OXIDATION-REDUCTION. SOLUTIONS TO REVIEW QUESTIONS. 1. Oxidation
of a metal occurs when the metal loses electrons. The easier it is for a metal to ...
CHAPTER 17
OXIDATION-REDUCTION SOLUTIONS TO REVIEW QUESTIONS 1. Oxidation of a metal occurs when the metal loses electrons. The easier it is for a metal to lose electrons, the more active the metal is. 2. (a) (b)
Iodine is oxidized. Its oxidation number increases from 0 to þ5. Chlorine is reduced. Its oxidation number decreases from 0 to 1.
3. The higher metal on the activity series list is more reactive. (a) Ca (b) Fe (c) Zn 4. If the free element is higher on the activity series than the ion with which it is paired, the reaction occurs. The higher metal on the activity series is more reactive. (a) 2 Al(s) þ 3 ZnCl2(aq) ! 3 Zn(s) þ 2 AlCl3(aq) (b) Sn(s) þ 2 HCl(aq) ! H2(g) þ SnCl2(aq) (c) Ag(s) þ H2SO4(aq) ! no reaction H2 is more reactive than Ag (d) Fe(s) þ 2 AgNO3(aq) ! 2 Ag(s) þ Fe(NO3)2(aq) (e) 2 Cr(s) þ 3 Ni2þ(aq) ! 3 Ni(s) þ 2 Cr3þ(aq) (f) Mg(s) þ Ca2þ(aq) ! no reaction Ca is more reactive than Mg (g) Cu(s) þ Hþ(aq) ! no reaction H2 is more reactive than Cu Al is more reactive than Ag (h) Ag(s) þ Al3þ(aq) ! no reaction 5. If a copper wire is placed into a solution of lead (II) nitrate, no reaction will occur. Lead is more active than copper and undergoes oxidation more readily than copper. Lead is already oxidized, therefore will stay oxidized in the presence of copper. 6. (a) (b) (c) (d)
2 Al þ Fe2O3 ! A12O3 þ 2 Fe þ Heat Al is above Fe in the activity series, which indicates Al is more active than Fe. No. Iron is less active than aluminum and will not displace aluminum from its compounds. Yes. Aluminum is above chromium in the activity series and will displace Cr3þ from its compounds.
7. (a)
2 Al(s) þ 6 HCl(aq) ! 2 AlCl3(aq) þ 3 H2(g) 2 Al(s) þ 3 H2SO4(aq) ! Al2(SO4)3(aq) þ 3 H2(g) 2 Cr(s) þ 6 HCl(aq) ! 2 CrCl3(aq) þ 3 H2(g) 2 Cr(s) þ 3 H2SO4(aq) ! Cr2(SO4)3(aq) þ 3 H2(g) Au(s) þ HCl(aq) ! no reaction Au(s) þ H2SO4(aq) ! no reaction Fe(s) þ 2 HCl(aq) ! FeCl2(aq) þ H2(g) Fe(s) þ H2SO4(aq) ! FeSO4(aq) þ H2(g) Cu(s) þ HCl(aq) ! no reaction Cu(s) þ H2SO4(aq) ! no reaction
(b) (c) (d) (e)
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- Chapter 17 Mg(s) þ 2 HCl(aq) ! MgCl2(aq) þ H2(g) Mg(s) þ H2SO4(aq) ! MgSO4(aq) þ H2(g) (g) Hg(l) þ HCl(aq) ! no reaction Hg(l) þ H2SO4(aq) ! no reaction (h) Zn(s) þ 2 HCl(aq) ! ZnCl2(aq) þ H2(g) Zn(s) þ H2SO4(aq) ! ZnSO4(aq) þ H2(g) (f)
8. The oxidation number for an atom in an ionic compound is the same as the charge of the ion that resulted when that atom lost or gained electrons to form an ionic bond. In a covalently bonded compound electrons are shared between the two atoms making up the bond. Those shared electrons are assigned to the atom in the bond with a higher electronegativity giving it a negative oxidation number. 9. In an electrolytic cell the anode is the positively charged electrode and attracts negatively charged ions (anions). The cathode is the negatively charged electrode and attracts positively charge ions (cations). In a voltaic cell the anode is the negatively changed electrode where oxidation occurs. The cathode is the positively charged electrode where reduction occurs. 10. (a)
Oxidation occurs at the anode. The reaction is 2 Cl ðaqÞ ! Cl2 ðgÞ þ 2 e
(b)
Reduction occurs at the cathode. The reaction is Ni2þ ðaqÞ þ 2 e ! NiðsÞ
(c)
The net chemical reaction is electrical
Ni2þ ðaqÞ þ 2 Cl ðaqÞ! NiðsÞ þ Cl2 ðgÞ energy
11. In Figure 17.3, electrical energy is causing chemical reactions to occur. In Figure 17.4, chemical reactions are used to produce electrical energy. 12. (a) (b)
It would not be possible to monitor the voltage produced, but the reactions in the cell would still occur. If the salt bridge were removed, the reaction would stop. Ions must be mobile to maintain an electrical neutrality of ions in solution. The two solutions would be isolated with no complete electrical circuit.
13. Oxidation and reduction are complementary processes because one does not occur without the other. The loss of e in oxidation is accompanied by a gain of e in reduction. 14.
Ca2þ þ 2 e ! Ca 2 Br ! Br2 þ 2 e
cathode reaction, reduction anode reaction, oxidation
15. During electroplating of metals, the metal is plated by reducing the positive ions of the metal in the solution. The plating will occur at the cathode, the source of the electrons. With an alternating current, the polarity of the electrode would be constantly changing, so at one instant the metal would be plating and the next instant the metal would be dissolving. 16. Since lead dioxide and lead(II) sulfate are insoluble, it is unnecessary to have salt bridges in the cells of a lead storage battery.
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- Chapter 17 17. The electrolyte in a lead storage battery is dilute sulfuric acid. In the discharge cycle, SO42, is removed from solution as it reacts with PbO2 and Hþ to form PbSO4(s) and H2O. Therefore, the electrolyte solution contains less H2SO4 and becomes less dense. 18. If Hg2þ ions are reduced to metallic mercury, this would occur at the cathode, because reduction takes place at the cathode. 19. In both electrolytic and voltaic cells, oxidation and reduction reactions occur. In an electrolytic cell an electric current is forced through the cell causing a chemical change to occur. In voltaic cells, spontaneous chemical changes occur, generating an electric current. 20. In some voltaic cells, the reactants at the electrodes are in solution. For the cell to function, these reactants must be kept separated. A salt bridge permits movement of ions in the cell. This keeps the solution neutral with respect to the charged particles (ions) in the solution.
