1. Complete the reaction equations (where necessary), select the coefficients using the electronic balance method. Calculate the equivalent weight of the oxidizer.
a) Cr 2 (SO 4) 3 + KClO 3 + NaOH = KCl + ...
b) Cu 2 S + O 2 + CaCO 3 = CuO + CaSO 3 + CO 2
c) Zn + H 2 SO 4 (conc) = H 2 S + ...
d) FeS + O 2 = Fe 2 O 3 + ...
e) NaMnO 4 + HI = I 2 + NaI + ...
f) NaMnO 4 + KNO 2 + H 2 SO 4 = ...
g) KMnO 4 + S = K 2 SO 4 + MnO 2
h) Cr(OH) 3 + Ag 2 O + NaOH → Ag + ...
i) Cr(OH) 3 + Br 2 + NaOH → NaBr + ...
j) NH 3 + KMnO 4 + KOH → KNO 3 + ...
2. Complete the OVR equation, select the coefficients using the electron-ion method, calculate the molar masses of the equivalents of the oxidizing agent and reducing agent in the reaction:
a) K 2 Cr 2 O 7 + H 2 S + H 2 SO 4 → Cr 2 (SO 4) + S + ...
b) Na 3 AsO 3 + KMnO 4 +KOH → Na 3 AsO 4 + K 2 MnO 4 + ...
c) NaNO 2 + KJ + H 2 SO 4 → J 2 + NO + ...
d) KMnO 4 + H 2 O 2 + H 2 SO 4 → MnSO 4 + ...
e) H 2 O 2 + KJO 3 + H 2 SO 4 → J 2 + O 2 + ...
f) Cr 2 (SO 4) 3 + KClO 3 + NaOH → Na 2 CrO 4 + KCl + ...
g) FeCl 2 + HClO 4 + HCl → Cl 2 + ...
h) NaNO 2 + K 2 Cr 2 O 7 + H 2 SO 4 → NaNO 3 + ...
i) KMnO 4 + MnSO 4 + H 2 O → H 2 SO 4 + ...
j) KMnO 4 + HCl → Cl 2 + ...
l) KMnO 4 + H 2 SO 4 + H 2 C 2 O 4 → CO 2 + ...
m) H 2 O 2 + CrCl 3 + KOH → K 2 CrO 4 + H 2 O + ...
3. Calculate the EMF of the process and determine in which direction this OVR proceeds spontaneously:
H 2 SO 4 + 2HCl ↔ Cl 2 + H 2 SO 3 + H 2 O?
(φ o (Cl 2 /2Cl -) \u003d + 1.36V, φº (SO 4 2 - /SO 3 2 -) \u003d +0.22 V)
4. In what direction does this OVR proceed spontaneously:
CuSO 4 + Zn ↔ ZnSO 4 + Cu?
(φ o (Zn 2+ / Zn) = -0.76V, φº (Cu 2+ / Cu) = +0.34 V)
5. In what direction does this OVR proceed spontaneously:
2NaCl + Fe 2 (SO 4) 3 ↔2FeSO 4 + Cl 2 + Na 2 SO 4
φº (Cl 2 / 2Cl -) \u003d + 1.36V, φº (Fe 3+ / Fe 2+) \u003d + 0.77V.
6. In what direction does this OVR proceed spontaneously:
2KMnO 4 + 5SnSO 4 + 8H 2 SO 4 ↔ 2MnSO 4 + 5Sn(SO 4) 2 + K 2 SO 4 + 8H 2 O?
φº (MnO 4 - / Mn 2+) \u003d + 1.51V, φº (Sn 4+ / Sn 2+) \u003d + 0.15V. Justify the answer.
7. Is it possible to simultaneously inject FeSO 4 and NaNO 2 into a patient, given that the environment in the stomach is acidic?
φºFe 3+ /Fe 2+ \u003d + 0.77V, φºNO 2 ─ / NO \u003d + 0.99V. Justify the answer.
8. Determine the redox properties of H 2 O 2 that it exhibits when interacting with K 2 Cr 2 O 7 in an acidic environment. φº (O 2 / H 2 O 2) \u003d + 0.68V, φº (Cr 2 O 7 2– / 2Cr 3+) \u003d + 1.33V. Justify the answer.
