Due to the appearance of a certain gas that causes an instant burning cough. This article is the identification of this gas. The article is replete with formulas; the number of formulas is due to the non-trivial nature of both the electrolysis process itself and the rust itself. Chemists and chemical engineers, help bring the article to full compliance with reality; it is your duty: to take care of your “little brothers” in case of chemical danger.

Let there be iron Fe 0:
- if there was no water on Earth, then oxygen would arrive and make oxide: 2Fe + O 2 = 2FeO (black). The oxide oxidizes further: 4FeO + O 2 = 2Fe 2 O 3 (red-brown). FeO 2 does not exist, it is an invention of schoolchildren; but Fe 3 O 4 (black) is quite real, but artificial: supplying superheated steam to iron or reducing Fe 2 O 3 with hydrogen at a temperature of about 600 degrees;
- but there is water on Earth - as a result, both iron and iron oxides tend to turn into the base Fe(OH) 2 (white?!. It gets dark quickly in air - isn’t it the point below): 2Fe + 2H 2 O + O 2 = 2Fe(OH) 2, 2FeO + H 2 O = 2Fe(OH) 2;
- it’s even worse: there is electricity on Earth - all the named substances tend to turn into the base Fe(OH) 3 (brown) due to the presence of moisture and potential difference (galvanic couple). 8Fe(OH) 2 + 4H 2 O + 2O 2 = 8Fe(OH) 3, Fe 2 O 3 + 3H 2 O = 2Fe(OH) 3 (slow). That is, if iron is stored in a dry apartment, it rusts slowly, but holds on; increasing the humidity or getting it wet will make it worse, but sticking it into the ground will be really bad.

Preparing a solution for electrolysis is also an interesting process:
- first, an analysis of the available substances for preparing solutions is carried out. Why soda ash and water? Soda ash Na 2 CO 3 contains the metal Na, which is much to the left of hydrogen in a number of electrical potentials - this means that during electrolysis the metal will not be reduced at the cathode (in solution, but not in the melt), and water will decompose into hydrogen and oxygen (in solution). There are only 3 variants of the solution reaction: metals strongly to the left of hydrogen are not reduced, metals weakly to the left of hydrogen are reduced with the release of H 2 and O 2, and metals to the right of hydrogen are simply reduced at the cathode. Here it is, the process of copper plating of the surface of parts in a CuSo 4 solution, galvanizing in ZnCl 2, nickel plating in NiSO 4 + NiCl 2, etc.;
- dilute soda ash in water in a calm place, slowly and without breathing. Do not tear the package with your hands, but cut it with scissors. After this, the scissors must be placed in water. Any of the four types of soda (baking soda, soda ash, washing soda, caustic soda) removes moisture from the air; its shelf life is essentially determined by the time of moisture accumulation and clumping. That is, in a glass jar the shelf life is forever. Also, any soda generates a solution of sodium hydroxide when mixed with water and electrolysis, differing only in the concentration of NaOH;
- soda ash is mixed with water, the solution becomes bluish. It would seem that a chemical reaction has taken place, but no: as in the case of table salt and water, the solution does not have a chemical reaction, but only a physical one: the dissolution of a solid substance in a liquid solvent (water). You can drink this solution and get mild to moderate poisoning - nothing fatal. Or evaporate and get the soda ash back.

The choice of anode and cathode is a whole undertaking:
- it is advisable to choose the anode as a solid inert material (so that it does not deteriorate, including from oxygen, and does not participate in chemical reactions) - that is why stainless steel acts as it (I read a lot of heresies on the Internet, and almost got poisoned);
- it is pure iron that is the cathode, otherwise rust will act as an excessively high resistance of the electrical circuit. To place the iron to be cleaned completely in the solution, you need to solder or screw it to some other iron. Otherwise, the metal of the iron holder itself will take part in the solution as a non-inert material and as a section of the circuit with the least resistance (parallel connection of metals);
- not yet specified, but there should be a dependence of the flowing current and the rate of electrolysis on the surface area of ​​the anode and cathode. That is, one M5x30 stainless steel bolt may not be enough to quickly remove rust from a car door (to realize the full potential of electrolysis).

Let us take an inert anode and cathode as an example: considering the electrolysis of only a blue solution. As soon as voltage is applied, the solution begins to transform to the final solution: Na 2 CO 3 + 4H 2 O = 2NaOH + H 2 CO 3 + 2H 2 + O 2 . NaOH - sodium hydroxide - crazy alkali, caustic soda, Freddy Krueger in a nightmare: the slightest contact of this dry substance with wet surfaces (skin, lungs, eyes, etc.) causes hellish pain and quick irreversible (but recoverable in case of mild burns) ) damage. Fortunately, sodium hydroxide is dissolved in carbonic acid H 2 CO 3 and water; when the water is finally evaporated by hydrogen at the cathode and oxygen at the anode, the maximum concentration of NaOH is formed in carbonic acid. You should absolutely not drink or smell this solution, and you should also not stick your fingers in (the longer the electrolysis, the more it burns). You can clean pipes with it, while understanding its high chemical activity: if the pipes are plastic, you can hold them for 2 hours, but if they are metal (grounded, by the way), the pipes will begin to eat: Fe + 2NaOH + 2H 2 O = Na 2 + H 2 , Fe + H 2 CO 3 = FeCO 3 + H 2.

This is the first of the possible causes of suffocating “gas”, a physical and chemical process: saturation of the air with a solution of concentrated caustic soda in carbonic acid (boiling bubbles of oxygen and hydrogen as carriers). In books of the 19th century, carbonic acid is used as a poisonous substance (in large quantities). This is why drivers who install a battery inside a car are damaged by sulfuric acid (essentially the same electrolysis): during the process of overcurrent on a heavily discharged battery (the car has no current limit), the electrolyte boils for a short time, the sulfuric acid comes out along with oxygen and hydrogen into the cabin. If the room is made completely sealed, due to the oxygen-hydrogen mixture (explosive gas), you can get a good bang with the destruction of the room. The video shows bang in miniature: water under the influence of molten copper decomposes into hydrogen and oxygen, and metal is more than 1100 degrees (I can imagine how a room completely filled with it stinks)... About the symptoms of inhaling NaOH: caustic, burning sensation, sore throat, cough, difficulty breathing, shortness of breath ; symptoms may be delayed. It feels quite suitable.
...at the same time, Vladimir Vernadsky writes that life on Earth is impossible without carbonic acid dissolved in water.

We replace the cathode with a rusty piece of iron. A whole series of funny chemical reactions begins (and here it is, borscht!):
- rust Fe(OH) 3 and Fe(OH) 2, as bases, begin to react with carbonic acid (released at the cathode), producing siderite (red-brown): 2Fe(OH) 3 + 3H 2 CO 3 = 6H 2 O + Fe 2 (CO3) 3, Fe(OH) 2 + H 2 CO 3 = FeCO 3 + 2 (H 2 O). Iron oxides do not participate in the reaction with carbonic acid, because there is no strong heat, and the acid is weak. Also, electrolysis does not reduce iron at the cathode, because these bases are not a solution, and the anode is not iron;
- caustic soda, as a base, does not react with bases. Necessary conditions for Fe(OH) 2 (amphoteric hydroxide): NaOH>50% + boiling in a nitrogen atmosphere (Fe(OH) 2 + 2NaOH = Na2). Necessary conditions for Fe(OH) 3 (amphoteric hydroxide): fusion (Fe(OH) 3 + NaOH = NaFeO 2 + 2H 2 O). Necessary conditions for FeO: 400-500 degrees (FeO+4NaOH=2H 2 O+Na 4 FeO 3). Or maybe there is a reaction with FeO? FeO + 4NaOH = Na 4 FeO 3 + 2H 2 O - but only at a temperature of 400-500 degrees. Okay, maybe the sodium hydroxide removes some of the iron and the rust just falls off? But here’s a bummer: Fe + 2NaOH + 2H 2 O = Na 2 + H 2 - but when boiling in a nitrogen atmosphere. Why the hell does a caustic soda solution without electrolysis remove rust? But it doesn’t remove it in any way (I drained the clear solution of caustic soda from the Auchan). It removes grease, and in my case, with a piece of Matiz, it dissolved the paint and primer (the resistance of the primer to NaOH is in its performance characteristics) - which exposed a clean iron surface, the rust simply disappeared. Conclusion: soda ash is needed only to produce acid by electrolysis, which cleans the metal, taking on rust at an accelerated pace; Sodium hydroxide seems to be of no use (but will react with debris in the cathode, cleaning it).