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- Chapter 17 -
SOLUTIONS TO EXERCISES 1. Oxidation numbers of each element in the compound Cu ¼ þ2 C ¼ þ4 O ¼ 2 (a) CuCO3 (b) CH4 C ¼ 4 H ¼ þl (c) IF I ¼ þ1 F ¼ l (d) CH2Cl2 C¼0 H ¼ þ1 Cl ¼ 1 (e) SO2 S ¼ þ4 O ¼ 2 Rb ¼ þ1 Cr ¼ þ6 O¼2 (f) Rb2CrO4 2. Oxidation numbers of each element in the compound. C ¼ þ2 H ¼ þl F ¼ 1 (a) CHF3 (b) P2O5 P ¼ þ5 O ¼ 2 S ¼ þ6 F ¼ 1 (c) SF6 (d) SnSO4 Sn ¼ þ2 S ¼ þ6 O ¼ 2 C ¼ 2 H ¼ þl O ¼ 2 (e) CH3OH (f) H3PO4 H ¼ þl P ¼ þ5 O ¼ 2 3. The oxidation number of the underlined element is indicated by the number following the formula. (a) P _ O33 þ 3 (c) NaHCO3 þ 4 (d) B _ rO4 þ 7 (b) CaSO4 þ 6 4. The oxidation number of the underlined element is indicated by the number following the formula. (c) NaH2PO4 þ 5 (a) C _ O32 þ 4 (b) H2SO4 þ 6 (d) C _ r 2 O72 þ 6 5. Balanced half-reaction (a) (b) (c) (d)
Na ! Naþ þ l e C2 O42 ! 2 CO2 þ 2 e 2 I ! I2 þ 2 e Cr2 O72 þ 14 Hþ þ 6 e ! 2 Cr3þ þ 7 H2 O
Element Changing Na C I Cr
6. Balanced half-reaction (a) (b) (c) (d) 7. (a) (b) 8. (a) (b)
Type of Reaction oxidation oxidation oxidation reduction
Element Changing Cu F I Mn
Cu2þ þ 1e ! Cu1þ F 2 þ 2 e ! 2 F 2 IO4 þ 16 Hþ þ 14 e ! I2 þ 8 H2 O Mn ! Mn2þ þ 2e Cu is oxidized, Ag is reduced; Cu is the reducing agent, AgNO3 is the oxidizing agent Zn is oxidized, H is reduced Zn is the reducing agent, HCl is the oxidizing agent C is oxidized, O is reduced CH4 is the reducing agent, O2 is the oxidizing agent Mg is oxidized, Fe is reduced Mg is the reducing agent, FeCl3 is the oxidizing agent
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Type of Reaction reduction reduction reduction oxidation
- Chapter 17 9. (a) (b) (c)
correctly balanced correctly balanced incorrectly balanced MgðsÞ þ 2 HClðaqÞ ! Mg2þ ðaqÞ þ 2 Cl ðaqÞ þ H2 ðgÞ
(d)
incorrectly balanced 3 CH3 OHðaqÞ þ Cr2 O72 ðaqÞ þ 8 Hþ ðaqÞ ! 2 Cr3þ ðaqÞ þ 3 CH2 OðaqÞ þ 7 H2 OðlÞ
10. (a)
incorrectly balanced 3 MnO2 ðsÞ þ 4 AlðsÞ ! 3 MnðsÞ þ 2 Al2 O3 ðsÞ
(b) (c) (d)
correctly balanced correctly balanced incorrectly balanced 8 H2 OðlÞ þ 2 MnO4 ðaqÞ þ 7 S2 ðaqÞ ! 2 MnSðsÞ þ 16 OH ðaqÞ þ 5 SðsÞ
11. Balancing oxidation-reduction equations using the change-in-oxidation number method: (a) Cu þ O2 ! CuO ox
Cu0 ! Cu2þ þ 2 e
red
2O þ 4e ! 2O 0
2
Multiply by 2; Add the equations; the 4 e cancel
2 Cu0 þ 2 O0 ! 2 Cu2þ þ 2 O2 Transfer the coefficients to the original equation appropriately. 2 Cu þ O2 ! 2 CuO KClO3 ! KCl þ O2
(b) ox red
3 O2 ! O0 þ 6 e 5þ
Cl
þ 6 e ! Cl
Multiply by 2; Multiply by 2, add; the 12 e cancel
2 Cl5þ þ 6 O2 ! 2 Cl þ 3 O2 Transfer the coefficients to the original equation appropriately. 2 KClO3 ! 2 KCl þ 3 O2 Ca þ H2 O ! CaðOHÞ2 þ H2
(c) ox
Ca ! Ca2þ þ 2 e
red
2 Hþ þ 2 e ! H2
Add equations together; the 2 e cancel:
Ca þ 2 Hþ ! Ca2þ þ H2 Balance the equation by inspection. Ca þ 2 H2 O ! CaðOHÞ2 þ H2 PbS þ H2 O2 ! PbSO4 þ H2 O
(d) ox
S2 ! Sþ6 þ 8 e
red
2O þ 2 e ! 2 O2 Multiply by 4, add; the 8 e cancel: S2 þ 8 O ! Sþ6 þ 4 O2
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- Chapter 17 Transfer the coefficients to the original equation and complete the balancing by inspection. PbS þ 4 H2 O2 ! PbSO4 þ 4 H2 O CH4 þ NO2 ! N2 þ CO2 þ H2 O
(e) ox
C4 ! C4þ þ 8 e
red
N4þ þ 4 e ! N0
Multiply by 2; add the 8 e cancel
C4 þ 2 N4þ ! C4þ þ 2 N0 Transfer the coefficients to the original equation and complete the balancing by inspection. CH4 þ 2 NO2 ! N2 þ CO2 þ 2 H2 O 12. Balancing oxidation-reduction equations using the change-in-oxidation number method: (a) Cu þ AgNO3 ! Ag þ CuðNO3 Þ2 ox
Cu0 ! Cu2þ þ 2 e
red
Agþ þ 1 e ! Ag0
Multiply by 2, add; the 2 e cancel
Cu0 þ 2 Agþ ! Cu2þ þ 2 Ag0 Transfer the coefficients to the original equation. Cu þ 2 AgNO3 ! 2 Ag þ CuðNO3 Þ2 MnO2 þ HCl ! MnCl2 þ Cl2 þ H2 O
(b) ox
Cl ! Cl0 þ 1 e
red
Mn4þ þ 2 e ! Mn2þ
Multiply by 2, add; the 2 e cancel
2 Cl þ Mn4þ ! 2 Cl0 þ Mn2þ Transfer the coefficients to the original equation and complete the balancing by inspection. MnO2 þ 4 HCl ! MnCl2 þ Cl2 þ 2 H2 O HCl þ O2 ! Cl2 þ H2 O
(c) red
2 O0 þ 4 e ! 2 O2
ox
2 Cl ! 2 Cl0 þ 2 e
Multiply by 2, add; the 4 e cancel
4 Cl þ 2 O0 ! 4 Cl0 þ 2 O2 Transfer the coefficients to the original equation and complete the balancing by inspection. 4 HCl þ O2 ! 2 Cl2 þ 2 H2 O
