3
A less laborious path to very fine iron dust (in fact, likely pyrophoric iron) perhaps follows from first purchasing or preparing iron(II) oxalate (a path is detailed in a reference provided below). Next, thermally decompose the salt and finally, stabilize the nano-iron product with say mineral oil.
The details on the thermal decomposition in an atmosphere ...
3
You have replicated electro-rafination of copper. Copper ions from the being dissolved anode are replacing the copper ions being deposited on the cathode.
Therefore overall electroneutrality of the solution is kept. The local neutrality is managed by electromigration of both ions $\ce{Cu^2+}$ and $\ce{SO4^2+}$.
2
The mathematical description of the cell constant ($l/A$) does not include factors like fringing fields (the ionic current does not travel precisely in a well-defined path between the electrodes) and the actual surface area for rough surfaces.
The cross section of a round electrode may have different fringing fields depending on its separation from its ...
2
There is a rather interesting explanation as to why only sodium ion can pass through a cation exchange membrane and chloride ion is repelled by it.
Let us recall some properties
a) Ion exchangers are very good electrical conductors. This is well studied in the 1940s. The current is carried, not by electrons but by mobile ions in ion exchange resins. In a ...
2
The electrode redox reactions of the Daniell cell:
$$\ce{Zn(s) <=> Zn^2+ + 2 e- }$$
$$\ce{Cu(s) <=> Cu^2+ + 2 e- }$$
maintain at the respective electrode a particular potential where both opposite reactions have the same rate, implying there are no external galvanic causes that affect this potential. Rates of oxidation electrode reactions ...
2
BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. ...
2
The capacity of a cell is measured toward the terminal cut off voltage. If e.g.a double current is drained from a cell, the external cell voltage is decrease by the doubled voltage drop on the cell internal resistance: $$U_\mathrm{e}=U_0 - R \cdot I$$
That caused reaching the terminal voltage sooner then in the half time.
It is additionally leading to the ...
2
With decreasing electrolyte concentration, either the needed voltage raises at constant current, either the passing current decreases at the constant voltage.
Both happens because at lower concentration, a higher potential gradient across the electrolyte is needed to maintain the same current, as the smaller ion count requires ions to move faster for the ...
2
Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.
So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving ...
1
By definition : $\ce{\Delta $G = - zEF$}$, where $z$ is the number of electrons in the half-reaction, and $F$ is the Faraday (about $96'500$ Cb)
In the first half-reaction , $z = 1$, and $E = +~ 0.98$ V. So $\Delta G_1 = - ~0.98~ F$ (in Joules)
In the second half-reaction, $z = 2$, and $E = +~1.59$ V. So $\Delta G_2$ = - $2ยท1.59 F$ = - $3.18 F$ (in Joules)
...
1
Lithium has a standard redox potential equal to -$3.04$ V. Coupled with any compound at the cathode, it could hardly produce cells yielding more than $4$ Volts, because part of the available energy is needed and lost to extract the external electron of the Lithium atom. The performance of a cell built with a Lithium anode could be more important if the ...
1
Reference electrodes must involve a well defined redox system with reproducible potential.
Mercury is an essential part of the redox system of the calomel reference electrode $\ce{Hg|Hg2Cl2(s)|Cl-}$:
$$\ce{2 Hg <=> Hg2^2+ + 2 e-}$$
$$\ce{Hg2^2+ + 2 Cl- <=> Hg2Cl2(s)}$$
and cannot be removed nor replaced, unless you want to use a difference ...
Only top voted, non community-wiki answers of a minimum length are eligible