About third-party substances after electrolysis:
- the solution changed its color and became “dirty”: with reacted bases Fe(OH) 3, Fe(OH) 2;
- black plaque on the iron. First thought: iron carbide Fe 3 C (triiron carbide, cementite), insoluble in acids and oxygen. But the conditions are not the same: to obtain it you need to apply a temperature of 2000 degrees; and in chemical reactions there is no free carbon to join the iron. Second thought: one of the iron hydrides (saturation of iron with hydrogen) - but this is also incorrect: the conditions for obtaining are not the same. And then it came: iron oxide FeO, the basic oxide does not react with either acid or sodium hydroxide; as well as Fe 2 O 3. And amphoteric hydroxides are located in layers above the main oxides, protecting the metal from further penetration of oxygen (they do not dissolve in water, preventing the access of water and air to FeO). You can put the cleaned parts in citric acid: Fe 2 O 3 + C 6 H 8 O 7 = 2FeO + 6CO + 2H 2 O + 2H 2 (special attention to the release of carbon monoxide and the fact that the acid and metal eat on contact) - and FeO is removed with a regular brush. And if you heat the higher oxide in carbon monoxide without getting burned, it will reduce iron: Fe 2 O 3 + 3CO = 2Fe + 3CO 2 ;
- white flakes in solution: certain salts that are insoluble during electrolysis in either water or acid;
- other substances: iron is initially “dirty”, water is not initially distilled, dissolution of the anode.

The second of the possible causes of the suffocating “gas” is a physical and chemical process: iron, as a rule, is not pure - with galvanization, primer and other foreign substances; and water - with minerals, sulfates, etc. Their reaction during electrolysis is unpredictable; anything can be released into the air. However, my piece was so small (0.5x100x5), and tap water (poorly mineralized) - this reason is unlikely. Also, the idea of ​​the presence of foreign substances in the soda ash itself has disappeared: only this is indicated on the packaging.

The third possible cause of asphyxiating gas is a chemical process. If the cathode is restored, then the anode must be destroyed by oxidation, if not inert. Stainless steel contains about 18% chromium. And this chromium, when destroyed, enters the air in the form of hexavalent chromium or its oxide (CrO 3, chromic anhydride, reddish - we will talk about it later), a strong poison and carcinogen with delayed catalysis of lung cancer. Lethal dose 0.08g/kg. Ignites gasoline at room temperature. Released when welding stainless steel. The scary thing is that the symptoms are the same as sodium hydroxide when inhaled; and sodium hydroxide already seems like a harmless animal. Judging by the descriptions of cases of at least bronchial asthma, you need to work as a roofer for 9 years, breathing this poison; however, a clear delayed effect is described - that is, it can shoot both 5 and 15 years after a single poisoning.

How to check whether chromium has been released from stainless steel (where - the question remains). After the reaction, the bolt became shinier compared to the same bolt from the same batch - a bad sign. As it turned out, stainless steel is such as long as chromium oxide exists in the form of a protective coating. If chromium oxide was destroyed by oxidation during electrolysis, it means that such a bolt will rust more intensely (free iron will react, and then the chromium in the untouched stainless steel will oxidize to CrO). Therefore, I created all the conditions for the two bolts to rust: salt water and a solution temperature of 60-80 degrees. Stainless steel grade A2 12Х18Н9 (Х18Н9): it contains 17-19% chromium (and in stainless iron-nickel alloys there is even more chromium, up to ~35%). One of the bolts has rusted in several places, all places in the area of ​​contact between the stainless steel and the solution! The reddest one is along the line of contact with the solution.

And my happiness is that the current strength was then only 0.15A during electrolysis, the kitchen was closed and the window in it was open. It was clearly imprinted in my mind: exclude stainless steel from electrolysis or do it in an open area and at a distance (there is no stainless steel without chromium, this is its alloying element). Because stainless steel is NOT an inert anode during electrolysis: it dissolves and releases toxic chromium oxide; couch chemists, hit the wall before someone dies from your advice! The question remains: in what form, how much and where; but taking into account the release of pure oxygen at the anode, CrO is already oxidized to the intermediate oxide Cr 3 O 2 (also toxic, MPC 0.01 mg/m 3), and then to the higher oxide CrO 3: 2Cr 2 O 3 + 3O 2 = 4CrO3. The latter remains an assumption (the required alkaline environment is present, but is high heat required for this reaction), but it is better to be on the safe side. Even blood and urine tests for chromium are difficult to do (not included in price lists, not even included in an extended general blood test).

Inert electrode - graphite. You need to go to the trolleybus depot and remove the discarded brushes. Because even on Aliexpress it’s 250 rubles per pin. And this is the cheapest of the inert electrodes.

And here is another real example when sofa electronics led to material losses. And to the right knowledge, really. As in this article. The benefits of sofa idle talk? - unlikely, they wreak havoc; and you have to wipe up after them.

I am inclined to the first reason for the suffocating “gas”: evaporation into the air of a solution of sodium hydroxide in carbonic acid. Because with chromium oxides they use hose gas masks with mechanical air supply - I would have suffocated in my pathetic RPG-67, but it was noticeably easier to breathe in it at the very epicenter.
How to check for chromium oxide in the air? Start the process of water decomposition in a pure solution of soda ash on a graphite anode (pick it out of a pencil, but not every pencil contains a pure graphite rod) and an iron cathode. And risk breathing the air in the kitchen again in 2.5 hours. Logical? Almost: the symptoms of caustic soda and hexavalent chromium oxide are identical - the presence of caustic soda in the air will not prove the absence of hexavalent chromium vapor. However, the absence of odor without stainless steel will clearly indicate the presence of hexavalent chromium. I checked, there was a smell - a phrase with hope "hurray! I breathed caustic soda, not hexavalent chromium!" You can tell jokes.

What else did you forget:
- How do acid and alkali exist together in one vessel? In theory, salt and water should appear. There is a very subtle point here that can only be understood experimentally (I haven’t tested it). If you decompose all the water during electrolysis and isolate the solution from the salts in the sediment - option 2: what remains is either a solution of caustic soda or caustic soda with carbonic acid. If the latter is in the composition, the release of salt will begin under normal conditions and the precipitation of... soda ash: 2NaOH + H 2 CO 3 = Na 2 CO 3 + 2H 2 O. The problem is that it will dissolve in water right away - it’s a pity, you cannot taste it and compare it with the original solution: suddenly the caustic soda has not completely reacted;
- Does carbonic acid interact with iron itself? The question is serious, because... The formation of carbonic acid occurs precisely at the cathode. You can check by creating a more concentrated solution and doing electrolysis until a thin piece of metal is completely dissolved (haven’t checked). Electrolysis is a more gentle method of removing rust than acid etching;
- What are the symptoms of inhaling detonating gas? No + no smell, no color;
- Do caustic soda and carbonic acid react with plastic? Carry out identical electrolysis in plastic and glass containers and compare the turbidity of the solution and the transparency of the container surface (I did not test it on glass). Plastic - has become less transparent in places of contact with the solution. However, these turned out to be salts that could be easily removed with a finger. Therefore, food grade plastic does not react with the solution. Glass is used to store concentrated alkalis and acids.

