, add distilled water, drop by drop, stirring each time and look for a change in conductivity. Measure the proportions, and you can graph the effect of ...

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$\ce{H2O}$ is not simply split apart by electricity, as you say. No ! What happens on one electrode is not related to what happens to the second electrode. Let's start by discussing what is happening on the negative electrode, the cathode. The negative electrode behaves as if it contains plenty of electrons ready to react with anything able to do it. It ...

4

Electron transfer between the 'species in the solution' and the 'electrode' take place at the surface of the electrode. So, considering water with some acid Oxidation at anode: $\ce{2 H2O(l) -> O2(g) + 4 H+(aq) + 4e−} \quad E^\circ = \pu{+1.23 V}$ (for the reduction half-equation) This oxidation (electron transfer from water to the electrode) occur at the ...

1

Being intensive or extensive property is not related to changing of system properties like concentrations. It is related to scalability of systems. If you change the system scale, like doubling or halving it, and if the property remains the same, like temperature, density, composition, equilibrium EMF, it is the intensive property. If you change the system ...

2

I like what Maurice said "When nobody understands a scientific phenomena, we give it a name. Here the name is overpotential. Here the hydrogen is said to have a big "over potential" on mercury cathode. That is a bright and remarkable way to hide our ignorance." This is little on the extreme side but people have spent their life on ...

1

I was learning about voltaic cells and came across salt bridges. If the purpose of the salt bridge is only to move electrons from an electrolyte solution to the other, then why can I not use a wire? If you connect the two electrodes with a wire, you will short-circuit anything else connected to the electrodes. If you connect the two electrolyte solutions ...

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Your aluminum half cell has a standard OCV (i.e., at 1 M $H^+$) of 1.676 volts. Ref. 1 You don't have standard conditions, but that doesn't matter; your conditions do not change for this half cell. The other half cell is a reduction of water, in alkaline solution, with the electrons from the aluminum to produce hydroxide ions. The CRC Handbook confirms -0....

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