New answers tagged

0

In addition to price advantage, CuSO4 also provides a source of sulfate ions, which are relatively stable. The sulfate anion is also a known scavenger of radicals that may be present in the solution. For example, a reaction between sulfate and the hydroxyl radical (a possible electrolysis product): $\ce{SO4^{2-} + .OH -> .SO4^- + OH^-}$ And, the ...


1

You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.


0

Alejandro, you are right, the electrolytes will make a difference. The difference is not in terms of hydrogen production per se, as Maurice mentions in the comments, it will be in terms of energy saving. Electrolytic reactions are "exact" in the sense that if 100 electrons reach the electrode from a battery, 100 hydrogen ions will be reduced. In other words, ...


0

I did find a work, 'Electrolyte Engineering toward Efficient Hydrogen Production Electrocatalysis with Oxygen-crossover Regulation under Densely Buffered Near-neutral pH Conditions', with comments of interest. I start with some background from the introduction: In recent decades, drastic progress in solar fuel production has occurred: photovoltaic cells ...


0

If you dip a Copper plate and a Zinc plate into a salt solution, you create a galvanic cell, where Zinc is oxidized and is the negative pole. If you want to apply an external tension from another cell, the result depends on the choice of the positive and the negative poles. If you connect positive poles to the negative poles, both cells produce courant, and ...


1

For an electrochemical reaction, you count the atoms / ions by mol, and use the coulomb as a counting unit of charge. For a more intuitive explaination of the $n$ factor in the Faraday equation, try this analogy: The summer olympics include swimming in a pool with lanes $50\,\pu{m}$ long. Among the typical competitions are runs about $50$, and $100\,\pu{m}$...


1

Suppose you have two galvanic cells A and B producing 1.14 V. If you connect them together, with both positive poles together, and both negative poles together, nothing will happen. No current will be produced. Now suppose that one of these two cells, say A, produces the same voltage 1.14 V, and B produces a little less than A, say 1.10 V. A works in a ...


2

You could create a constant-current source instead of using an off-the-shelf power supply. There are also off-the-shelf constant-current supplies, so ask your lab manager if that's a resource. An EE colleague can set this up for you if it's too far outside your domain. Here are a few articles as starting points. Total cost should be < $15. http://www....


14

You can't control both voltage and current electrically: Your power supply will hold one or the other constant, but the properties of the electrolytic cell will then determine the other one. So you need to hold all of the chemical parts of the experiment constant: Have a large ion bath so that the concentration doesn't change much as the plating takes ...


4

Try to change the geometry of your electrodes. For example, increase or decrease the distance between the electrodes. This will modify the internal resistance of the solution. So if you have a generator producing a constant voltage, you may reach the desired currant, simply by changing the distance between the electrodes. Of course the electrodes must be ...


Top 50 recent answers are included