If you have inhaled a lot of scalding gas, regardless of whether it is NaOH or CrO 3, you need to take “unithiol” or a similar drug. And the general rule applies: no matter what poisoning occurs, no matter what the strength and origin of it, drink a lot of water in the next 1-2 days, if your kidneys allow. The task: to remove the toxin from the body, and if vomiting or expectoration does not do this, give additional opportunities to the liver and urinary system to do this.

The most annoying thing is that this is all the 9th grade school curriculum. Damn, I'm 31 years old - and I won't pass the Unified State Exam...

Electrolysis is interesting because it turns back time:
- a solution of NaOH and H 2 CO 3 under normal conditions will lead to the formation of soda ash, but electrolysis inverts this reaction;
- iron is oxidized under natural conditions, but is reduced during electrolysis;
- hydrogen and oxygen tend to combine in any way: mix with air, burn and become water, be absorbed or react with something; electrolysis, on the contrary, generates gases of various substances in their pure form.
A local time machine, no less: it returns the position of the molecules of substances to their original state.

According to the reaction formulas, a solution of powdered caustic soda is more dangerous during its creation and electrolysis, but more effective in certain situations:
- for inert electrodes: NaOH + 2H 2 O = NaOH + 2H 2 + O 2 (the solution is a source of pure hydrogen and oxygen without impurities);
- reacts more intensely with organic materials, there is no carbonic acid (a quick and cheap degreaser);
- if you take iron as an anode, it will begin to dissolve at the anode and be reduced at the cathode, thickening the layer of iron on the cathode in the absence of carbonic acid. This is a way to restore the cathode material or coat it with another metal when there is no solution with the desired metal at hand. Removal of rust, according to experimenters, also goes faster if the anode is made of iron in the case of soda ash;
- but the concentration of NaOH in the air during evaporation will be higher (you still need to decide what is more dangerous: carbonic acid with caustic soda or moisture with caustic soda).

Earlier I wrote about education that a lot of time is wasted in school and university. This article does not change this opinion, because the average person will not need matan, organic chemistry or quantum physics in life (only at work, and when I needed matan 10 years later, I learned it again, I didn’t remember anything at all). But inorganic chemistry, electrical engineering, physical laws, Russian and foreign languages ​​- this is what should be a priority (we should also introduce the psychology of gender interaction and the foundations of scientific atheism). Now, I didn’t study at the Faculty of Electronics; and then bam, it happened - and I learned to use Visio, and I learned MultiSim and some of the element symbols, etc. Even if I had studied at the Faculty of Psychology, the result would have been the same: stuck in life - bit into it - figured it out. But if at school the emphasis on natural sciences and languages ​​was strengthened (and young people were explained why it was strengthened), life would be easier. Both at school and at the institute in chemistry: they talked about electrolysis (theory without practice), but not about the toxicity of the vapors.

Finally, an example of producing pure gases (using inert electrodes): 2LiCl + 2H 2 O = H 2 + Cl 2 + 2LiOH. That is, first we poison ourselves with pure chlorine, and then we explode with hydrogen (again to the issue of the safety of the released substances). If there was a solution of CuSO 4, and the cathode was iron, the metal would leave the base and leave an oxygen-containing acidic residue SO4 2-, it does not participate in the reactions. If the acidic residue did not contain oxygen, it would decompose into simple substances (as can be seen in the example of C 1 - released as Cl 2).

(added 05/24/2016) If you need to boil NaOH with rust for their mutual reaction - why not? Nitrogen in the air is 80%. The efficiency of rust removal will increase significantly, but then this process must be done outdoors.

About hydrogenation of metal (increased fragility): I did not find any formulas or adequate opinions on this topic. If possible, I will electrolyze the metal for several days, adding a reagent, and then knock with a hammer.

(added 05/27/2016) Graphite can be removed from a used salt battery. If it stubbornly resists disassembly, deform it in a vice.

(added 06/10/2016) Hydrogenation of metal: H + + e - = H adc. H ads + H ads = H 2, where ADS is adsorption. If a metal has the ability, under the necessary conditions, to dissolve hydrogen in itself (that’s the number!), then it dissolves it in itself. The conditions of occurrence for iron have not been discovered, but for steel they are described in the book by A.V. Schrader. "The influence of hydrogen on chemical and petroleum equipment." In Figure 58 p. 108 there is a graph of brand 12Х18Н10Т: at a pressure comparable to atmospheric pressure and a temperature of 300-900 degrees: 30-68 cm 3 /kg. Figure 59 shows the dependencies for other steel grades. The general formula for hydrogenation of steel: K s = K 0 e -∆H/2RT, where K 0 is the pre-exponential factor 1011 l/mol s, ∆H is the heat of dissolution of steel ~1793K), R is the universal gas constant 8.3144598 J/(mol ·K), T - medium temperature. As a result, at room temperature 300 K we have K s = 843 L/mol. The number is not correct, you need to double-check the parameters.

(added 06/12/2016) If caustic soda does not interact with metals without high temperature, it is a safe (for metal) degreaser for pallets, pots and other things (iron, copper, stainless steel - but not aluminum, Teflon, titanium, zinc).

With inspiration - clarifications. The pre-exponential factor K 0 lies in the range of 2.75-1011 l/mol s; this is not a constant value. Its calculation for stainless steel: 10 13 · C m 2/3, where C m is the atomic density of steel. Atomic density of stainless steel 8 · 10 22 at/cm 3 - K 0 = 37132710668902231139280610806.786 at/cm 3 = - and then everything is stuck.

If you look closely at Schrader’s graphs, you can make an approximate conclusion about the hydrogenation of steel in HC (a decrease in temperature by 2 times slows down the process by 1.5 times): approximately 5.93 cm 3 /kg at 18.75 degrees Celsius - but the time of penetration into the metal of such a volume is not indicated. In the book by Sukhotin A.M., Zotikov V.S. "Chemical resistance of materials. Handbook" on page 95 in table 8 indicates the effect of hydrogen on the long-term strength of steels. It makes it possible to understand that the hydrogenation of steels with hydrogen under a pressure of 150-460 atmospheres changes the long-term strength limit by a maximum of 1.5 times over a period of 1000-10000 hours. Therefore, one should not consider the hydrogenation of steels during electrolysis in HC as a destructive factor.

(added 06/17/2016) A good way to disassemble the battery: do not flatten the case, but open it up like a tulip bud. From the positive input, bend down parts of the cylinder piece by piece - the positive input is removed, the graphite rod is exposed - and smoothly unscrewed with pliers.

(added 06/22/2016) The easiest batteries to disassemble are Ashanov batteries. And then in some models there are 8 circles of plastic to fix the graphite rod - it becomes difficult to pull it out and begins to crumble.

(added 07/05/2016) Surprise: the graphite rod breaks down much faster than the metal anode: literally in a few hours. Using stainless steel as an anode is the optimal solution if you forget about toxicity. The conclusion from this whole story is simple: electrolysis should be carried out only in the open air. If there is an open balcony in this role, do not open the windows, but pass the wires through the rubber door seal (just press the wires with the door). Taking into account the current during electrolysis up to 8A (Internet opinion) and up to 1.5A (my experience), as well as the maximum voltage of the PSU PC 24V, the wire should be rated at 24V/11A - this is any insulated wire with a cross-section of 0.5mm 2.