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- Chapter 17 Ag þ H2 S þ O2 ! Ag2 S þ H2 O
(d) red
2 O0 þ 4 e ! 2 O2
ox
Ag ! Agþ þ 1 e
Multiply by 4, add; the 4 e cancel
2 Ag þ 2 O0 ! 2 Agþ þ 2 O2 Transfer the coefficients to the original equation and complete the balancing by inspection. 4 Ag þ 2 H2 S þ O2 ! 2 Ag2 S þ 2 H2 O KMnO4 þ CaC2 O4 þ H2 SO4 ! K2 SO4 þ MnSO4 þ CaSO4 þ CO2 þ H2 O
(e) red ox
Mn7þ þ 5 e ! Mn2þ 2C
3þ
! 2C
4þ
þ 2e
Multiply by 2; Multiply by 5, add; the 10 e cancel
10 C3þ þ 2 Mn7þ ! 10 C4þ þ 2 Mn2þ Transfer the coefficients to the original equation and complete the balancing by inspection. 2 KMnO4 þ 5 CaC2 O4 þ 8 H2 SO4 ! K2 SO4 þ 2 MnSO4 þ 5 CaSO4 þ 10 CO2 þ 8 H2 O 13. Balancing ionic redox equations (a) Zn þ NO3 ! Zn2þ þ NH4þ ðacidic solutionÞ Step 1
Write half-reaction equations. Balance except H and O. Zn ! Zn2þ NO3 ! NH4þ
Step 2
Balance H and O using H2O and Hþ Zn ! Zn2þ
Step 3
10 Hþ þ NO3 ! NH4þ þ 3 H2 O Balance electrically with electrons Zn ! Zn2þ þ 2 e
Step 4
10 Hþ þ NO3 þ 8 e ! NH4þ þ 3 H2 O Equalize the loss and gain of electrons 4ðZn ! Zn2þ þ 2 e Þ
Step 5
10 Hþ þ NO3 þ 8 e ! NH4þ þ 3 H2 O Add the half-reactions-electrons cancel 10 Hþ þ 4 Zn þ NO3 ! 4 Zn2þ þ NH4þ þ 3 H2 O
(b)
NO3 þ S ! NO2 þ SO42 Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. S ! SO42 NO3 ! NO2
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- Chapter 17 Step 2
Balance H and O using H2O and Hþ 4 H2 O þ S ! SO42 þ 8 Hþ
Step 3
2 Hþ þ NO3 ! NO2 þ H2 O Balance electrically with electrons 4 H2 O þ S ! SO42 þ 8 Hþ þ 6 e
Step 4 and 5
2 Hþ þ NO3 þ e ! NO2 þ H2 O Equalize the loss and gain of electrons; add the half-reactions 4 H2 O þ S ! SO42 þ 8 Hþ þ 6 e 6 ð2 Hþ þ NO3 þ e ! NO2 þ H2 OÞ 4 Hþ þ S þ 6 NO3 ! 6 NO2 þ SO42 þ 2 H2 O 4 H2O, 8 Hþ and 6e canceled from each side
(c)
PH3 þ I2 ! H3 PO2 þ I Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. PH3 ! H3 PO2
Step 2
I2 ! 2 I Balance H and O using H2O and Hþ 2 H2 O þ PH3 ! H3 PO2 þ 4 Hþ
Step 3
I2 ! 2 I Balance electrically with electrons 2 H2 O þ PH3 ! H3 PO2 þ 4 Hþ þ 4 e
Step 4 and 5
I2 þ 2 e ! 2 I Equalize the loss and gain of electrons; add the half-reactions 2 H2 O þ PH3 ! H3 PO2 þ 4 Hþ þ 4 e 2 ðI2 þ 2 e ! 2 I Þ PH3 þ 2 H2 O þ 2 I2 ! H3 PO2 þ 4 I þ 4 Hþ
(d)
Cu þ NO3 ! Cu2þ þ NO
ðacidic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
Step 2
Cu ! Cu2þ NO3 ! NO Balance H and O using H2O and Hþ Cu ! Cu2þ 4 Hþ þ NO3 ! NO þ 2 H2 O
Step 3
Balance electrically with electrons Cu ! Cu2þ þ 2 e 4 Hþ þ NO3 þ 3 e ! NO þ 2 H2 O
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- Chapter 17 Step 4 and 5
Equalize the loss and gain of electrons; add the half-reactions 3 ðCu ! Cu2þ þ 2 e Þ 2 ð4 Hþ þ NO3 þ 3 e ! NO þ 2 H2 OÞ 3 Cu þ 8 Hþ þ 2 NO3 ! 3 Cu2þ þ 2 NO þ 2 H2 O
(e)
ClO3 þ Cl ! Cl2 Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. Cl ! Cl0
Step 2
ClO3 ! Cl0 Balance H and O using H2O and Hþ Cl ! Cl0
Step 3
6 Hþ þ ClO3 ! Cl0 þ 3 H2 O Balance electrically with electrons Cl ! Cl0 þ e
Step 4 and 5
6 Hþ þ ClO3 þ 5 e ! Cl0 þ 3 H2 O Equalize the loss and gain of electrons; add the half-reactions 5ðCl ! Cl0 þ e Þ 6 Hþ þ ClO3 þ 5 e ! Cl0 þ 3 H2 O 6 Hþ þ ClO3 þ 5 Cl ! 3 Cl2 þ 3 H2 O
14. (a)
ClO3 þ I ! I2 þ Cl Step 1
Step 2
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. 2 I ! I2 ClO3 ! Cl Balance H and O using H2O and Hþ 2 I ! I2
Step 3
Step 4 and 5
6 Hþ þ ClO3 ! Cl þ 3 H2 O Balance electrically with electrons 2 I ! I2 þ 2 e 6 Hþ þ ClO3 þ 6 e ! Cl þ 3 H2 O Equalize the loss and gain of electrons; add the half-reactions 3 ð2 I ! I2 þ 2 e Þ
(b)
6 Hþ þ ClO3 þ 6 e ! Cl þ 3 H2 O 6 Hþ þ ClO3 þ 6 I ! 3 I2 þ Cl þ 3 H2 O Cr2 O72 þ Fe2þ ! Cr3þ þ Fe3þ ðacidic solutionÞ Step 1
Write half-reaction equations. Balance except H and O. Fe2þ ! Fe3þ Cr2 O72 ! 2 Cr3þ
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- Chapter 17 Step 2
Balance H and O using H2O and Hþ Fe2þ ! Fe3þ
Step 3
14 Hþ þ Cr2 O72 ! 2 Cr3þ þ 7 H2 O Balance electrically with electrons Fe2þ ! Fe3þ þ e
Step 4 and 5
14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O Equalize the loss and gain of electrons; add the half-reactions 6ðFe2þ ! Fe3þ þ e Þ 14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O 14 Hþ þ Cr2 O72 þ 6 Fe2þ ! 2 Cr3þ þ 6 Fe3þ þ 7 H2 O
(c)
MnO4 þ SO2 ! Mn2þ þ SO42 Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. SO2 ! SO42
