18

Yes, you could have multiple electrodes in the same electrolyte, but to some extent, that would short-circuit the battery. For example, if you stack copper and silver coins with blotting paper (bp) between them, in the order: Cu bp Ag Cu bp Ag ... Cu bp Ag and immerse the whole in an electrolyte, rather than just wetting each piece of blotting paper, some of ...


14

Yes, you can have same the electrolyte and a pair of two different metals, but the key point is that if you wish to increase voltage difference, you need to connect them in series and use separate containers for each pair, and of course each pair must be connected. Your postulate in the comment is correct. If we use a large bucket, only the pair connected to ...


13

Aluminum redox potential has nothing to do with the voltage of the cell you have build with it. Let me explain why, because it is not obvious. Electrolytic cells made with aluminum anodes always yield rather low voltages. The reason is that usual pieces of aluminum are always covered by a thin, continuous and colorless layer of aluminum oxide $\ce{Al2O3}$. ...


11

$\ce{H2O}$ 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 ...


11

When aluminum is the anode (connected to the positive battery terminal), a thin layer of insulating aluminum oxide is produced via the “anodization”. It serves as a dielectric, so basically you have a leaky capacitor: current is low. But when the aluminum is connected to the negative terminal, it is the cathode and then you get reduction of hydrogen ions to ...


9

Interesting question but keep in mind that it is normal to dislike a subject. You don't have to like chemistry. Many people loathe physics and mathematics. The world is not affected. I am sure your experience is limited to general chemistry and Oxtoby. In statistical terms, one cannot and should not trust a single sample (=experience). However, some of your ...


8

The $\pu{3.6-3.8 V}$ range is a good general choice, but it may be battery-specific. The particular voltage for 40% charge may differ for different cell technologies, e.g. various deviations of electrode materials and due to cell aging. The optimal storage conditions, as you mentioned, are more often expressed as charge/capacity % ratio. Usually, the optimal ...


7

There are a couple factors here: What you are using for an anode is probably the most important. If you’re not using copper, you’re depleting your copper sulfate electrolyte of copper ions. Use a copper anode and you’ll be continuously replenishing your electrolyte with copper ions. Concentration of your electrolyte is another factor. If the electrolyte is ...


7

If the hydrogen and oxygen molecules take up the same amount of space, then why did the tube over the negative terminal have much more than twice the gas as the other one? Good questions but keep in mind that you did not mention the concentration of sodium chloride. If you have too much salt in water, chlorine gas is produced (eventually bleach) instead of ...


7

The short answer is: The ending stage uses much smaller charging current, so it lasts longer. TCharging of any cell in the contant voltage mode, not limited to lithium cells, leads to asymptotically decreasing of the charging current and progressively slowing down charging proces. The reason behind is the charging voltage must not cross the maximal allowed ...


7

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's ...


7

The figures in the OP's post are nicely drawn, as expected in a modern textbook, but figure 3.2(c) has the external battery reversed, which is incorrect. Here is how it works, without the needless complication of the potentiometer. First, start with a standard Daniell cell with standard assumptions, i.e., negligible internal resistance, unimolar ...


6

Start with a simple thought experiment: pour 100 mL of 1 M nickel (II) sulfate solution into a beaker and very carefully layer 100 mL of 0.01 M nickel (II) sulfate solution on top of the more concentrated layer. Then, even without convection or deliberate mixing, diffusion will, sooner or later, result in the solution having concentration of 0.55 M. In what ...


6

Electrolyzing a NaCl solution does produce hydrogen at the cathode, but no oxygen at the anode. Chlorine $\ce{Cl2}$ is produced at the anode, with maybe a small proportion of oxygen as an impurity. Unfortunately this $\ce{Cl2}$ gas is relatively soluble in water. That is why you have obtained relatively few gas at the anode. The mas of the atoms has no ...


6

You simply started your data range at too high a concentration to see the initial increase. Semantic Scholar gives data for the conductivities of several aqueous electrolyte solutions at 25°C, including magnesium chloride. The strongly increasing portion of the magnesium chloride curve goes up to about 10%, which corresponds to about one molar (where your ...


6

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 ...


6

Your teacher is correct. I will take the example that you gave in the question. You wish to analyze the reaction: $$3\ce{MnO4^{2-}}\to 2\ce{MnO4^{-}}+\ce{MnO2}...(1)$$ This reaction can be split into two parts, the oxidation half reaction, which forms the anode: $$\ce{MnO4^{2-}}\to \ce{MnO4^{-}}+\ce{e-}...(2)$$ And the reduction half-cell, which forms ...


5

If there is no published resistance for that carbon sheet, then the best choice is to do exactly what you have done, i.e., measure the resistance per square. However, that specification is in given in the Strem catalog! It might be difficult o find if you do not spell the company's name correctly. Two notes: Resistance of a uniform, homogeneous sheet is ...


5

Sometimes the electrolyte directly takes part in the chemistry, by being oxidized or reduced at an electrode. In those cases, it is easy to understand why the concentration would have an influence on the kinetics of the redox reactions (and with that, on the observed electrical current). For water electrolysis, the "active" electrolytes are ...


5

If you want energy as heat, the best way to generate energy is combustion. If you want energy as electricity, the best way is a fuel cell (which also produce some heat). If you want mechanical work, using hydrogen like gas or oil in a usual combustion engine, the yield would be a maximum of $30$%. The best yield would be to produce electricity in a fuel cell,...


5

I've only basic training in chemistry, since my competences are in Electrical Engineering, but I'd like to tackle your question from another angle. I'm assuming your question is sort of an X-Y problem and you are not really interested in increasing the efficiency of an actual lemon-based cell. From an engineering POV, what you want from a power source is, ...


4

Here the iron electrode is made by dipping an iron plate or wire in a $\ce{FeCl2}$ or $\ce{FeSO4}$ solution. It is an anode, where $\ce{Fe}$ is oxidized according to $$\ce{Fe -> Fe^{2+} + 2 e-}$$ The chlorine electrode is a cathode made by dipping a platinum electrode in a solution where some chlorine gas $\ce{Cl2}$ is arriving from under the platinum ...


4

I can't present you a movie of electron interactions, but maybe a fuzzy picture. Imagine a metal body (start with copper): the atoms are held in a solid structure - in a sea of electrons! Those loose electrons can be pushed around easily, so we can make wires out of copper and transmit electricity thru them. If you put two different metals together, the sea &...


4

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 ...


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 ...


4

Ions leave the salt bridge and move further, without accumulating at the bridge end. For all salt bridges where there is no other reactions involved, so all simple salts in solution or molten salts. Ions move all the way through. There is some concentration difference in salt bridge and main half cell volume. It is true for simple salt solutions and molten ...


4

Here are the answers : The ions form the salt bridge must get out of this bridge. But they usually stays at the border, because they push other ions already present in the solution to compensate for the new charge having appeared on the electrode, like a locomotive or an engine pushes its trucks on the other end of the train. It is the same in all bridges. ...


4

No ! It is not possible to deposit aluminum by electrolysis of a water solution. It may be done in a solution of $\ce{AlCl3}$ in an organic solvent like toluene, but not in water. And even in toluene, the conductivity is low. Metallic aluminum reacts with water, when this metal is not protected by an outer layer of aluminum oxide $\ce{Al2O3}$ as it is as ...


4

I think, it is a result of the word having multiple definitions in different contexts. Specifically, the original definition in context of chemistry Electrolyte is a solid, causing electrical conductivity when dissolved in a liquid. Later works extended it to non-solids, of course, but the root "lyte" (AFAIK, Greek for "earth") remained....


3

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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