Now about iron oxide on an already processed part. There are parts that are difficult to reach into to wipe off black deposits (or an object under restoration, when you cannot rub the surface with an iron brush). While analyzing chemical processes, I came across a method for removing it with citric acid and tried it out. Indeed, it also works with FeO - the plaque disappeared/crumbled within 4 hours at room temperature, and the solution turned green. But this method is considered less gentle, because acid and metal eat away (cannot be overexposed, constant monitoring). Plus, a final rinse with a soda solution is required: otherwise the remaining acid will eat away at the metal in the air, and you will get an undesirable coating (an awl for soap). And you need to be careful: if as much as 6CO is released with Fe 2 O 3, then what is released with FeO is difficult to predict (an organic acid). It is assumed that FeO + C 6 H 8 O 7 = H 2 O + FeC 6 H 6 O 7 (formation of iron citrate) - but I also release gas (3Fe + 2C 6 H 8 O 7 → Fe 3 (C 6 H 5 O 7) 2 + 3H 2). They also write that citric acid decomposes in light and temperature - I can’t find a correct reaction.

(added 07/06/2016) I tried citric acid on a thick layer of rust on nails - it dissolved in 29 hours. As I expected: citric acid is suitable specifically for post-purification of metal. To clean thick rust: use a high concentration of citric acid, high temperature (up to boiling), frequent stirring - to speed up the process, which is inconvenient.

In practice, a soda ash solution after electrolysis is difficult to regenerate. It’s not clear: add water or add soda. Adding table salt as a catalyst completely killed the solution + the graphite anode collapsed in just an hour.

Total: coarse rust is removed by electrolysis, FeO is etched with citric acid, the part is washed with a soda solution - and almost pure iron is obtained. Gas when reacting with citric acid - CO 2 (decarboxylation of citric acid), a darkish coating on iron - iron citrate (it is easy to clean off, does not perform any protective functions, soluble in warm water).

In theory, these methods of removing oxides are ideal for restoring coins. Unless weaker proportions of reagents are needed for lower solution concentrations and lower currents.

(added 07/09/2016) Conducted experiments with graphite. It is during the electrolysis of soda ash that it is destroyed extremely quickly. Graphite is carbon; when dissolved at the time of electrolysis, it can react with steel and precipitate iron carbide Fe 3 C. The condition of 2000 degrees is not met, but electrolysis is not NU.

(added 07/10/2016) When electrolyzing soda ash using graphite rods, the voltage should not be increased above 12V. A lower value may be needed - keep an eye on the graphite breakdown time at your voltage.

(added 07/17/2016) Discovered a method for local rust removal.

(added 07/25/2016) Instead of citric acid, you can use oxalic acid.

(added 07/29/2016) Steel grades A2, A4 and others are written in English letters: imported and from the word “austenitic”.

(added 10/11/2016) It turns out that there is 1 more type of rust: iron metahydroxide FeO(OH). Formed when iron is buried in the ground; in the Caucasus they used this method of rusting strip iron to saturate it with carbon. After 10-15 years, the resulting high-carbon steel became sabers.

Creating iron (read cast iron and steel) by electrolysis rather than conventional smelting could prevent the release of a billion tons of carbon dioxide into the atmosphere every year. So says Donald Sadoway from the Massachusetts Institute of Technology (MIT), who developed and tested a “green” method of producing iron by electrolysis of its oxides.

If the process demonstrated in the laboratory were scaled up, it could eliminate the need for conventional smelting, which releases nearly a ton of carbon dioxide into the atmosphere for every ton of steel produced.

In conventional technology, iron ore is combined with coke. The coke reacts with the iron, producing CO 2 and carbon monoxide, leaving an iron-carbon alloy called cast iron, which can then be smelted into steel.

In Sadoway's method, iron ore is mixed with a solvent - silica and quicklime - at a temperature of 1600 degrees Celsius - and an electric current is passed through the mixture.

Negatively charged oxygen ions migrate to the positively charged anode, from where the oxygen escapes. Positively charged iron ions migrate to the negatively charged cathode, where they are reduced to iron, which is collected at the base of the cell and pumped out.

A similar process is used in the production of aluminum (and requires a fair amount of electricity), the oxide of which is so stable that it cannot actually be reduced by carbon in a blast furnace, which, for example, produces cast iron. And it is clear that the steel industry has never had any reason to switch to electrolysis of iron ore, since it is easily reduced by carbon.

But if governments around the world begin to impose higher taxes on greenhouse gas emissions—carbon dioxide in particular—then a new method of producing cast iron could become more attractive. True, from laboratory installations of this kind to industrial installations, as scientists estimate, it will take 10-15 years.

The author of the paper says the biggest obstacle is finding a practical material for the anode. In his experiments, he used an anode made of graphite. But, unfortunately, the carbon reacts with oxygen, releasing as much carbon dioxide into the air as during normal iron smelting.

Ideal platinum anodes, for example, are too expensive for large-scale production. But there may be a way out - in the selection of some resistant metal alloys that form an oxide film on their outer surface, but still conduct electricity. Conductive ceramics can also be used.

Another problem is that the new process uses a lot of electricity - approximately 2 thousand kilowatt-hours per ton of iron produced. So, economic and even environmental sense in a new method of producing cast iron will appear only on the condition that this electricity will be generated in some environmentally friendly, and at the same time cheap, way, without emitting carbon dioxide. The author of the method himself admits this.