Step 2
MnO4 ! Mn2þ Balance H and O using H2O and Hþ 2 H2 O þ SO2 ! SO42 þ 4 Hþ
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O Balance electrically with electrons 2 H2 O þ SO2 ! SO42 þ 4 Hþ þ 2 e
Step 4 and 5
8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 O Equalize the loss and gain of electrons; add the half-reactions 5ð2 H2 O þ SO2 ! SO42 þ 4 Hþ þ 2 e Þ 2ð8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 OÞ 2 H2 O þ 2 MnO4 þ 5 SO2 ! 4 Hþ þ 2 Mn2þ þ 5 SO42 8 H2O, 16 Hþ, and 10 e canceled from each side
(d)
H3 AsO3 þ MnO4 ! H3 AsO4 þ Mn2þ Step 1
Step 2
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. H3 AsO3 ! H3 AsO4 MnO4 ! Mn2þ Balance H and O using H2O and Hþ H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O Balance electrically with electrons H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e 8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 O
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- Chapter 17 Step 4 and 5
Equalize the loss and gain of electrons; add the half-reactions 5ðH2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e Þ 2ð8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 OÞ 6 Hþ þ 5 H3 AsO3 þ 2 MnO4 ! 5 H3 AsO4 þ 2 Mn2þ þ 3 H2 O 5 H2O, 10 Hþ, and 10 e canceled from each side
(e)
Cr2 O72 þ H3 AsO3 ! Cr3þ þ H3 AsO4 ðacidic solutionÞ Step 1
Step 2
Write half-reaction equations. Balance except H and O. H3 AsO3 ! H3 AsO4 Cr2 O72 ! 2 Cr3þ Balance H and O using H2O and Hþ H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4
Step 3
14 Hþ þ Cr2 O72 ! 2 Cr3þ þ 7 H2 O Balance electrically with electrons H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e
Step 4 and 5
14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O Equalize the loss and gain of electrons; add the half-reactions 3 ðH2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e Þ 14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O 8 Hþ þ Cr2 O72 þ 3 H3 AsO3 ! 2 Cr3þ þ 3 H3 AsO4 þ 4 H2 O 3 H2O, 6 Hþ, and 6 e canceled from each side
15. (a)
Cl2 þ IO3 ! Cl þ IO4
ðbasic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
Step 2
IO3 ! IO4 Cl2 ! 2 Cl Balance H and O using H2O and Hþ H2 O þ IO3 ! IO4 þ 2 Hþ
Step 3
Cl2 ! 2 Cl Add OH ions to both sides (same number as Hþ ions) 2 OH þ H2 O þ IO3 ! IO4 þ 2 Hþ þ 2 OH
Step 4
Cl2 ! 2 Cl Combine Hþ and OH to form H2O; cancel H2O where possible 2 OH þ H2 O þ IO3 ! IO4 þ 2 H2 O Cl2 ! 2 Cl 2 OH þ IO3 ! IO4 þ H2 O
Cl2 ! 2 Cl
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ð1 H2 O cancelledÞ
- Chapter 17 Step 5
Step 6 Step 7
(b)
Balance electrically with electrons 2 OH þ IO3 ! IO4 þ H2 O þ 2 e Cl2 þ 2 e ! 2 Cl Electron loss and gain is balanced Add half-reactions 2 OH þ IO3 þ Cl2 ! IO4 þ 2 Cl þ H2 O
MnO4 þ ClO2 ! MnO2 þ ClO4 Step 1
Step 2
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. ClO2 ! ClO4 MnO4 ! MnO2 Balance H and O using H2O and Hþ 2 H2 O þ ClO2 ! ClO4 þ 4 Hþ
Step 3
MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O Add OH ions to both sides (same number as Hþ ions) 4 OH þ 2 H2 O þ ClO2 ! ClO4 þ 4 Hþ þ 4 OH
Step 4
4 OH þ MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O þ 4 OH Combine Hþ and OH to form H2O; cancel H2O where possible 4 OH þ 2 H2 O þ ClO2 ! ClO4 þ 4 H2 O 4 H2 O þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH 4 OH þ ClO2 ! ClO4 þ 2 H2 O MnO4
Step 5
Step 6 and 7
(c)
2 H2 O þ ! MnO2 þ 4 OH ð2 H2 O cancelledÞ Balance electrically with electrons 4 OH þ ClO2 ! ClO4 þ 2 H2 O þ 4 e 2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH Equalize gain and loss of electrons; add half-reactions 3 ð4 OH þ ClO2 ! ClO4 þ 2 H2 O þ 4 e Þ 4 ð2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH Þ 2 H2 O þ 4 MnO4 þ 3 ClO2 ! 4 MnO2 þ 3 ClO4 þ 4 OH 6 H2O, 12 OH, and 12 e canceled from each side
Se ! SeO32 þ Se2 Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. Se ! SeO32
Step 2
ð2 H2 O cancelledÞ
Se ! Se2 Balance H and O using H2O and Hþ 3 H2 O þ Se ! SeO32 þ 6 Hþ Se ! Se2
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- Chapter 17 Step 3
Add OH ions to both sides (same number as Hþ ions) 6 OH þ 3 H2 O þ Se ! SeO32 þ 6 Hþ þ 6 OH
Step 4
Se ! Se2 Combine Hþ and OH to form H2O; cancel H2O where possible 6 OH þ 3 H2 O þ Se ! SeO32 þ 6 H2 O Se ! Se2
Step 5
6 OH þ Se ! SeO32 þ 3 H2 O Balance electrically with electrons
