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0

In a (standard) Zn/Zn++ half cell, no oxidation (and no reduction) is taking place. A piece of zinc is sitting in a 1M solution of Zn++ ions (with their accompanying anions). It is not imaginary, and on a micro scale, an equilibrium is occurring (could be occurring!) where Zn + Zn++ --> Zn++ + Zn. No overall change occurs; this is a reference point. If it ...

3

Nernst equation is the only way of solving this problem. Let's consider the example of the reduction of permanganate. The half-reaction is: $$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}$$ The potential can be written as: $$E = E^\circ + \frac{0.059}{5}\log \left(\frac{\ce{[MnO4-][H+]^8}}{\ce{[Mn^2+]}} \right)$$ If you want to see explicitly ...

3

The answer is: it depends. Dissolution of a salt implies that the entropy gained exceeds the cost of breaking lattice interactions (the solution enthalpy, assuming it is positive). Electrostatic interactions compete with kT (thermal jostling). Under physiological conditions, long range interactions are strongly screened by intervening solvent molecules and ...

1

You could try considering the Cottrell equation or the mass transfer limiting case of the butler-volmer equation $$i = \frac {nFAc_{j}^{0}\sqrt{D_{j}}}{\sqrt{\pi t}}$$ i = current, in unit A n = number of electrons (to reduce/oxidize one molecule of analyte j, for example) F = Faraday constant, 96485 C/mol A = area of the (planar) electrode in cm2 ...

0

You are confusing several electrochemical conventions. First of all you have to accept like the rest of the electrochemists that all potentials will be written as reduction potentials. You will not change their sign whether is oxidation or reduction half cell. Only then, $$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$$ is valid....

0

I just add a good mnemonic tool how to remember the naming convention : Anode = anabasis ( from Greek ana = "upward", bainein = "to step or march") , electrons would be going upwards from the electrode to the wire = oxidation. Xenophon, Anabasis, "The March Up Country" Cathode = cathabasis (the journey downwards), electrons would be going downwards from ...

2

The good mnemonic tool to remember is : Anode = anabasis, electrons would be going upwards from the electrode to the wire = oxidation, ( Xenophon, Anabasis, 404BC, "The journey upwards(to north)) Cathode = cathabasis(the journey downwards), electrons would be going downwards from the wire to the electrode = reduction That means a cathode/anode is the ...

5

Who was responsible for this naming system and how can we change it? Michael Faraday was responsible for the terms anode and cathode more than hundred years ago. All the confusion regarding the nomenclature will vanish if you do not associate electrostatic signs with these two terms. One should identify the electrode labels with the redox processes rather ...

0

You can imagine the cell electrodes as capacitors, that are charged by chemical way. But the capacitance of these capacitors $$C=\frac {\mathrm{d}q}{\mathrm{d}E}$$ is very small, so does the accumulated charge $q$. The charge loses due pushing electrons through the wires of a closed circuit are balanced by simultaneous recharging by ongoing redox ...

3

I am sorry, this is not an answer but a comment, too long to be edited in the comments section. I don't agree with "straightforward" in the sentence : " It is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. " This would be an ...

2

It is known that an electrochemical impedance spectra can be modelized with series-parallel combination of resistors and capacitors elements. What is more, for a given impedance spectra they are an infinity of different combinations of R-C elements which accurately model the same complex impedance. Similar observations are commonly done in the field of ...

2

is it true that if I connect an aqueous oxidant to an aqueous reductant via a wire, will a current flow along the wire? In particular, does current flow in the following scenario? No true. Nothing will happen as the circuit is incomplete. The oxidant or the reductant species have to see each other "face to face". Guess what is the most powerful reducing ...

0

For example, a silver electrode is immersed in a silver solution and so on, but is why is this necessary? Surely any positive metal ion will do? There is a fundamental reason, which is the Nernst equation. It requires that the metal be in contact with its own ions. What is the functional meaning of the electrode potential? It is the tendency of an ion to ...

0

Not at all, Zero Resistance Ammeters are dedicated testing equipment provided with internal feedback to automatically compensate for voltage losses caused by the measurement. They actually have "infinite" input resistance, so don't load the source of current in any way, but are backed by an op-amp based circuit which acts like a zero impedance current source....

0

If you're choosing the voltage, then you won't get to choose the current. This is because the resistance is already chosen for you, as it is a physical property of your electrochemical cell. Therefore, two of three variables are known in Ohm's Law, which means the third, in this case current, is fixed by the other two, voltage and resistance. For example, ...

0

For these metals to deposit, they would have to be present in solution such that they can come into contact with the cathode and be reduced. However, this would first require them to be oxidized, so that they dissolve in solution. Unless you are actually electrolytically refining for gold, the voltage will not be enough to oxidize it.

-1

I think it has to do with the equation heat dq=TdS so since gibbs dG=SdT+VdP+mudN we know S or entropy is partialG partialT and also gibbs= -nFE so S=partialE partialT and heat is TdS so TpartialE partialT* nF. In short TdS is the reversible heat

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