Industrial grades of technically pure iron (Armco type), obtained by pyrometallurgical method, have a purity degree of 99.75-99.85%. Further removal of mainly non-metallic impurities (C, O, S, P, N) contained in this iron is possible by special remelting in a high vacuum or annealing in an atmosphere of dry hydrogen. However, even after such treatment, the impurity content reaches 2000-1500 ppm iron, with the main impurities being C, P, S, Mn and O.
Iron of a higher degree of purity is obtained by electrolytic and chemical methods, but it also requires additional complex purification.
By electrolytic methods, iron is obtained from moderately concentrated or concentrated solutions of ferrous chloride or sulfate, respectively, at low current densities and room temperatures or high densities and temperatures of the order of 100°.
According to one of the methods, iron was precipitated from a solution of the following composition, g/l: 45-60 Fe2+ (in the form of FeCl2), 5-10 BaCl2 and 15 NaHCOs. Armco iron or Ural roofing iron plates were used as anodes, and pure aluminum as cathodes. Electrolysis was carried out at room temperature and a current density of 0.1 A/dm2. A precipitate with a coarse-crystalline structure was obtained, containing about 0.01% C, traces of phosphorus and free of sulfur.
The purity of electrolytic iron depends on the purity of the electrolyte and the purity of the anode metal. During precipitation, impurities more noble than iron, such as tin, zinc, and copper, can be removed. Nickel, cobalt, and manganese cannot be removed. The total content of impurities in electrolytic iron is approximately the same as in commercially pure iron. It usually contains a significant amount of oxygen (up to 0.1-0 2%), as well as sulfur (0.015-0.05%), if precipitation was carried out from sulfate baths.
Removal of oxygen from electrolytic iron is carried out by reduction processes: treatment of liquid or solid metal with hydrogen or deoxidation of the melt in a vacuum with carbon. By annealing in a stream of dry hydrogen at 900-1400° it is possible to reduce the oxygen content to 0.003%.
To obtain high-purity iron on a semi-industrial scale, a method of reduction with hydrogen in a vacuum melting plant is used. The electrolytic iron is first desulfurized by manganese doping in a lime-fluorspar crucible under a carbon monoxide atmosphere (sulfur content reduced from 0.01 to 0.004%), then the melt is reduced with hydrogen by blowing or purging in an alumina crucible. At the same time, it was possible to reduce the oxygen content to 0.004-0.001%. Metal desulfurization can also be carried out in high vacuum using additives to the melt of metals (tin, antimony, bismuth) that form volatile sulfides. By deoxidizing the melt with carbon in high-vacuum furnaces, it is possible to obtain iron with an oxygen and carbon content of up to 0.002% each.
The production of iron with a lower oxygen content by deoxidation in a high vacuum is complicated by the interaction of the metal with the crucible material, which is accompanied by the transition of oxygen into the metal. The best crucible materials that ensure minimal oxygen transfer are ZrO2 and ThO2.
High-purity iron is also obtained by the carbonyl method from pentacarbonyl Fe(CO)5 by decomposing it at 200-300°. Carbonyl iron does not contain the impurities that usually accompany iron - sulfur, phosphorus, copper, manganese, nickel, cobalt, chromium, molybdenum, zinc, silicon. Specific impurities in it are carbon and oxygen. The presence of oxygen is due to secondary reactions between the resulting carbon dioxide and iron. Carbon content reaches 1%; it can be reduced to 0.03% by adding a small amount of ammonia to the iron carbonyl vapor or treating the iron powder in hydrogen. Removal of carbon and oxygen is achieved by the same vacuum smelting methods used for electrolytic iron.
The purest iron can be obtained chemically, but this method is very complicated and makes it possible to obtain the metal in small quantities. In chemical methods for purifying iron salts from impurities Co, Ni, Cu, Cr, Mn, recrystallization, precipitation reactions, or extraction of impurities by precipitation are used.
One of the chemical methods that makes it possible to obtain iron of a very high degree of purity (less than 30-60 parts per million of impurities) includes the following sequential stages:
1) extraction of the FeCl3 complex with ether from a 6-N HCl solution with regeneration of the aqueous solution and subsequent extraction of the ether;
2) reduction of FeCls to FeCl2 with high-purity iron;
3) additional purification of FeCl2 from copper by treatment with a sulfur reagent and then with ether;
4) electrolytic deposition of metal from FeCl2 solution;
5) annealing of metal grains in hydrogen (to remove oxygen and carbon);
6) production of compact iron using powder metallurgy (pressing into rods and sintering in hydrogen)
The last stage can be carried out by crucibleless zone melting, which eliminates the disadvantage of vacuum processing - the transfer of oxygen from the crucible to the metal.

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When electric current passes through metals (conductors of the 1st kind), chemical reactions do not occur, and the metals remain unchanged. If an electric current passes through a melt or solution of an electrolyte (conductors of the 2nd type), various chemical reactions (electrolysis) occur at the boundary between the electrolyte and the metal conductor (electrode) and new compounds are formed.

Electrolysis is a set of processes that occur when an electric current passes through an electrochemical system consisting of two electrodes and a melt or electrolyte solution.

During electrolysis, cations move to the negative electrode (cathode) and anions move to the positive electrode (anode). In this case, however, the cations and anions of the electrolyte are not always discharged, accepting or donating electrons. Electrolysis reactions often involve a solvent-electrolyte, such as water.

The fundamental difference between reactions in a galvanic cell and an electrolyzer lies only in their direction and spontaneity. In a closed circuit of a galvanic cell, the electrochemical reaction occurs spontaneously, but in an electrolyzer it occurs only under the influence of electric current from an external source.

You should pay attention to the name of the electrodes: in a galvanic cell, the negative electrode is the anode, and the positive electrode is the cathode; in an electrolyzer, on the contrary, the negative electrode is the cathode, and the positive electrode is the anode.

It should be remembered that the terms “negative” and “positive” always refer to the poles of the current source, which is how they refer to the electrolyzer electrodes. What these processes have in common is that both in the galvanic cell and in the electrolyzer, an excess of electrons is created on the negative electrode, and a deficiency of electrons on the positive electrode. At the cathode, ions or molecules are reduced under the action of electrons; at the anode, particles are oxidized, giving up their electrons to the electrode.

In the electrolyzer cations (M n+) move to the cathode (–), and anions (A n–) - to the anode (+).

Decomposition voltage electrolyte during electrolysis is called the minimum voltage (external emf) that must be applied to the electrodes. For example, for a solution of zinc chloride under standard conditions:

Zn 2+ + 2 ē = Zn φ° = – 0.76 V,

Cl2+2 ē = 2Cl – φ° = + 1.36 V,

and the decomposition voltage is equal (in absolute value) to the sum of the standard electrode potentials of both electrodes: 0.76 + 1.36 = 2.12 V, i.e. the decomposition voltage cannot be lower than the emf of the corresponding galvanic cell.

Decomposition voltage is made up of the potentials of two electrodes - the ion discharge potentials.

Vacuum potential cation is sometimes called deposition potential metal This is the minimum potential that must be applied to the electrode in order for the cation to lose charge and metal deposition to occur. For some ions (Cu 2+, Ag +, Cd 2+) the deposition potential is close to the electrode potential, while for other ions (Fe 2+, Co 2+, Ni 2+) the deposition potentials significantly exceed the electrode potentials of metals - for electrolysis it is necessary certain overvoltage.

Distinguish electrolysis of solutions And electrolysis of melts. Electrolysis of solutions is divided into electrolysis with inert electrodes And electrolysis with soluble anode. Metal (Pt, Au) and non-metallic (graphite) electrodes can be inert. Anodes made of Cr, Ni, Cd, Zn, Ag, Cu, etc. are used as soluble anodes.

Some metals are practically insoluble due to high anodic polarization, for example Ni and Fe in an alkaline solution, Pb in H 2 SO 4.

Electrolysis of solutions with inert electrodes. During the electrolysis of aqueous solutions of electrolytes, hydrogen, rather than metal, is often released at the cathode instead of metal. In acidic environments, hydrogen is formed by the reaction:

2H + + 2 ē = H2.

In neutral and alkaline environments, hydrogen is formed by a reaction involving water molecules:

2H2O+2 ē = H 2 + OH – .

Cations such as Na + or K + are not discharged at all in an aqueous solution, but hydrogen is released.

Cations can be grouped according to their ability to discharge, ranging from non-dischargeable to easily dischargeable. At the same time, the electrolysis products also change. For some cations, simultaneous formation of metal and hydrogen is possible.

Below are the cations in order of decreasing difficulty of their discharge and the products of electrolysis:

Cations Products of electrolysis

Li + , K + , Na + , Mg 2+ , Al 3+ , H + (redirect) H 2

Mn 2+ , Zn 2+ , Cr 3 + , Fe 2 + , H + (pH 7) M + H 2

Co 2+, Ni 2+, Sr 2+, Pb 2+, H + (pH 0) M + H 2

Cu 2+ , Ag + , Au 3 + M

The different positions of hydrogen in this series are explained by the following reasons. The position of hydrogen between lead and copper corresponds to the numerical values ​​of standard electrode potentials at WITH M n+ = WITH H + = 1 mol/l, i.e. at pH=0. The position of hydrogen between iron and cobalt corresponds to the electrode potential of hydrogen in water at pH = 7 (φº H 2 / H + = –0.414 V). Under these conditions, all metals can be precipitated from solutions, the value of φ ° which are greater than –0.414 V. However, in practice, in addition to cobalt, nickel, tin and lead, it is also possible to precipitate zinc, chromium and iron from aqueous solutions. This is explained by the fact that the release of hydrogen gas at the cathode is hampered by hydrogen overvoltage.