ð3 H2 O cancelledÞ
6 OH þ Se ! SeO32 þ 3 H2 O þ 4 e Step 6 and 7
Se þ 2 e ! Se2 Equalize gain and loss of electrons; add half-reactions 6 OH þ Se ! SeO32 þ 3 H2 O þ 4 e 2ðSe þ 2 e ! Se2 Þ 6 OH þ 3 Se ! SeO32 þ 2 Se2 þ 3 H2 O
(d)
Fe3 O4 þ MnO4 ! Fe2 O3 þ MnO2 Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. 2 Fe3 O4 ! 3 Fe2 O3
Step 2
MnO4 ! MnO2 Balance H and O using H2O and Hþ
Step 3
H2 O þ 2 Fe3 O4 ! 3 Fe2 O3 þ 2 Hþ 4 Hþ þ MnO4 ! MnO2 þ 2 H2 O Add OH ions to both sides (same number as Hþ ions)
Step 4
2 OH þ H2 O þ 2 Fe3 O4 ! 3 Fe2 O3 þ 2 Hþ þ 2 OH 4 OH þ 4 Hþ þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH Combine Hþ and OH to form H2O; cancel H2O where possible 2 OH þ H2 O þ 2 Fe3 O4 ! 3 Fe2 O3 þ 2 H2 O 4 H2 O þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH 2 OH þ 2 Fe3 O4 ! 3 Fe2 O3 þ H2 O MnO4
Step 5
Step 6 and 7
ð1 H2 O cancelledÞ
2 H2 O þ ! MnO2 þ 4 OH ð2 H2 O cancelledÞ Balance electrically with electrons 2 OH þ 2 Fe3 O4 ! 3 Fe2 O3 þ H2 O þ 2 e 2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH Equalize gain and loss of electrons; add half-reactions 3ð2 OH þ 2 Fe3 O4 ! 3 Fe2 O3 þ H2 O þ 2 e Þ 2ð2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH Þ H2 O þ 6 Fe3 O4 þ 2 MnO4 ! 9 Fe2 O3 þ 2 MnO2 þ 2 OH Þ 3 H2O, 6 OH, and 6 e canceled from each side
- 238 -
- Chapter 17 (e)
BrO þ CrðOHÞ4 ! Br þ CrO42 Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. CrðOHÞ4 ! CrO42
Step 2
BrO ! Br Balance H and O using H2O and Hþ CrðOHÞ4 ! CrO42 þ 4 Hþ
Step 3
2 Hþ þ BrO ! Br þ H2 O Add OH ions to both sides (same number as Hþ ions) 4 OH þ CrðOHÞ4 ! CrO42 þ 4 Hþ þ 4 OH
Step 4
2 OH þ 2 Hþ þ BrO ! Br þ H2 O þ 2 OH Combine Hþ and OH to form H2O; cancel H2O where possible 4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O 2 H2 O þ BrO ! Br þ H2 O þ 2 OH
Step 5
H2 O þ BrO ! Br þ 2 OH Balance electrically with electrons
ð1 H2 O cancelledÞ
4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e Step 6 and 7
H2 O þ BrO þ 2 e ! Br þ 2 OH Equalize gain and loss of electrons; add half-reactions 2 ð4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e Þ 3 ðH2 O þ BrO þ 2 e ! Br þ 2 OH Þ 2 OH þ 3BrO þ 2 CrðOHÞ4 ! 3Br þ 2 CrO42 þ 5 H2 O 3 H2O, 6 OH and 6 e canceled from each side
16. (a)
MnO4 þ SO32 ! MnO2 þ SO42 Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. SO32 ! SO42
Step 2
MnO4 ! MnO2 Balance H and O using H2O and Hþ H2 O þ SO32 ! SO42 þ 2 Hþ
Step 3
MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O Add OH ions to both sides (same number as Hþ ions) 2 OH þ H2 O þ SO32 ! SO42 þ 2 Hþ þ 2 OH 4 OH þ MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O þ 4 OH
- 239 -
- Chapter 17 Step 4
Combine Hþ and OH to form H2O; cancel H2O where possible 2 OH þ H2 O þ SO32 ! SO42 þ 2 H2 O MnO4 þ 4 H2 O ! MnO2 þ 2 H2 O þ 4 OH 2 OH þ SO32 ! SO42 þ H2 O ð1 H2 O cancelledÞ
Step 5
MnO4 þ 2 H2 O ! MnO2 þ 4 OH Balance electrically with electrons
ð2 H2 O cancelledÞ
2 OH þ SO32 ! SO42 þ H2 O þ 2 e Step 6 and 7
3 e þ MnO4 þ 2 H2 O ! MnO2 þ 4 OH Equalize gain and loss of electrons; add half-reactions 3 ð2 OH þ SO32 ! SO42 þ H2 O þ 2 e Þ 2 ðMnO4 þ 2 H2 O þ 3 e ! MnO2 þ 4 OH Þ H2 O þ 2 MnO4 þ 3 SO32 ! 2 MnO2 þ 3 SO42 þ 2 OH 3 H2O, 4 OH, and 6 e canceled from each side
(b)
ClO2 þ SbO2 ! ClO2 þ SbðOHÞ6 þ 2 H2 O ðbasic solutionÞ Step 1
Write half-reaction equations. Balance except H and O. SbO2 ! SbðOHÞ6
Step 2
ClO2 ! ClO2 Balance H and O using H2O and Hþ 4 H2 O þ SbO2 ! SbðOHÞ6 þ 2 Hþ
Step 3
ClO2 ! ClO2 Add OH ions to both sides (same number as Hþ ions) 2 OH þ 4 H2 O þ SbO2 ! SbðOHÞ6 þ 2 Hþ þ 2 OH
Step 4
ClO2 ! ClO2 Combine Hþ and OH to form H2 O; cancel H2 O where possible 2 OH þ 4 H2 O þ SbO2 ! SbðOHÞ6 þ 2 H2 O ClO2 ! ClO2
Step 5
Step 6 and 7
2 OH þ 2 H2 O þ SbO2 ! SbðOHÞ6 ð2 H2 O cancelledÞ Balance electrically with electrons 2 OH þ 2 H2 O þ SbO2 ! SbðOHÞ6 þ 2 e ClO2 þ e ! ClO2 Equalize gain and loss of electrons; add half-reactions 2 H2 O þ 2 OH þ SbO2 ! SbðOHÞ6 þ 2 e 2 ClO2 þ e ! ClO2 2 H2 O þ 2 ClO2 þ 2 OH þ SbO2 ! 2 ClO2 þ SbðOHÞ6
- 240 -
- Chapter 17 (c)
Al þ NO3 ! NH3 þ AlðOHÞ4 Step 1
Step 2
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. Al ! AlðOHÞ4 NO3 ! NH3 Balance H and O using H2 O and Hþ 4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ
Step 3
