Fundamentals of General Chemistry. Terms and Problems in Tests In 2 parts. P.1. Terms and Examples in Tasks. Study guide

5. Окраска индикатора фенолфталеина в водном растворе карбоната

6. Лакмус приобретает красную окраску в водном растворе соли:

А) NaNO3; Б) KI; В) ZnSO4; Г) K3PO4

7. Среда водного раствора становится кислой при растворении соли:

9. Среда водного раствора становится щелочной при растворении соли:

гидролиза иона Fe3+ по второй ступени составляет:

81

А) 5.5 10−2; Б) 7.4 10−3; В) 1.82 10−3; Г) 5.5 10−4.

12. Для Н3PO4 Kа1 = 7.1 10−3, Kа2 = 6.2 10−8, Kа3 = 5.0 10−13. Значение константы гидролиза иона PO34– по второй ступени составляет:

A) 1.4 10−12; Б) 6.2 10−4; В) 2.0 10−2; Г) 1.6 10−7.

82

1 0 . O X I D A T I O N – R E D U C T I O N R E A C T I O N S

83

1 0 . 2 . W o r k e d e x a m p l e s

Example 1. Balance the equation of redox reaction by the electronic balance method:

NH3 + O2 → NO + H2O.

Answer

Determine the oxidation states of each atom in reactants and products оf reaction:

NH3 + O2 → NO + H2O.

Make up the electronic balance and determine the oxidizing agent and the reducing agent:

In NH3 the oxidation state of N is −3. It is oxidized to nitrogen(II) in NO. The oxidation state of nitrogen varies from –3 to +2, formally will be equal to five electrons moles lost. So NH3 is the reducing agent. The oxidation state of oxygen varies from 0 to –2 which formally will be equal to four electrons moles gained per two atom of oxygen. The oxidizing agent is O2. Choose coefficients for the species containing the atom oxidized and the atom reduced such that the number electrons moles lost is equal to the number electrons moles gained. Such coefficients are 5 and 4 in the reaction equation before the oxidizing agent and the reducing agent, respectively:

4NH3 + 5O2 → 4NO + H2O.

Then balance the number of hydrogen atoms moles:

4NH3 + 5O2 → 4NO + 6H2O.

Check the number of oxygen atoms moles in the equation: on the left-hand side of the equation: 5 2 = 10 moles.

on the right-hand side of the equation: 4 + 6 = 10 moles.

84

Therefore, the equation is balanced.

Example 2. Using the half-reaction method write balanced ionic equation for the reaction:

Answer

Make up the unbalanced equations of the two half-reactions:

Fe2+ → Fe3+

ClO3– → Cl–.

To balance the charge in the first half-reaction 1 mole of electrons is subtracted on the left-hand side of the equation:

Fe2+ – e– → Fe3+.

The equation of the second half-reaction for an acidic solution is not balanced by oxygen atoms:

ClO3– → Cl–.

To balance of oxygen atoms, need to add 3 moles of H2O on the right-hand side of the equation and 6 moles of H+ on the left-hand side of the equation:

ClO3– + 6H+ → Cl– + 3H2O.

To balance of the charge, need to add 6 moles of electrons on the left-hand side of the equation:

ClO3– + 6H+ + 6 e–→ Cl– + 3H2O.

To derive the overall equation, the first half-reaction must be multiplied by 6 and the second half-reaction by 1 so that the number of moles of electrons lost will be equal to the number of moles of electrons gained:

Then two half-reactions are summarized:

85

6Fe2+ – 6 e– + ClO3– + 6H+ + 6 e– = 6Fe3+ + Cl– + 3H2O.

The final balanced ionic equation is:

6Fe2+ + ClO3– + 6H+ = 6Fe3+ + Cl– + 3H2O.

Example 3. Based on the values of the standard reduction potentials of the half-reactions:

determine, which metal is dissolved in an aqueous nitric acid solution. Calculate the value of the standard cell potential Eocell and the standard Gibbs energy change for the possible reaction.