Thus, in the series of cations from Li + up to A1 3+ no metal is formed, and during electrolysis hydrogen is released due to the reduction of water. In the series of cations from Mn 2+ to Pb 2+ During electrolysis, metal and hydrogen are formed simultaneously, and, finally, in the series Cu 2+ - Au 3+ only metal is formed.

Consequently, the further to the left (closer to the beginning) a metal is in the series of standard electrode potentials (voltage series), the more difficult it is to isolate this metal by electrolysis of an aqueous solution.

If a gradually increasing voltage is applied to a solution containing several cations, electrolysis begins when the deposition potential of the cation with the highest electrode potential (most positive) is reached. During electrolysis of a solution containing zinc ions (φ °= –0.76 V) and copper (φ ° = +0.34 V), copper is first released at the cathode, and only after almost all Cu 2+ ions are discharged does zinc begin to be released. Thus, if the solution simultaneously contains various cations, then during electrolysis they can be isolated sequentially in accordance with the values ​​of their electrode potentials. It is assumed that the overvoltage of metal release for them is approximately the same (and small).

Concerning anion discharge potentials, then the picture here is much more complicated due to the ability of water to participate in the electrolysis process. In general, we can say that the anions with the lowest potential (least positive) are discharged at the anode first. If the solution contains ions Cl - (φº = 1.36 V), Br - (φ° = 1.09 V) and I - (φº = 0.54 V), then iodine will be formed first, then bromine and, finally, chlorine. Fluoride ions in an aqueous solution cannot be discharged at all (φ ° = 2.87 V).

Most oxygen-containing anions (except for acetate ion) are not discharged in an aqueous solution; instead, water decomposes in acidic and neutral solutions:

2H 2 O – 4 ē = O 2 + 4H + ,

and in alkaline solutions - discharge of hydroxide ions:

2OH – – 2 ē = 1/2 O 2 + H 2 O.

Anions, according to their ability to be discharged during the electrolysis of aqueous solutions, are located in the following row from anions of oxygen-containing acids such as SO 4 2–, NO 3 – that are not discharged in an aqueous solution to those that are easily discharged:

Anions Products of electrolysis

SO 4 2–, NO 3 – etc., OH – O 2

Cl – , Br – , I – Cl 2 (ClO – , ClO 3 –), Br 2 , I 2 (+O 2)

S 2– S, SO 2 (+ O 2)

Thus, we can formulate the following main rules for electrolysis of aqueous solutions of electrolytes with insoluble electrodes:

1. Of the anions of electrolytes, the anions of oxygen-free acids (Cl–, Br–, S2–, etc.) are first discharged at the anode.

2. Anions of oxygen-containing acids (SO 4 2–, NO 3 –, CO 3 2–, etc.) are not discharged in the presence of water; instead, water is oxidized according to the reaction:

2H 2 O – 4 ē = O 2 + 4H + .

3. Active metals located in the voltage range up to Al (inclusive) are not reduced at the cathode; water is reduced instead:

2H2O+2 ē = H 2 + 2OH – .

4. Metals located in the voltage series after aluminum, but before hydrogen, are reduced at the cathode along with water molecules:

K: 1) Zn 2+ + 2 ē = Zn

2) 2H 2 O + 2 ē = H 2 + 2OH – .

5. Metals with a positive electrode potential are reduced at the cathode first:

Cu 2+ + 2 ē = Cu

For example, during the electrolysis of sulfuric acid (graphite electrodes), the following processes occur:

at the cathode 2H + + 2 ē = H 2,

at the anode 2H 2 O – 4 ē = O 2 + H + .

Summary equation:

2H 2 O = 2H 2 + O 2,

those. During the electrolysis of a sulfuric acid solution, hydrogen and oxygen are released due to the decomposition of water molecules. Electrolysis products: hydrogen and oxygen.

Electrolysis of copper sulfate solution:

at the cathodeСu 2 + + 2 ē = Cu,

at the anode 2H 2 O – 4 ē = O 2 + 4H +

Summary equation:

2Cu 2+ + 2H 2 O = 2Cu + O 2 + 4H +

2CuSO 4 + 2H 2 O = 2Сu + O 2 + 2H 2 SO 4.

Electrolysis products: copper, oxygen, sulfuric acid.

The possibility of anion discharge depends on its concentration. Thus, the electrolysis products of concentrated and dilute NaCl solutions are chlorine and oxygen, respectively.

Electrolysis of a dilute solution of sodium chloride occurs without discharge of Cl – ions (and, accordingly, Na + ions), i.e. water decomposition occurs. As the salt concentration at the anode increases, together with oxygen, the release of chlorine begins, and chlorine (with an admixture of oxygen) is formed in concentrated solutions:

at the cathode 2H2O+2 ē = H 2 + 2OH –

at the anode 2Сl – ​​– 2 ē = Cl2 .

Summary equation:

2Cl – + 2H 2 O = H 2 + Cl 2 + 2OH –

2NaCl + 2H 2 O = H 2 + Cl 2 + 2NaOH.

Electrolysis products: hydrogen, chlorine and sodium hydroxide.

In the case of the release of chlorine during the electrolysis of chloride solutions, the main process of chlorine formation is superimposed on the reactions of interaction of chlorine with water (hydrolysis) and subsequent transformations of the resulting substances. Hydrolysis of chlorine occurs with the formation of weak hypochlorous acid and chloride ions (hydrochloric acid):

Cl 2 + H 2 O = H + + Cl – + HC1O.

Hypochlorous acid with the alkali formed during electrolysis (more precisely, Na + +OH –) gives sodium hypochlorite NaClO as a product. In an alkaline environment, the overall reaction equation has the form:

Cl 2 + 2NaOH = NaCl + NaClO + H 2 O.

At elevated temperatures (boiling water), chlorine hydrolysis occurs with the formation of chlorate ion. Possible reaction equations:

3Cl 2 + 3H 2 O = ClO 3 – + 5 Cl – + 6H +,

3HClO = СlO 3 – + 2Сl – ​​+ 3Н +,

3СlО – = СlO 3 – + 2Сl – ​​.

In an alkaline environment, the overall equation has the form

3Cl 2 + 6NaOH = NaClO 3 + 5NaCl + 3H 2 O.

Electrolysis with diaphragm. During electrolysis of a dilute solution of sodium chloride, Na + ions move to the cathode, but hydrogen is released:

2H 2 O+2 ē = H 2 + OH –

and the sodium hydroxide solution is concentrated.

Chloride ions move to the anode, but due to their low concentration, oxygen, rather than chlorine, is mainly formed:

2H 2 O – 4 ē = O 2 + 4H +

and the hydrochloric acid solution is concentrated.

If electrolysis is carried out in a beaker or other similar vessel, the alkali and acid solutions are mixed and electrolysis is reduced to the formation of hydrogen and oxygen due to the decomposition of water. If the anode and cathode spaces are separated by a partition (diaphragm) that allows current-carrying ions to pass through but prevents the mixing of near-electrode solutions, then acid and alkali solutions can be obtained as electrolysis products.

During the electrolysis of a sodium chloride solution, hydroxide ions formed at the cathode by the reaction:

2H2O+2 ē = H 2 + 2OH –

immediately begin to participate in the transfer of electricity and, together with C1 ions, move to the anode, where both ions are discharged and a mixture of oxygen and chlorine is formed. Therefore, the chlorine yield decreases. If the anode is made of coal (graphite), then it is oxidized by oxygen and carbon oxides CO and CO 2 are formed, which pollute chlorine. Next, the chlorine formed at the anode interacts with hydroxide ions:

C1 2 + OH – = H + + Cl – + OCl – .