9 Hþ þ NO3 ! NH3 þ 3 H2 O Add OH ions to both sides (same number as Hþ ions) 4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ þ 4 OH
Step 4
9 OH þ 9 Hþ þ NO3 ! NH3 þ 3 H2 O þ 9 OH Combine Hþ and OH to form H2 O; cancel H2 O where possible 4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 H2 O 9 H2 O þ NO3 ! NH3 þ 3 H2 O þ 9 OH 4 OH þ Al ! AlðOHÞ4 NO3
Step 5
Step 6 and 7
ð4 H2 O cancelledÞ
6 H2 O þ ! NH3 þ 9 OH ð3 H2 O cancelledÞ Balance electrically with electrons 4 OH þ Al ! AlðOHÞ4 þ 3 e 6 H2 O þ NO3 þ 8 e ! NH3 þ 9 OH Equalize gain and loss of electrons; add half-reactions 8 4 OH þ Al ! AlðOHÞ4 þ 3 e 3 6 H2 O þ NO3 þ 8 e ! NH3 þ 9 OH 8 Al þ 3 NO3 þ 18 H2 O þ 5 OH ! 3 NH3 þ 8 AlðOHÞ4 27 OH and 24 e canceled from each side
(d)
P4 ! HPO32 þ PH3 Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O. P4 ! 4 HPO32
Step 2
P4 ! 4 PH3 Balance H and O using H2 O and Hþ 12 H2 O þ P4 ! 4 HPO32 þ 20 Hþ
Step 3
12 Hþ þ P4 ! 4 PH3 Add OH ions to both sides (same number as Hþ ions) 20 OH þ 12 H2 O þ P4 ! 4 HPO32 þ 20 Hþ þ 20 OH 12 OH þ 12 Hþ þ P4 ! 4 PH3 þ 12 OH
- 241 -
- Chapter 17 Step 4
Combine Hþ and OH to form H2 O; cancel H2 O where possible 20 OH þ 12 H2 O þ P4 ! 4 HPO32 þ 20 H2 O 12 H2 O þ P4 ! 4 PH3 þ 12 OH
Step 5
20 OH þ P4 ! 4 HPO32 þ 8 H2 O ð12 H2 O cancelledÞ Balance electrically with electrons 20 OH þ P4 ! 4 HPO32 þ 8 H2 O þ 12 e
Step 6 and 7
12 H2 O þ P4 þ 12 e ! 4 PH3 þ 12 OH Loss and gain of electrons are equal; add half-reactions 8 OH þ 4 H2 O þ 2 P4 ! 4 HPO32 þ 4 PH3 Divide equation by 2 4 OH þ 2 H2 O þ P4 ! 2 HPO32 þ 2 PH3
(e)
Al þ OH ! AlðOHÞ4 þ H2 ðbasic solutionÞ Step 1
Step 2
Write half-reaction equations. Balance except H and O. Al ! AlðOHÞ4 OH ! H2 Balance H and O using H2O and Hþ 4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ
Step 3
4 Hþ þ OH ! H2 þ H2 O Add OH ions to both sides (same number as Hþ ions) 4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ þ 4 OH
Step 4
3 OH þ 3 Hþ þ OH ! H2 þ H2 O þ 3 OH Combine Hþ and OH to form H2O; cancel H2O where possible 4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 H2 O 3 H2 O þ OH ! H2 þ H2 O þ 3 OH 4 OH þ Al ! AlðOHÞ4
ð4 H2 O cancelledÞ
Step 5
2 H2 O þ OH ! H2 þ 3 OH ð1 H2 O cancelledÞ Balance electrically with electrons 4 OH þ Al ! AlðOHÞ4 þ 3 e
Step 6 and 7
2 H2 O þ OH þ 2 e ! H2 þ 3 OH Equalize gain and loss of electrons; and half-reactions 2 4 OH þ Al ! AlðOHÞ4 þ 3 e 3 ð2 H2 O þ OH þ 2 e ! H2 þ 3 OH Þ 2 Al þ 6 H2 O þ 2 OH ! 2 AlðOHÞ4 þ 3 H2 9 OH and 6 e canceled on each side
- 242 -
- Chapter 17 17. (a)
IO3 þ I ! I2 Step 1
Step 2
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O. 2 IO3 ! I2 2 I ! I2 Balance H and O using H2O and Hþ 12 Hþ þ 2 IO3 ! I2 þ 6 H2 O
Step 3
2 I ! I2 Balance electrically with electrons 12 Hþ þ 2 IO3 þ 10 e ! I2 þ 6 H2 O
Step 4 and 5
2 I ! I2 þ 2 e Equalize the loss and gain of electrons; add the half-reaction. 12 Hþ þ 2 IO3 þ 10 e ! I2 þ 6 H2 O 5 ð2 I ! I2 þ 2 e Þ 12 Hþ þ 2 IO3 þ 10 I ! 6 I2 þ 6 H2 O
(b)
Mn2þ þ S2 O82 ! MnO4 þ SO42 Step 1
(acid solution)
Write half-reaction equations. Balance except H and O Mn2þ ! MnO4
Step 2
S2 O82 ! 2 SO42 Balance H and O using H2O and Hþ 4 H2 O þ Mn2þ ! MnO4 þ 8 Hþ
Step 3
S2 O82 ! 2 SO42 Balance electrically with electrons 4 H2 O þ Mn2þ ! MnO4 þ 8 Hþ þ 5 e
Step 4 and 5
(c)
2 e þ S2 O82 ! 2 SO42 Equalize the loss and gain of electrons; add the half-reactions 2 4 H2 O þ Mn2þ ! MnO4 þ 8 Hþ þ 5 e 5 2 e ! S2 O82 ! 2 SO42 2 Mn2þ þ 5 S2 O82 þ 8 H2 O ! 2 MnO4 þ 10 SO42 þ 16 Hþ Each side has 2 Mn, 10 S, 16 H, and 48 O and a 6 charge.
2þ CoðNO2 Þ63 þ MnO þ Mn2þ þ NO3 4 ! Co
Step 1
(acidic solution)
Write half-reaction equations. Balance except H and O. CoðNO2 Þ63 ! Co2þ þ 6 NO3 MnO4 ! Mn2þ
- 243 -
- Chapter 17 Step 2
Balance H and O using H2O and Hþ 6 H2 O þ CoðNO2 Þ63 ! Co2þ þ 6 NO3 þ 12 Hþ
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O Balance electrically with e 6 H2 O þ CoðNO2 Þ63 ! Co2þ þ 6 NO3 þ 12 Hþ þ 11 e
Step 4
5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O Equalize the loss and gain of electrons. 5 ð6 H2 O þ CoðNO2 Þ63 ! Co2þ þ 6 NO3 þ 12 Hþ þ 11 e Þ
Step 5
11 ð5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 OÞ Add the half-reactions 5 CoðNO2 Þ63 þ 11 MnO4 þ 28 Hþ ! 5 Co2þ þ 30 NO3 þ 11 Mn2þ þ 14 H2 O Each side has 5 Co, 30 N, 11 Mn, 28 H, 104 O and a þ 2 charge.