9. What halogens oxidize Fe 2+ to Fe 3+? Which of the halide ions can reduce Fe 3+ ? Write the equations for the corresponding reactions. Calculate the EMF of each of the reactions and determine the sign of DG. When calculating, use the following values of redox potentials:
φºFe 3+ /Fe 2+ \u003d + 0.77V;
φº (F 2 / 2F -) \u003d + 2.87V;
φº (Cl 2 / 2Cl -) \u003d + 1.36V;
φº (Br 2 / 2Br -) \u003d + 1.07 V;
φº (I 2 / 2I -) \u003d + 0.54V.
10. How many grams of KMnO 4 must be taken to prepare 100 ml of a 0.04 N solution for titration in an acidic medium?
12. The titer of H 2 C 2 O 4 2H 2 O is 0.0069 g / ml. For titration of 30 ml of this solution, 25 ml of KMnO 4 solution are consumed. Calculate the normality of this solution.
13. 1 liter of ferrous sulfate solution contains 16 g (FeSO 4 7H 2 O). What volume of this solution can be oxidized by 25 ml of 0.1 N KMnO 4 solution in an acid medium?
In the 20th task of the OGE in chemistry, it is necessary to provide a solution in full. Solution 20 of the task - drawing up an equation for a chemical reaction using the electronic balance method.
Theory for task No. 20 OGE in chemistry
We have already talked about redox reactions in. Now we will consider the electronic balance method using a typical example, but before that we will find out what kind of method it is and how to use it.
Electronic balance method
The electron balance method is a method for equalizing chemical reactions based on changing the oxidation states of atoms in chemical compounds.
The algorithm of our actions is as follows:
- We calculate the change in the oxidation state of each element in the chemical reaction equation
- We select only those elements that have changed their oxidation state
- For the elements found, we draw up an electronic balance, which consists in counting the number of acquired or donated electrons
- Finding the Least Common Multiple of Transferred Electrons
- The values obtained are the coefficients in the equation (with rare exceptions)
Using the electronic balance method, arrange the coefficients in the reaction equation, the scheme of which
HI + H2SO4 → I2 + H2S + H2O
Determine the oxidizing agent and reducing agent.
So, we make an electronic balance. In this reaction, we change the oxidation state sulfur and iodine .
Sulfur was in the oxidation state +6, and in the products - -2. Iodine had an oxidation state of -1, and became 0.
If you have any difficulties with the calculation, then remember.
1 | S +6 + 8ē → S –2
4 | 2I –1 – 2ē → I 2
Sulfur takes 8 electrons, and iodine gives only two - a common multiple of 8, and additional factors of 1 and 4!
We arrange the coefficients in the reaction equation according to the data obtained:
8HI + H2SO4 = 4I2 + H2S + 4H2O
Do not forget to indicate that sulfur in the +6 oxidation state is oxidizing agent , a iodine in oxidation state –1 – reducing agent.
321–340 . For this reaction, select the coefficients using the electronic balance method. Specify the oxidizing agent and reducing agent.
