So2 i2 h2o h2so4 hi electronic balance

1. Complete the reaction equations (where necessary), select the coefficients using the electronic balance method. Calculate the equivalent mass of the oxidizing agent.

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 + ...

e) 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 ORR 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 +…

e) 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 ORR occurs spontaneously:

H 2 SO 4 +2HCl ↔ Cl 2 +H 2 SO 3 +H 2 O?

(φ o (Cl 2 /2Cl ―)=+1.36V, φº(SO 4 2― /SO 3 2―) = +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 –)=+1.36V, φº(Fe 3+ /Fe 2+)=+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+)=+1.51V, φº(Sn 4+ /Sn 2+)=+0.15V. Justify your answer.

7. Is it permissible to simultaneously administer FeSO 4 and NaNO 2 orally to a patient, given that the environment in the stomach is acidic?

φºFe 3+ /Fe 2+ =+0.77V, φºNO 2 ─ /NO=+0.99V. Justify your answer.

8. Determine the redox properties of H 2 O 2, which it exhibits when interacting with K 2 Cr 2 O 7 in acidic environment. φº(O 2 /H 2 O 2) = +0.68V, φº(Cr 2 O 7 2– /2Cr 3+) = +1.33V. Justify your answer.

9. What halogens oxidize Fe 2+ to Fe 3+? Which halide ions can reduce Fe 3+? Write the equations for the corresponding reactions. Calculate the emf of each reaction and determine the sign of DG. When calculating, use the following values ​​of redox potentials:

φºFe 3+ /Fe 2+ =+0.77V;

φº(F 2 /2F –)=+2.87V;

φº(Cl 2 /2Cl –)=+1.36V;

φº(Br 2 /2Br –)=+1.07V;

φº(I 2 /2I –)=+0.54V.

10. How many grams of KMnO 4 are needed to prepare 100 ml of 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. To titrate 30 ml of this solution, 25 ml of KMnO 4 solution is 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 solution of KMnO 4 in an acidic medium?

321–340 . For this reaction, select the coefficients using the electronic balance method. Specify the oxidizing agent and the reducing agent.

321. KClO 3 + Na 2 SO 3 + = KCl + Na 2 SO 4.

322. Au + HNO 3 + HCl = AuCl 3 + NO + H 2 O.

323. P + HNO 3 + H 2 O = H 3 PO 4 + NO.

324. Cl 2 + I 2 + H 2 O = HCl + HIO 3.

325. MnS + HNO 3 = MnSO 4 + NO 2 + H 2 O.

326. HCl + HNO 3 = Cl 2 + NO + H 2 O.

327. H 2 S + HNO 3 = S + NO + H 2 O.

328. HClO 4 + SO 2 + H 2 O = HCl + H 2 SO 4.

329. As + HNO 3 = H 3 AsO 4 + NO 2 + H 2 O.

330. KI + KNO 2 + H 2 SO 4 = I 2 + NO + K 2 SO 4 + H 2 O.

331. KNO 2 + S = K 2 S + N 2 + SO 2.

332. HI + H 2 SO 4 = I 2 + H 2 S + H 2 O.

333. H 2 SO 3 + H 2 S = S + H 2 O.

334. H 2 SO 3 + H 2 S = 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 = H 3 PO 4 + SO 2 + H 2 O.

337. H 2 S + Cl 2 + H 2 O = H 2 SO 4 + HCl.

338. P + HIO 3 + H 2 O = 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 = CrCl 3 + SnCl 4 + KCl + H 2 O.

341. Make a galvanic circuit using Cu, Pb, CuCl 2 and Pb(NO 3) 2. Write the equations 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 =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 =1.294 V.

345. From the four metals Ag, Cu, Al and Sn, select those pairs that give the lowest and highest emf of the galvanic cell composed of them.

Answer: a pair of Cu and Ag has a minimum emf,

pair of Al and Ag – maximum emf.

346. Draw a diagram of two galvanic cells, in one of which lead would be the cathode, and in the other, the anode. Write equations for electrode processes and calculate the emf of each element.

347. Draw 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. Draw 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 =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. Draw 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. Draw a diagram of a galvanic cell consisting of two aluminum plates immersed in 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 the 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. Draw a diagram of a galvanic cell consisting of two aluminum plates immersed in 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. Draw up a diagram of a galvanic cell consisting of two copper electrodes immersed in 0.001 M and 0.1 M solutions of Cu(NO 3) 2. Write the equations of electrode processes and calculate the EMF of this element.

Answer: EMF = 0.059 V.

359. Draw 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 solutions of Pb(NO 3) 2. 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 current strength. Write the equations for the reactions occurring on inert electrodes during the electrolysis of ZnSO 4.

Answer: I= 5.74 A.

362. In what sequence will metal ions be discharged at the cathode during the electrolysis of molten mixtures of salts KCl, ZnCl 2, MgCl 2. Explain your answer.

Answer: ZnCl2(D E=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. Define atomic mass this metal.

Answer: M(Cd) = 112.5 g/mol.

364. How many grams of copper will be deposited at 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 occurring on inert electrodes during the electrolysis of potassium chloride located: 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 Cl.

367. What mass of cadmium is released at the cathode if a current of 3.35 A is passed through a solution of cadmium sulfate for 1 hour?

Answer: m(Cd) = 7 g.

368. What mass of silver is released at the cathode if silver nitrate is passed through a solution? electricity force 0.67 A for 20 hours?