The formation of hypochlorite ions is also an undesirable process (if obtaining a sodium hypochlorite solution is not the goal). All these undesirable consequences can be avoided if you use a diaphragm that separates the cathode and anode spaces and retains OH - ions, but allows Cl - ions to pass through. Finally, the diaphragm prevents the diffusion of gases and allows for purer hydrogen to be produced.

If the solution contains several anions, it is more difficult to predict the sequence of their discharge at the anode than cations, but, generally speaking, the rule is that the anion characterized by the lowest potential value (or the highest negative value of the electrode potential of the reaction taking place at the anode) is discharged first. anode).

Electrolysis of solutions with a soluble anode. Electrolysis with a soluble anode is possible when the metal gives up electrons more easily than Cl–, OH– ions or water molecules. For example, on a copper anode in a solution of copper chloride or sulfate, chlorine or oxygen are not released, but Cu 2+ ions pass into the solution. At the same time, the same ions are discharged at the cathode and metallic copper is deposited. Thus, electrolysis with a soluble anode is reduced to the transfer of copper from the anode to the cathode.

The reaction at the anode is in most cases complicated by numerous side and often undesirable processes. For example, the resulting ions can form oxides, hydroxides and their films:

M 2+ + 2OH – = MO + H 2 O.

Oxygen may also be released at the anode:

2H 2 O – 4 ē = O 2 + 4H + ,

which can participate in a wide variety of reactions in the electrolytic system.

When gaseous products, especially oxygen, are formed, in most cases the decomposition potentials do not correspond to the electrode potentials due to high overvoltage values . Overvoltage is the difference between the actual decomposition voltage and the EMF of the corresponding reaction theoretically calculated from the electrode potentials. The nature of the released substance (for chlorine, bromine and iodine, the overvoltage is very small) and the material of the electrode have a particularly strong influence on the magnitude of the overvoltage. Below are data on overvoltage during the evolution of hydrogen and oxygen at various cathodes and anodes.

Electrode Overvoltage, V

Hydrogen Oxygen

Pt blackened 0.00 0.2–0.3

Pt shiny 0.1 0.4–0.5

Fe 0.1–0.2 0.2–0.3

Ni 0.1–0.2 0.1–0.3

Сu 0.2 0.2–0.3

Pb 0.4–0.6 0.2–0.3

Overvoltage also depends on the shape of the electrodes, the state of their surface, current density, solution temperature, intensity of solution stirring and other factors.

The overvoltage of hydrogen on iron is ~ 0.1 V, and oxygen on the same material is ~ 0.3 V. Therefore, the overvoltage during electrolysis on iron electrodes will be 0.1 + 0.3 = 0.4 V. The sum of this value and theoretically calculated will be the minimum value of the discharge voltage of the corresponding electrolyte.

The attitude towards overvoltage is ambivalent. On the one hand, overvoltage leads to increased energy consumption; on the other hand, thanks to overvoltage, it is possible to precipitate from aqueous solutions many metals that, according to the values ​​of their standard electrode potentials, should not be precipitated. These are Fe, Pb, Sn, Ni, Co, Zn , Cr. It is thanks to the overvoltage, as well as the influence of the solution concentration on the electrode potential, that electrolytic chromium and nickel plating of iron products is possible, and even sodium can be obtained from an aqueous solution on a mercury electrode.

The discharge of Cl – ions in an aqueous solution rather than OH – in solutions with high electrolyte concentrations is also explained by oxygen overvoltage. However, this overvoltage is not enough to cause the discharge of F ions and the release of free fluorine.

The magnitude of the overvoltage is influenced by many other kinetic factors - the rate of transfer of particles to the electrodes and removal of electrolysis products, the rate of destruction of hydration and other shells of discharged ions, the rate of combination of atoms into diatomic gas molecules, etc.

Solving chemical problems
aware of Faraday's law
high school

Author's development

Among the great variety of different chemical problems, as teaching practice at school shows, the greatest difficulties are caused by problems whose solution, in addition to solid chemical knowledge, requires a good command of the physics course material. And although not every high school pays attention to solving even the simplest problems using the knowledge of two courses - chemistry and physics, problems of this type are sometimes found on entrance exams in universities where chemistry is a major discipline. Therefore, without examining problems of this type in class, a teacher may unintentionally deprive his student of the chance to enter a university to major in chemistry.
This author's development contains over twenty tasks, one way or another related to the topic “Electrolysis”. To solve problems of this type, it is necessary not only to have a good knowledge of the topic “Electrolysis” of the school chemistry course, but also to know Faraday’s law, which is studied in the school physics course.
Perhaps this selection of problems will not be of interest to absolutely all students in the class or accessible to everyone. Nevertheless, it is recommended that tasks of this type be discussed with a group of interested students in a circle or elective lesson. It can be noted with confidence that problems of this type are complicated and, at least, are not typical for a school chemistry course (we are talking about a secondary school), and therefore problems of this type can be safely included in the versions of the school or district chemistry Olympiad for the 10th or 11th grade.
Having a detailed solution for each problem makes the development a valuable tool, especially for beginning teachers. Having gone through several problems with students during an elective lesson or a club lesson, a creative teacher will certainly assign several similar problems at home and use this development in the process of checking homework, which will significantly save invaluable teacher time.

Theoretical information on the problem

Chemical reactions occurring under the influence of electric current on electrodes placed in a solution or molten electrolyte are called electrolysis. Let's look at an example.

In a glass at a temperature of about 700 ° C there is a melt of sodium chloride NaCl, electrodes are immersed in it. Before an electric current is passed through the melt, the Na + and Cl – ions move chaotically, but when an electric current is applied, the movement of these particles becomes ordered: the Na + ions rush towards the negatively charged electrode, and the Cl – ions towards the positively charged electrode.

And he– a charged atom or group of atoms that has a charge.

Cation– a positively charged ion.

Anion– negatively charged ion.

Cathode– a negatively charged electrode (positively charged ions – cations) move towards it.

Anode– a positively charged electrode (negatively charged ions – anions) move towards it.

Electrolysis of sodium chloride melt on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of sodium chloride on carbon electrodes

Total reaction:

or in molecular form:

Electrolysis of an aqueous solution of copper(II) chloride on carbon electrodes

Total reaction:

In the electrochemical series of metal activities, copper is located to the right of hydrogen, therefore copper will be reduced at the cathode, and chlorine will be oxidized at the anode.

Electrolysis of an aqueous solution of sodium sulfate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of potassium nitrate occurs similarly (platinum electrodes).

Electrolysis of an aqueous solution of zinc sulfate on graphite electrodes

Total reaction:

Electrolysis of an aqueous solution of iron(III) nitrate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of silver nitrate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of aluminum sulfate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of copper sulfate on copper electrodes - electrochemical refining

The concentration of CuSO 4 in the solution remains constant, the process comes down to the transfer of anode material to the cathode. This is the essence of the electrochemical refining process (obtaining pure metal).

When drawing up electrolysis schemes for a particular salt, you need to remember that:

– metal cations that have a higher standard electrode potential (SEP) than hydrogen (from copper to gold inclusive) are almost completely reduced at the cathode during electrolysis;

– metal cations with small SEP values ​​(from lithium to aluminum inclusive) are not reduced at the cathode, but instead water molecules are reduced to hydrogen;

– metal cations, whose SEP values ​​are less than those of hydrogen, but greater than those of aluminum (from aluminum to hydrogen), are reduced simultaneously with water during electrolysis at the cathode;

– if an aqueous solution contains a mixture of cations of various metals, for example Ag +, Cu 2+, Fe 2+, then in this mixture silver will be reduced first, then copper and iron last;

– on the insoluble anode during the electrolysis process, oxidation of anions or water molecules occurs, and the anions S 2–, I–, Br–, Cl– are easily oxidized;

– if the solution contains anions of oxygen-containing acids , , , , then water molecules are oxidized to oxygen at the anode;

– if the anode is soluble, then during electrolysis it itself undergoes oxidation, that is, it sends electrons to the external circuit: when electrons are released, the equilibrium between the electrode and the solution shifts and the anode dissolves.