321. KClO 3 + Na 2 SO 3 + = KCl + Na 2 SO 4.
322. Au + HNO 3 + HCl \u003d AuCl 3 + NO + H 2 O.
323. P + HNO 3 + H 2 O \u003d H 3 PO 4 + NO.
324. Cl 2 + I 2 + H 2 O \u003d HCl + HIO 3.
325. MnS + HNO 3 \u003d MnSO 4 + NO 2 + H 2 O.
326. HCl + HNO 3 \u003d Cl 2 + NO + H 2 O.
327. H 2 S + HNO 3 \u003d S + NO + H 2 O.
328. HClO 4 + SO 2 + H 2 O \u003d HCl + H 2 SO 4.
329. As + HNO 3 \u003d H 3 AsO 4 + NO 2 + H 2 O.
330. KI + KNO 2 + H 2 SO 4 \u003d I 2 + NO + K 2 SO 4 + H 2 O.
331. KNO 2 + S \u003d K 2 S + N 2 + SO 2.
332. HI + H 2 SO 4 \u003d I 2 + H 2 S + H 2 O.
333. H 2 SO 3 + H 2 S \u003d S + H 2 O.
334. H 2 SO 3 + H 2 S \u003d S + H 2 O.
335. Cr 2 (SO 4) 3 + Br 2 + KOH = K 2 CrO 4 + KBr + K 2 SO 4 + H 2 O.
336. P + H 2 SO 4 \u003d H 3 PO 4 + SO 2 + H 2 O.
337. H 2 S + Cl 2 + H 2 O \u003d H 2 SO 4 + HCl.
338. P + HIO 3 + H 2 O \u003d H 3 PO 4 + HI.
339. NaAsO 2 + I 2 + NaOH = Na 3 AsO 4 + HI.
340. K 2 Cr 2 O 7 + SnCl 2 + HCl \u003d CrCl 3 + SnCl 4 + KCl + H 2 O.
341. Make a galvanic circuit with Cu, Pb, CuCl 2 and Pb(NO 3) 2 at your disposal. Write the equations of electrode processes and calculate the EMF of this element (solution concentrations are 1 mol/l).
Answer: EMF = 0.463 V.
342. Draw a diagram of a galvanic cell consisting of iron and tin plates immersed in solutions of iron (II) and tin (II) chlorides, respectively. Write the equations of electrode processes and calculate the EMF of this element (solution concentrations are 1 mol/l).
Answer: EMF = 0.314 V.
343. The galvanic cell is composed according to the scheme: Ni | NiSO 4 (0.1 M) || AgNO 3 (0.1 M) | Ag. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF \u003d 1.019 V.
344. Draw a diagram of a galvanic cell consisting of iron and mercury plates immersed in solutions of their salts. Write the equations of electrode processes and calculate the EMF of this element (solution concentrations are 1 mol/l).
Answer: EMF \u003d 1.294 V.
345. From the four metals Ag, Cu, Al and Sn, select those pairs that give the smallest and largest EMF of the galvanic cell composed of them.
Answer: a pair of Cu and Ag has a minimum EMF,
a pair of Al and Ag - the maximum EMF.
346. Draw a diagram of two galvanic cells, in one of which the lead would be the cathode, and in the other the anode. Write the equations of electrode processes and calculate the EMF of each element.
347. Make a diagram of a galvanic cell consisting of lead and zinc plates immersed in solutions of their salts, where = = 0.01 mol / l. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.637 V.
348. Make a diagram of a galvanic cell consisting of aluminum and zinc plates immersed in solutions of their salts, where = = 0.1 mol/l. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.899 V.
349.
Answer: EMF \u003d 0.035 V.
350. Draw a diagram of a galvanic cell consisting of a zinc plate immersed in a 0.1 M solution of zinc nitrate and a lead plate immersed in a 1 M solution of lead nitrate. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.666 V.
351. Make a diagram of a galvanic cell, in which one electrode is nickel with = 0.1 mol / l, and the second is lead with = 0.0001 mol / l. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.035 V.
352. Draw a diagram of a galvanic cell consisting of a cadmium plate immersed in a 0.1 M solution of cadmium sulfate and a silver plate immersed in a 0.01 M solution of silver nitrate. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 1.113 V.
353. Make a diagram of a galvanic cell consisting of two aluminum plates dipped into solutions of its salt with a concentration of = 1 mol/l at one electrode and = 0.1 mol/l at the other electrode. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.029 V.
354. Draw a diagram of a galvanic cell consisting of two silver electrodes immersed in 0.0001 mol/l and 0.1 mol/l AgNO 3 solutions. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.563 V.
355. Write the equations of electrode processes, the total reaction and calculate the EMF of the galvanic cell Ni | NiSO 4 (0.01 M) || Cu(NO 3) 2 (0.1 M) | Cu.
Answer: EMF = 0.596 V.
356. Draw a diagram of a galvanic cell consisting of a cadmium plate immersed in a 0.1 M solution of cadmium nitrate and a silver plate immersed in a 1 M solution of silver nitrate. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 1.233 V.
357. Make a diagram of a galvanic cell consisting of two aluminum plates dipped into solutions of its salt with a concentration of = 1 mol/l at one electrode and = 0.01 mol/l at the other electrode. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.059 V.
