(aq) 5VO2 + (aq) + Mn 2+ (aq) + 10H + + 4H 2 O. (aq) 5VO2 + (aq) + Mn 2+ (aq) + 2H + (aq) basic solution. MnO2 + 2H 2 O) 3H 2 O + I IO 3
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1 Chem 1515 Section 2 Problem Set #15 Spring 1998 Name ALL work must be shown to receive full credit. Due Due in lecture at 1:30 p.m. Friday, May 1st. PS15.1. Balance the following oxidation-reduction reactions using the half-reaction method. (a) VO 2+ (aq) + MnO 4 (aq) VO2 + (aq) + Mn 2+ (aq) acidic solution (5e + 8H + + MnO 4 Mn H 2 O) 5(H 2 O + VO 2+ VO H + + e ) 8H + + 5H 2 O+ 5VO 2+ (aq) + MnO 4 (aq) 5VO2 + (aq) + Mn 2+ (aq) + 10H + + 4H 2 O H 2 O+ 5VO 2+ (aq) + MnO 4 (aq) 5VO2 + (aq) + Mn 2+ (aq) + 2H + (b) NO 3 (aq) + Cu(s) Cu 2+ (aq) + NO(g) acidic solution 2(3e + 4H + + NO 3 NO + 2H 2 O) 3(Cu Cu e ) 3Cu + 8H + + 2NO 3 2NO + 4H 2 O + 3Cu 2+ (c) Fe 3+ (aq) + NH 3 OH + (g) Fe 2+ (aq) + N 2 O(g) acidic solution 4(Fe 3+ (aq) + 1e Fe 2+ (aq) ) 2NH 3 OH + (g) N 2 O(g) + H 2 O + 6H + + 4e 4Fe 3+ (aq) + 2NH 3 OH + (g) N 2 O(g) + H 2 O + 6H + + 4Fe 2+ (aq) (d) MnO 4 (aq) + I (aq) MnO2 (s) + IO 3 (aq) basic solution 2(3e + 4H + + MnO 4 MnO2 + 2H 2 O) 3H 2 O + I IO 3 + 6H + + 6e 3H 2 O + I + 8H + + 2MnO 4 2MnO2 + 4H 2 O + IO 3 + 6H + I + 2H + + 2MnO 4 2MnO2 + H 2 O + IO 3 + 2OH + 2OH I + 2H 2 O + 2MnO 4 2MnO2 + H 2 O + IO 3 + 2OH I + H 2 O + 2MnO 4 2MnO2 + IO 3 + 2OH (e) As 2 S 3 (aq) + H 2 O 2 (aq) AsO 4 3 (aq) + SO4 2 (aq) basic solution 4(2e + 2H + + H 2 O 2 2H 2 O) 20H 2 O + As 2 S 3 (aq) 2AsO 4 3 (aq) + 3SO4 2 (aq) + 20H + + 8e 8H H 2 O + As 2 S 3 (aq) + 4H 2 O 2 (aq) 2AsO 4 3 (aq) + 3SO4 2 (aq) + 8H2 O + 20H + 12H 2 O + As 2 S 3 (aq) + 4H 2 O 2 (aq) 2AsO 4 3 (aq) + 3SO4 2 (aq) + 12H OH + 12OH 12H 2 O + 12OH + As 2 S 3 (aq) + 4H 2 O 2 (aq) 2AsO 4 3 (aq) + 3SO4 2 (aq) + 12H2 O 12OH + As 2 S 3 (aq) + 4H 2 O 2 (aq) 2AsO 4 3 (aq) + 3SO4 2 (aq) May 2, Spring 1998
2 (f) S 2 O 3 2 (aq) + Cl 2 (g) SO 4 2 (aq) + Cl (aq) acidic solution 4(2e + Cl 2 (g) 2Cl (aq) ) S 2 O 3 2 (aq) + 5H 2 O 2SO 4 2 (aq) + 10H + + 8e S 2 O 3 2 (aq) + 4Cl 2 (g) + 5H 2 O 2SO 4 2 (aq) + 8Cl (aq) + 10H + (g) Bi(OH) 3 (aq) + Sn(OH) 3 (aq) Sn(OH)6 2 + Bi(s) basic solution 2(3e + 3H + + Bi(OH) 3 Bi + 3H 2 O) 3(3H 2 O + Sn(OH) 3 Sn(OH) H + + 2e ) 9H 2 O + 3Sn(OH) 3 + 6H + + 2Bi(OH) 3 2Bi + 6H 2 O + 3Sn(OH) H + 3H 2 O + 3Sn(OH) 3 + 2Bi(OH) 3 2Bi + 3Sn(OH) H + + 3OH + 3OH 3OH + 3H 2 O + 3Sn(OH) 3 + 2Bi(OH) 3 2Bi + 3Sn(OH) H 2 O 3OH + 3Sn(OH) 3 + 2Bi(OH) 3 2Bi + 3Sn(OH) 6 2 (h) H 2 O 2 (aq) + MnO4 (aq) O 2 (g) + MnO 2 (s) basic solution 3(H 2 O 2 O 2 + 2H + + 2e ) 2(3e + 4H + + MnO 4 MnO2 + 2H 2 O) 3H 2 