Answer

The overall redox reaction can be written by combining the two halfreactions. Compare the Еo for half-reactions (10.1) and (10.3):

Еo(Au3+/Au) > Еo(NO3−/ NO),

i.e. the ions Au3+ are a more powerful oxidizing agent than the ions NO3−. This means that the half-reaction (10.1) occurs from left to right and halfreaction (10.3) occurs from right to left

Au3+ + 3e– = Au

NO3− + 4H+ + 3e– = NO + 2H2O.

Oxidation–reduction reaction is

Au3+ + NO + 2H2O = Au + NO3− + 4H+.

The ions Au3+ will oxidize NO and gold is not dissolved in an aqueous nitric acid solution.

Compare the Еo for half-reactions (10.2) and (10.3):

Еo(Cu2+/Cu) < Еo(NO3−/ NO),

86

i.e. the ions Cu2+ are a better reducing agent than the ions NO3−. This means that the half-reaction (10.2) occurs from right to left (oxidation process) and half-reaction (10.3) occurs from left to right (reduction process). In order that the number electrons moles lost is equal to the number electrons moles gained the half-reaction (10.2) must be multiplied by 3 and half-reaction (10.3) must be multiplied by 2:

3Cu + 2NO3− + 8H+ + 6e– = 3Cu2+ + 2NO + 4H2O + 6e–. (10.4)

The final balanced equation is:

3Cu + 2NO3− + 8H+ = 3Cu2+ + 2NO + 4H2O.

Copper is dissolved in an aqueous nitric acid solution.

The thermodynamic favorability of this reaction can be assessed by

calculating Eocell and Go . The standard cell potential is given by the equation:

Eocell = Eoreduction process − Eooxidation process.

Eocell = +0.960 V − 0.340 V = 0,62 V.

The change in standard Gibbs energy of an electrochemical cell is:

Go = − zFEocell,

where Go is in J, z is the number of moles of electrons involved in the re-

action, F is Faraday constant, F 96500 C mol−1, Eocell is in volts. For the reaction (4) z is 6, then

Go = –6 · 96500 · 0.62 = –358980 J = –359 kJ.

87

1 0 . 3 . T a s k s f o r s e l f - c o n t r o l

1. The sum of the coefficients in the equation of the redox reaction KMnO4 + Na2SO3 + H2SO4 → MnSO4 + …

2. The sum of the coefficients in the equation of the redox reaction Mg + HNO3→ NH4NO3 + …

3. The sum of the coefficients in the equation of the redox reaction K2Cr2O7 + HCl → CrCl3 + …

4. The sum of the coefficients in the equation of the redox reaction Fe + KOH + KNO3 → K2FeO4 + …

5. The sum of the coefficients in the equation of the redox reaction H2O2 + H2SO4 + K2S → S + …

6. The sum of the coefficients in the equation of the redox reaction Cu + HNO3 → NO2 + …

7. The sum of the coefficients in the equation of the redox reaction NH3 + Cl2 → N2 + …

88

8. The sum of the coefficients in the equation of the redox reaction K2Cr2O7 + (NH4)2S + H2O → Cr(OH)3 + …

9. According to the standard reduction potentials of the half-reactions:

10. Based on the values of the standard reduction potentials of the halfreactions:

11. According to the standard reduction potentials of the half-reactions:

12. According to the standard reduction potentials of the half-reactions:

89

13. Based on the standard reduction potentials of the half-reactions:

determine, whether a cadmium plate immersed in an aqueous solution of copper (II) sulfate will be covered with copper. The calculated value of Eocell for the redox process is:

14. Based on the standard reduction potentials of the half-reactions

for reaction

Cr2O27– + NO–2 + H+ → Cr3+ + NO–3 + H2O

the standard Gibbs energy change is: А) Go = 301 kJ;

B)Go = −602 kJ;

C)Go = −301 kJ;

D)Go = −100 kJ.

90