358. Make a diagram of a galvanic cell consisting of two copper electrodes immersed in 0.001 M and 0.1 M Cu(NO 3) 2 solutions. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.059 V.
359. Make a diagram of a galvanic cell consisting of two nickel plates immersed in solutions of nickel salt with a concentration of = 1 mol/l at one electrode and = 0.01 mol/l at the other electrode. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.059 V.
360. Draw a diagram of a galvanic cell consisting of two lead electrodes immersed in 0.001 mol/l and 1 mol/l Pb(NO 3) 2 solutions. Write the equations of electrode processes and calculate the EMF of this element.
Answer: EMF = 0.088 V.
361. As a result of passing current through an aqueous solution of zinc sulfate for 5 hours, 6 liters of oxygen were released. Determine the strength of the current. Write the equations for the reactions that take place on inert electrodes during the electrolysis of ZnSO 4 .
Answer: I= 5.74A.
362. In what sequence will metal ions be discharged at the cathode during the electrolysis of melts of a mixture of salts KCl, ZnCl 2 , MgCl 2 . Explain the answer.
Answer: ZnCl 2 (D E\u003d 2.122 B), MgCl 2 (D E= 3.72 V),
KCl(D E= 4.28 V).
363. As a result of passing a current of 1.2 A through an aqueous solution of a divalent metal salt for 1 hour, 2.52 g of metal was released. Determine the atomic mass of this metal.
Answer: M(Cd) = 112.5 g/mol.
364. How many grams of copper will be released on the cathode when a current of 5 A is passed through a solution of copper sulfate for 10 minutes?
Answer: m(Cu) = 0.987 g.
365. Write the equations for the reactions that take place on inert electrodes during the electrolysis of potassium chloride, which is: a) in the melt; b) in solution.
366. During the electrolysis of a solution of copper sulfate with copper electrodes, the mass of the cathode increased by 40 g. What amount of electricity (in coulombs) was passed through the solution?
Answer: Q= 121574.8 C.
367. What mass of cadmium was released at the cathode if a current of 3.35 A was passed through a solution of cadmium sulfate for 1 hour?
Answer: m(Cd) = 7 g.
368. What mass of silver was precipitated at the cathode if an electric current of 0.67 A was passed through a solution of silver nitrate for 20 hours?
Answer: m(Ag) = 53.9 g.
369. Write the equations for the reactions that take place on the electrodes during the electrolysis of an aqueous solution of CuCl 2: a) with an inert anode; b) with a copper anode.
370. Write the equations of the reactions occurring on the electrodes during the electrolysis of an aqueous solution of Zn(NO 3) 2: a) with an inert anode; b) with zinc anode.
371. What amount of chlorine will be released at the anode as a result of passing a current of 5 A through a solution of silver chloride for 1 hour?
Answer: V(Cl 2) \u003d 2 l.
372. What amount of nickel will be released when a current of 5 A is passed through a solution of nickel nitrate for 5.37 hours? Write the equations for reactions occurring on inert electrodes.
Answer: m(Ni) = 29.4 g.
373. Electrolysis of a nickel sulfate solution releases 4.2 L of oxygen (N.O.). How many grams of nickel will be released at the cathode?
Answer: m(Ni) = 22 g.
374. What amount of electricity is required to produce 44.8 liters of hydrogen by electrolysis of an aqueous solution of potassium chloride? Write the equations for reactions occurring on inert electrodes.
Answer: Q= 386000 C.
375. Calculate the mass of silver released at the cathode when a current of 7 A is passed through a solution of silver nitrate for 30 minutes.
Answer:m(Ag) = 14 g.
376. How long does it take to completely decompose 2 moles of water with a current of 2 A?
Answer:53.6 hours
377. Find the volume of oxygen (N.O.) that is released when a current of 6 A is passed through an aqueous solution of KOH for 30 minutes.
Answer: V(O 2) = 627 ml.
378. Find the volume of hydrogen (n.o.) that will be released when a current of 3 A is passed through an aqueous solution of H 2 SO 4 for 1 hour.
Answer: V(H 2) \u003d 1.25 l.