18. (a)
Mo2 O3 þ MnO4 ! MoO3 þ Mn2þ Step 1
Step 2
(acid solution)
Write half-reactions equations. Balance except H and O Mo2 O3 ! 2 MoO3 MnO4 ! Mn2þ Balance H and O using H2O and Hþ 3 H2 O þ Mo2 O3 ! 2 MoO3 þ 6 Hþ
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O Balance electrically with electrons 3 H2 O þ Mo2 O3 ! 2 MoO3 þ 6 Hþ þ 6 e
Steps 4 and 5
5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O Equalize the loss and gain of electrons; add the half-reactions. 5 ð3 H2 O þ Mo2 O3 ! 2 MoO3 þ 6 Hþ þ 6 e Þ 6 ð5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 OÞ 5 Mo2 O3 þ 6 MnO4 þ 18 Hþ ! 10 MoO3 þ 6 Mn2þ þ 9 H2 O
(b)
BrO þ CrðOHÞ4 ! Br þ CrO42 Step 1
(basic solution)
Write half-reaction equation. Balance except H and O BrO ! Br
Step 2
CrðOHÞ4 ! CrO42 Balance H and O using H2O and Hþ 2 Hþ þ BrO ! Br þ H2 O 2 CrðOHÞ þ 4 Hþ 4 ! CrO4
- 244 -
- Chapter 17 Step 3
Add OH ions to both sides (same number as Hþ) 2 OH þ 2 Hþ þ BrO ! Br þ H2 O þ 2 OH
Step 4
4 OH þ CrðOHÞ4 ! CrO42 þ 4 Hþ þ 4 OH Combine Hþ and OH to form H2O; cancel H2O where possible 2 H2 O þ BrO ! Br þ H2 O þ 2 OH 4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O
Step 5
H2 O þ BrO ! Br þ 2 OH ð1 H2 O cancelledÞ Balance electrically with electrons 2 e þ H2 O þ BrO ! Br þ 2 OH
Steps 6 and 7
4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e Equalize loss and gain of electrons; add the half-reactions 3 ð2 e þ H2 O þ BrO ! Br þ 2 OH Þ 2 ð4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e Þ 3 BrO þ 2 CrðOHÞ4 þ 2 OH ! 3 Br þ 2 CrO42 þ 5 H2 O
(c)
S2 O32 þ MnO4 ! SO42 þ MnO2 Step 1
(basic solution)
Write half-reaction equations. Balance except H and O. S2 O32 ! 2 SO42
Step 2
MnO4 ! MnO2 Balance H and O using H2O and Hþ 5 H2 O þ S2 O32 ! 2 SO42 þ 10 Hþ
Step 3
4 Hþ þ MnO4 ! MnO2 þ 2 H2 O Add OH ions to both sides (same number as Hþ) 10 OH þ 5 H2 O þ S2 O32 ! 2 SO42 þ 10 Hþ þ 10 OH
Step 4
4 OH þ 4 Hþ þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH Combine Hþ and OH to form H2O; cancel H2O where possible 10 OH þ 5 H2 O þ S2 O32 ! 2 SO42 þ 10 H2 O 4 H2 O þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH 10 OH þ S2 O32 ! 2 SO42 þ 5 H2 O ð5 H2 O cancelledÞ
Step 5
2 H2 O þ MnO4 ! MnO2 þ 4 OH Balance electrically with electrons
ð2 H2 O cancelledÞ
10 OH þ S2 O32 ! 2 SO42 þ 5 H2 O þ 8 e Step 6 and 7
3 e þ 2 H2 O þ MnO4 ! MnO2 þ 4 OH Equalize loss and gain of electrons; add half-reactions. 3 ð10 OH þ S2 O32 ! 2 SO42 þ 5 H2 O þ 8 e Þ 8 ð3 e þ 2 H2 O þ MnO4 ! MnO2 þ 4 OH Þ 3 S2 O32 þ 8 MnO4 þ H2 O ! 6 SO42 þ 8 MnO2 þ 2 OH Each side has 6 S, 8 Mn, 2 H, 42 O and a 14 charge.
- 245 -
- Chapter 17 19. – Voltage source
+ Anode (+)
Cathode (–)
Br– H3O+ Solution of HBr
20. (a) (b) (c) 21. (a) (b) (c)
Pb þ SO42 ! PbSO4 þ 2 e PbO2 þ SO42 þ 4 Hþ þ 2 e ! PbSO4 þ 2 H2 O The first reaction is oxidation (Pb0 is oxidized to Pb2þ). The second reaction is reduction (Pb4þ is reduced to Pb2þ). The first reaction (oxidation) occurs at the anode of the battery. The oxidizing agent is KMnO4. The reducing agent is HCl. 5 moles of electrons 5 e þ Mn7þ ! Mn2þ 5 mol e 6:022 1023 e electrons ¼ 3:011 1024 mol KMnO4 mol e mol KMnO4
22. Zinc is a more reactive metal than copper so when corrosion occurs the zinc preferentially reacts. Zinc is above hydrogen in the Activity series of metals; copper is below hydrogen. 23. 3 Cu þ 8 HNO3 ! 3 Cu(NO3)2 þ 2 NO þ 4 H2O 1 mol Cu 2 mol NO 22:4 L ¼ 17:7 L NO ð75:5 g CuÞ 63:55 g 3 mol Cu mol 1 mol HNO3 2 mol NO 22:4 L ð55:0 g HNO3 Þ ¼ 4:89 L NO 63:02 g 8 mol HNO3 mol 4.89 L of NO gas at STP will be produced; HNO3 is limiting. Cu is oxidized; N is reduced. 24. K2S2O8 þ H2C2O4 ! K2SO4 þ H2SO4 þ 2 CO2 1 mol K2 S2 O8 2 mol CO2 22:4 L ð25:5 g K2 S2 O8 Þ ¼ 4:23 L CO2 270:3 g 1 mol HNO3 mol 1 mol H2 C2 O4 2 mol CO2 22:4 L ð35:5 g H2 C2 O4 Þ ¼ 17:7 L CO2 90:04 g 1 mol H2 C2 O4 mol 4.23 L CO2 will be produced; K2S2O8 is limiting. C is oxidized; S is reduced.