If from the entire series of electrode processes we select only those that correspond to the general equation

M z+ + ze= M,

then we get metal stress range. Hydrogen is also always placed in this row, which allows you to see which metals are capable of displacing hydrogen from aqueous solutions of acids and which are not (table).

Table

Metal stress range

The equation electrode process	Standard electrode potential at 25 °C, V	The equation electrode process	Standard electrode potential at 25 °C, V
Li + + 1 e= Li 0	–3,045	Co 2+ + 2 e= Co 0	–0,277
Rb + + 1 e= Rb 0	–2,925	Ni 2+ + 2 e= Ni 0	–0,250
K + + 1 e= K 0	–2,925	Sn 2+ + 2 e= Sn 0	–0,136
Cs + + 1 e= Cs 0	–2,923	Pb 2+ + 2 e= Pb 0	–0,126
Ca 2+ + 2 e= Ca 0	–2,866	Fe 3+ + 3 e= Fe 0	–0,036
Na + + 1 e= Na 0	–2,714	2H + + 2 e=H2	0
Mg 2+ + 2 e= Mg 0	–2,363	Bi 3+ + 3 e= Bi 0	0,215
Al 3+ + 3 e= Al 0	–1,662	Cu 2+ + 2 e= Cu 0	0,337
Ti 2+ + 2 e= Ti 0	–1,628	Cu + +1 e= Cu 0	0,521
Mn 2+ + 2 e= Mn 0	–1,180	Hg 2 2+ + 2 e= 2Hg 0	0,788
Cr 2+ + 2 e= Cr 0	–0,913	Ag + + 1 e= Ag 0	0,799
Zn 2+ + 2 e= Zn 0	–0,763	Hg 2+ + 2 e= Hg 0	0,854
Cr 3+ + 3 e= Cr 0	–0,744	Pt 2+ + 2 e= Pt 0	1,2
Fe 2+ + 2 e= Fe 0	–0,440	Au 3+ + 3 e= Au 0	1,498
Cd 2+ + 2 e= Cd 0	–0,403	Au + + 1 e= Au 0	1,691

In a simpler form, the series of metal stresses can be represented as follows:

To solve most electrolysis problems, knowledge of Faraday's law is required, the formula for which is given below:

m = M I t/(z F),

Where m– mass of substance released on the electrode, F– Faraday number equal to 96,485 A s/mol, or 26.8 A h/mol, M– molar mass of the element reduced during electrolysis, t– time of the electrolysis process (in seconds), I– current strength (in amperes), z– the number of electrons participating in the process.

Problem conditions

1. What mass of nickel will be released during the electrolysis of a nickel nitrate solution for 1 hour at a current of 20 A?

2. At what current strength is it necessary to carry out the process of electrolysis of a silver nitrate solution in order to obtain 0.005 kg of pure metal within 10 hours?

3. What mass of copper will be released during the electrolysis of a melt of copper(II) chloride for 2 hours at a current of 50 A?

4. How long does it take to electrolyze an aqueous solution of zinc sulfate at a current of 120 A in order to obtain 3.5 g of zinc?

5. What mass of iron will be released during the electrolysis of a solution of iron(III) sulfate at a current of 200 A for 2 hours?

6. At what current strength is it necessary to carry out the process of electrolysis of a solution of copper(II) nitrate in order to obtain 200 g of pure metal within 15 hours?

7. How long does it take to electrolyze a melt of iron(II) chloride at a current of 30 A in order to obtain 20 g of pure iron?

8. At what current strength is it necessary to carry out the process of electrolysis of a solution of mercury(II) nitrate in order to obtain 0.5 kg of pure metal within 1.5 hours?

9. At what current strength is it necessary to carry out the process of electrolysis of molten sodium chloride in order to obtain 100 g of pure metal within 1.5 hours?

10. The potassium chloride melt was subjected to electrolysis for 2 hours at a current of 5 A. The resulting metal reacted with water weighing 2 kg. What concentration of alkali solution was obtained?

11. How many grams of a 30% hydrochloric acid solution will be required to completely react with iron obtained by electrolysis of a solution of iron(III) sulfate for 0.5 hours at current strength
10 Huh?

12. In the process of electrolysis of molten aluminum chloride, carried out for 245 minutes at a current of 15 A, pure aluminum was obtained. How many grams of iron can be obtained by the aluminothermic method by reacting a given mass of aluminum with iron(III) oxide?

13. How many milliliters of a 12% KOH solution with a density of 1.111 g/ml will be required to react with aluminum (to form potassium tetrahydroxyaluminate) obtained by electrolysis of an aluminum sulfate solution for 300 minutes at a current of 25 A?

14. How many milliliters of a 20% sulfuric acid solution with a density of 1.139 g/ml will be required to react with zinc obtained by electrolysis of a solution of zinc sulfate for 100 minutes at a current of 55 A?

15. What volume of nitrogen(IV) oxide (n.o.) will be obtained by reacting an excess of hot concentrated nitric acid with chromium obtained by electrolysis of a solution of chromium(III) sulfate for 100 min at a current of 75 A?

16. What volume of nitrogen(II) oxide (n.o.) will be obtained by the interaction of an excess nitric acid solution with copper obtained by electrolysis of a copper(II) chloride melt for 50 minutes at a current of 10.5 A?

17. How long does it take to electrolyze a melt of iron(II) chloride at a current of 30 A to obtain the iron necessary for complete reaction with 100 g of a 30% hydrochloric acid solution?

18. How long does it take to electrolyze a solution of nickel nitrate at a current of 15 A to obtain the nickel required for complete reaction with 200 g of a 35% solution of sulfuric acid when heated?

19. The sodium chloride melt was electrolyzed at a current of 20 A for 30 minutes, and the potassium chloride melt was electrolyzed for 80 minutes at a current of 18 A. Both metals were dissolved in 1 kg of water. Find the concentration of alkalis in the resulting solution.

20. Magnesium obtained by electrolysis of magnesium chloride melt for 200 minutes at current strength
10 A, dissolved in 1.5 l of 25% sulfuric acid solution with a density of 1.178 g/ml. Find the concentration of magnesium sulfate in the resulting solution.

21. Zinc obtained by electrolysis of a solution of zinc sulfate for 100 minutes at current strength

17 A, dissolved in 1 liter of 10% sulfuric acid solution with a density of 1.066 g/ml. Find the concentration of zinc sulfate in the resulting solution.

22. Iron, obtained by electrolysis of a melt of iron(III) chloride for 70 minutes at a current of 11 A, was turned into powder and immersed in 300 g of an 18% solution of copper(II) sulfate. Find the mass of copper that precipitated.

23. Magnesium obtained by electrolysis of magnesium chloride melt for 90 minutes at current strength
17 A, was immersed in a solution of hydrochloric acid taken in excess. Find the volume and amount of hydrogen released (n.s.).

24. A solution of aluminum sulfate was subjected to electrolysis for 1 hour at a current of 20 A. How many grams of a 15% solution of hydrochloric acid will be required to completely react with the resulting aluminum?

25. How many liters of oxygen and air (n.o.) will be required to completely burn magnesium obtained by electrolysis of magnesium chloride melt for 35 minutes at a current of 22 A?

For answers and solutions, see the following issues