379. During the electrolysis of an aqueous solution of SnCl 2 at the anode, 4.48 l of chlorine (n.o.) were released. Find the mass of tin released on the cathode.
Answer:m(Sn) = 23.7 g.
380. When a current of 1.5 A was passed through a solution of a salt of a trivalent metal for 30 minutes, 1.071 g of metal was released at the cathode. Calculate the atomic mass of the metal.
Answer:A r(In) = 114.8 amu
test questions
1. What is a galvanic cell? Describe the principle of its work.
2. What is the standard electrode potential?
3. What is the electromotive force of a galvanic cell? How is the EMF of a galvanic cell calculated for standard and non-standard conditions?
4. What is the difference between metal and concentration galvanic cells?
5. What processes take place during the operation of a galvanic cell consisting of iron and silver electrodes immersed in solutions of their salts?
6. Make diagrams of galvanic cells in which the mercury electrode is: a) the anode; b) cathode.
7. What is electrolysis?
8. Name the products of electrolysis of an aqueous solution of copper nitrate on an insoluble anode.
9. Define the phenomenon of overvoltage. When does it occur?
Corrosion of metals
Corrosion – this is a spontaneous process of destruction of materials and products from them as a result of the physical and chemical effects of the environment, in which the metal passes into an oxidized (ionic) state and loses its inherent properties.
Metals and alloys, coming into contact with the environment (gaseous or liquid), are subject to destruction. The rate of corrosion of metals and metal coatings in atmospheric conditions is determined by the complex effect of a number of factors: the presence of adsorbed moisture on the surface, air pollution with corrosive substances, changes in air and metal temperature, the nature of corrosion products, etc.
According to the laws of chemical thermodynamics, corrosion processes arise and proceed spontaneously only under the condition of a decrease in the Gibbs energy of the system (∆ G<0).
91.1. Classification of corrosion processes
1. By type of destruction corrosion is continuous and local. With a uniform distribution of corrosion damage, it does not pose a danger to structures and apparatus, especially in cases where the loss of metals does not exceed technically justified standards. Local corrosion is much more dangerous, although metal loss can be small. The danger lies in the fact that, by reducing the strength of individual sections, it sharply reduces the reliability of structures, structures, and apparatus.
2. According to flow conditions There are: atmospheric, gas, liquid, underground, marine, soil corrosion, stray current corrosion, stress corrosion, etc.
3 . According to the mechanism of the corrosion process distinguish chemical and electrochemical corrosion.
Chemical corrosion can occur when interacting with dry gaseous oxidizers and solutions of non-electrolytes. Most metals interact with gases at elevated temperatures. At the same time, two processes take place on the surface: the oxidation of the metal and the accumulation of oxidation products, which sometimes prevent further corrosion. In general, the reaction equation for the oxidation of metals with oxygen is as follows:
x M+ y/2 O 2 \u003d M x O y. (1)
The Gibbs energy of metal oxidation is equal to the Gibbs energy of oxide formation, since ∆ G formation of simple substances is equal to 0. For the oxidation reaction (1) it is equal to
∆G=∆G 0-ln p O 2 ,
where ∆ G 0 is the standard Gibbs energy of the reaction; p O 2 is the relative pressure of oxygen.
Methods of protection against gas corrosion: alloying metals, creating protective coatings on the surface and changing the properties of the gaseous medium.
Electrochemical corrosion of metals develops upon contact of metal with electrolyte solutions (all cases of corrosion in aqueous solutions, since even pure water is a weak electrolyte, and sea water is strong). The main oxidizing agents are water, dissolved oxygen and hydrogen ions.
Cause of electrochemical corrosion consists in the fact that the metal surface is always energetically inhomogeneous due to the presence of impurities in metals, differences in the chemical and phase composition of the alloy, etc. This leads to the formation of microgalvanic cells on the surface in a humid atmosphere. In areas of the metal with a more negative potential value, the process of oxidation of this metal occurs:
M 0 + ne– =M n+ (anodic process).
Oxidizers that accept electrons at the cathode are called cathode depolarizers. Cathodic depolarizers are: hydrogen ions (hydrogen depolarization), oxygen molecules (oxygen depolarization).
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