- 246 -
- Chapter 17 25. 5 H2O2 þ 2 KMnO4 þ 3 H2SO4 ! 5 O2 þ 2 MnSO4 þ K2SO4 þ 8 H2O mL H2O2 ! g H2O2 ! mol H2O2 ! mol KMnO4 ! g KMnO4 1:031 g 9:0 g H2 O2 1 mol ð100: mL H2 O2 solutionÞ 100: g H2 O2 solution 34:02 g mL 2 mol KMnO4 158:0 g ¼ 17 g KMnO4 5 mol H2 O2 mol 26. 3 Zn þ 2 Fe3þ ! 3 Zn2þ þ 2 Fe 1:2 mol FeCl3 3 mol Zn 65:39 g ð0:0250 L FeCl3 Þ ¼ 2:94 g Zn L 2 mol FeCl3 mol 27. Cr2 O72 þ 6 Fe2þ þ 14 Hþ ! 2 Cr3þ þ 6 Fe3þ þ 7 H2 O mL FeSO4 ! mol FeSO4 ! mol Cr2 O72 ! mL Cr2 O72 0:200 mol 1 mol Cr2 O72 1000 mol ð60:0 mL FeSO4 Þ 6 mol FeSO4 1000 mL 0:200 mol ¼ 10:0 mL of 0:200 M K2 Cr2 O7 28. 2 Al þ 2 OH þ 6 H2 O ! 2 AlðOHÞ4 þ 3 H2
g Al ! mol Al ! mol H2 1 mol Al 3 mol H2 ð100:0 g AlÞ ¼ 5:560 mol H2 2 mol Al 26:98 g 29. (a)
Cuþ ! Cu2þ is an oxidation, but when electrons are gained reduction should occur. Cuþ þ e ! Cu0
(b)
or Cuþ ! Cu2þ þ e
When Pb2þ is reduced, it requires two individual electrons. Pb2þ þ 2 e ! Pb0. An electron has only a single negative charge (e).
30. The electrons lost by the species undergoing oxidation must be gained (or attracted) by another species which then undergoes reduction. 31. A(s) þ B2þ(aq) ! NR B2þ cannot take e from A A(s) þ Cþ(aq) ! NR Cþ cannot take e from A þ 2þ Cþ takes e from D D(s) þ 2 C (aq) ! 2C(s) þ D (aq) 2þ 2þ B(s) þ D (aq) ! D(s) þ B (aq) D2þ takes 2 e from B 2þ 2þ Therefore, B is least able to attract e , then D , then Cþ, then Aþ 32.
Sn4þ can only be an oxidizing agent. Sn0 can only be a reducing agent. Sn2þ can be both oxidizing and reducing.
Sn4þ þ 2 e ! Sn2þ Sn4þ þ 4 e ! Sn0 Sn0 ! Sn2þ þ 2 e Sn0 ! Sn4þ þ 4 e Sn2þ þ 2 e ! Sn0 Sn2þ ! Sn4þ þ 2 e
- 247 -
(oxidizing) (reducing)
- Chapter 17 33. Mn(OH)2 MnF3 MnO2 K2MnO4 KMnO4
þ2 þ3 þ4 þ6 þ7
KMnO4 is the best oxidizing agent of the group, since its greater oxidation number (þ7) makes it very attractive to electrons.
34. Equations (a) and (b) represent oxidation (a) Mg ! Mg2þ þ 2 e (b) SO2 ! SO3; (S4þ ! S6þ þ 2 e) 35. (a) (b)
(c)
36. (a) (b) (c) (d)
MnO2 þ 2 Br þ 4 Hþ ! Mn2þ þ Br2 þ 2 H2O mL Mn2þ ! mol Mn2þ ! mol MnO2 ! g MnO2 0:05 mol 1 mol MnO2 86:94 g 2þ ð100:0 mL Mn Þ ¼ 0:4 g MnO2 1 mol Mn2þ 1000 mL mol 0:05 mol 1 mol Br2 ¼ 0:005 mol Br2 ð100:0 mL Mn2þ Þ 1000 mL 1 mol Mn2þ nRT PV ¼ nRT V¼ P 0:005 mol 0:0821 L atm ð323 KÞ ¼ 0:09 L Br2 vapor V¼ 1:4 atm mol K F2 þ 2 Cl ! 2 F þ Cl2 Br2 þ Cl ! NR I2 þ Cl ! NR Br2 þ 2 I ! 2 Br þ I2
37. Mn(s) þ 2 HCl(aq) ! Mn2þ(aq) þ H2(g) þ 2 Cl(aq) 38. 4 Zn þ NO3 þ 10 Hþ ! 4 Zn2þ þ NH4þ þ 3 H2 O
See Exercise 13(a).
39. Equation 1
2
3
4
5
a
C oxidized
S oxidized
N oxidized
S oxidized
O22 oxidized
b
O2 reduced
N reduced
Cu reduced
O22 reduced
O22 reduced
c
O2, O.A.
HNO3, O.A.
CuO, O.A.
H2O2,O.A.
H2O2, O.A.
d
C3H8, R.A.
H2S, R.A.
NH3, R.A.
Na2SO3, R.A.
H2O2, R.A.
e
C
S2 ! S0
N3 ! N20
S4þ ! S6þ
O22 ! O20
f
O0 ! O2
N5þ ! N2þ
Cu2þ ! Cu0
O22 ! O2
O22 ! O2
2 3
2 þ
! C4þ
O.A. ¼ Oxidizing agent R.A. ¼ Reducing agent
- 248 -
- Chapter 17 40. Pb þ 2 Agþ ! 2 Ag þ Pb2þ (a) Pb is the anode (b) Ag is the cathode (c) Oxidation occurs at Pb (anode) (d) Reduction occurs at Ag (cathode) (e) Electrons flow from the lead electrode through the wire to the silver electrode. (f) Positive ions flow through the salt bridge towards the negatively charged strip of silver; negative ions flow toward the positively charged strip of lead. Salt bridge Ag
Pb
Pb2+ NO–3
Ag+ NO3–
41. 8 KI þ 5 H2 SO4 ! 4 I2 þ H2 S þ 4 K2 SO4 þ 4 H2 O start with grams I2 and work towards g KI g I2 ! mol I2 ! mol KI ! g KI 1 mol 8 mol KI 166:0 g ¼ 3:65 g KI in sample ð2:79 g I2 Þ 253:8 g 4 mol I2 mol 3:65 g KI ð100Þ ¼ 91:3% KI 4:00 g sample 42. 3 Ag þ 4 HNO3 ! 3 AgNO3 þ NO þ 2 H2 O mol Ag ! mol NO 1 mol NO ð0:500 mol AgÞ ¼ 0:167 mol NO 3 mol Ag nRT PV ¼ nRT V¼ P 1 atm ¼ 0:979 atm P ¼ ð744 torrÞ 760: torr T ¼ 301 K ð0:167 mol NOÞð0:0821 L atm=mol KÞð301 KÞ ¼ 4:22 L NO V¼ ð0:979 atmÞ
- 249 -