Answer: m(Ag) = 53.9 g.

369. Write the equations for the reactions occurring 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 for 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 a 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) = 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 the reactions occurring on inert electrodes.

Answer: m(Ni) = 29.4 g.

373. During electrolysis of a nickel sulfate solution, 4.2 liters of oxygen (n.o.) are released. How many grams of nickel will be deposited at the cathode?

Answer: m(Ni) = 22 g.

374. How much electricity will be required to produce 44.8 liters of hydrogen by electrolysis of an aqueous solution of potassium chloride? Write the equations for the reactions occurring on inert electrodes.

Answer: Q= 386000 Cl.

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 will it take to completely decompose 2 moles of water using a current of 2 A?

Answer:53.6 hours

377. Find the volume of oxygen (no.) that will be 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.s.) 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) = 1.25 l.

379. During the electrolysis of an aqueous solution of SnCl 2 at the anode, 4.48 liters of chlorine (no.) were released. Find the mass of tin released at the cathode.

Answer:m(Sn) = 23.7 g.

380. When a current of 1.5 A was passed through a solution of a trivalent metal salt 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

Control questions

1. What is a galvanic cell called? Describe the principle of its operation.

2. What is standard electrode potential?

3. What is the electromotive force of a galvanic cell? How is the emf of a galvanic cell calculated for standard conditions and conditions other than standard?

4. What is the difference between metal and concentration galvanic cells?

5. What processes occur during the operation of a galvanic cell consisting of iron and silver electrodes immersed in solutions of their salts?

6. Draw up 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?


Metal corrosion

Corrosionis a spontaneous process of destruction of materials and products made from them as a result of physical and chemical exposure to the environment, in which the metal passes into an oxidized (ionic) state and loses its inherent properties.

Metals and alloys coming into contact with 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 influence of a number of factors: the presence of adsorbed moisture on the surface, air pollution with corrosive substances, changes in air and metal temperatures, the nature of corrosion products, etc.

According to the laws of chemical thermodynamics, corrosion processes arise and proceed spontaneously only if the Gibbs energy of the system decreases (∆ G<0).

91.1. Classification of corrosion processes

1. By type of destruction Corrosion can be continuous or local. If corrosion damage is evenly distributed, it does not pose a danger to structures and apparatus, especially in cases where metal losses do not exceed technically justified standards. Local corrosion is much more dangerous, although metal loss may be small. The danger is that, by reducing the strength of individual sections, it sharply reduces the reliability of structures, structures, and devices.

2. According to the flow conditions distinguish: atmospheric, gas, liquid, underground, sea, soil corrosion, corrosion by stray currents, corrosion under voltage, etc.

3 . According to the mechanism of the corrosion process differentiate chemical And electrochemical corrosion.

Chemical corrosion can occur when interacting with dry gaseous oxidizing agents and solutions of non-electrolytes. Most metals interact with gases at elevated temperatures. In this case, two processes occur on the surface: metal oxidation and 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 = 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 O2,

where ∆ G 0 – standard Gibbs energy of the reaction; p O 2 – relative oxygen pressure.

Methods of protection against gas corrosion: alloying metals, creating protective coatings on the surface and changing the properties of the gas environment.

Electrochemical corrosion of metals develops when metal comes into contact with electrolyte solutions (all cases of corrosion in aqueous solutions, since even pure water is a weak electrolyte, and sea water is a strong one). The main oxidizing agents are water, dissolved oxygen and hydrogen ions.

Cause of electrochemical corrosion is that the surface of a metal is always energetically inhomogeneous due to the presence of impurities in the metals, differences in the chemical and phase composition of the alloy, etc. This leads to the formation of microgalvanic elements on the surface in a humid atmosphere. In areas of the metal that have a more negative potential value, the process of oxidation of this metal occurs:

M 0 + ne– =M n+ (anodic process).

Oxidizing agents that accept electrons at the cathode are called cathodic depolarizers. The cathodic depolarizers are: hydrogen ions (hydrogen depolarization), oxygen molecules (oxygen depolarization).

In task 20 of the OGE in chemistry, you must provide a complete solution. Solution to task 20 - drawing up a chemical reaction equation using the electronic balance method.

Theory for task No. 20 OGE in chemistry

We have already talked about redox reactions in. Now let’s look at the electronic balance method using a typical example, but before that we’ll find out what this method is and how to use it.

Electronic balance method

The electron balance method is a method for equalizing chemical reactions, based on changes in 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 found elements, we draw up an electronic balance, which consists of counting the number of acquired or given electrons
  • Finding the least common multiple of the transferred electrons
  • The resulting values ​​are the coefficients in the equation (with rare exceptions)

Using the electronic balance method, arrange the coefficients in the reaction equation, the diagram of which

HI + H2SO4 → I2 + H2S + H2O

Identify the oxidizing agent and the reducing agent.

So, let's create an electronic balance. In this reaction we change the oxidation states sulfur And iodine .

Sulfur was in the oxidation state +6, and in the products -2. Iodine had an oxidation state of -1, but became 0.

If you have difficulties with the calculation, then remember.

1 | S +6 + 8ē → S –2
4 | 2I –1 – 2ē → I 2

Sulfur takes 8 electrons, but iodine gives up only two - a total 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 point out that sulfur in the oxidation state +6 is oxidizing agent , A iodine in oxidation state –1 – reducing agent.

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