O 2 + 8H + + 2MnO 4 2MnO2 + 4H 2 O + 3O 2 + 6H + 3H 2 O 2 + 2H + + 2MnO 4 2MnO2 + 4H 2 O + 3O 2 + 2OH + 2OH 3H 2 O 2 + 2H 2 O + 2MnO 4 2MnO2 + 4H 2 O + 3O 2 + 2OH 3H 2 O 2 + 2MnO 4 2MnO2 + 2H 2 O + 3O 2 + 2OH (i) Fe(CrO 2 ) 2 + Na 2 CO 3 + O 2 Fe 2 O 3 + Na 2 CrO 4 + CO 2 acidic solution 2(7H 2 O + 2Fe(CrO 2 ) 2 + 4Na 2 CO 3 Fe 2 O 3 + 4Na 2 CrO 4 + 4CO H e ) 7(4e + 4H + + O 2 2H 2 O) 28H H 2 O + 4Fe(CrO 2 ) 2 + 7O 2 + 8Na 2 CO 3 2Fe 2 O 3 + 8Na 2 CrO 4 + 8CO H H 2 O 4Fe(CrO 2 ) 2 + 7O 2 + 8Na 2 CO 3 2Fe 2 O 3 + 8Na 2 CrO 4 + 8CO 2 (j) S 8 (aq) S 2 O 3 2 (aq) + S 2 (aq) basic solution (12H 2 O + S 8 4S 2 O H e ) (16e + S 8 8S 2 ) 12H 2 O + 2S 8 4S 2 O H + + 8S 2 6H 2 O + S 8 2S 2 O H + + 4S OH + 12OH 6H 2 O + S OH 2S 2 O H 2 O + 4S 2 S OH 2S 2 O H 2 O + 4S 2 May 2, Spring 1998
3 PS15.2. Draw a diagram of the cells in which the following reactions occur. In each case, label the anode and cathode, the anode and cathode electrode material, the half-reaction at each electrode, the ions in the anode and cathode compartments and salt bridge, the direction of electron flow, and the direction of ion movement. (a) Al(s) + Sn 2+ (aq) Al 3+ (aq) + Sn(s) Al(s) Al 3+ (aq) + 3e 2e + Sn 2+ (aq) Sn(s) (b) Al(s) + 3H + (aq) Al 3+ (aq) + 3/2H 2 (g) (c) Al(s) Al 3+ (aq) + 3e 2e + 2H + (aq) H 2 (g) NiO 2 (s) + 4H + (aq) + 2Ag(s) Ni 2+ (aq) + 2Ag + (aq) + 2H 2 O(l) May 2, Spring 1998
4 PS15.3. Write the balanced chemical equation for the overall cell reaction represented in each of the following cell notations. (a) Sn(s) Sn 2+ (aq) Ag + (aq) Ag(s) (b) Pt(s) Fe 2+ (aq), Fe 3+ (aq) (Pt(s))Cl 2 (g) Cl (aq) (c) Pb(s),PbSO 4 (s) SO 4 2 (aq) H + (aq), SO4 2 (aq) PbO2 (s),pbso 4 PS15.4. Which of the following species are reduced by Ag? Cl 2 Fe 3+ Pb 2+ I 2 NO 3 (in H + ) Ag Ag + + 1e e + Cl 2 2Cl e + Fe 3+ Fe e + Pb 2+ Pb e + I 2 2I e + NO 3 + 4H + NO + 2H 2 O To determine which combination will produce a spontaneous reaction, the E for the reduction reaction must be more positive than volts. The only species to fulfill that criteria are Cl 2 and NO 3 (in 1 M H + ). So silver will be oxidized when combined with either Cl 2 or NO 3 (in 1 M H + ). PS15.5. Which of the following species are oxidized by nitrate in acidic solution? Cl Fe 2+ Cu Au H + 3e + NO 3 + 4H + NO + 2H 2 O Cl 2e + Cl Fe 2+ 1e + Fe Cu Cu e Au Au e H + no possible reaction To determine which combination will produce a spontaneous reaction, the E for the oxidation reaction must be more positive than volts. The only species to fulfill that criteria are Fe 2+ and Cu. So nitrate in 1 M H + will be reduced when combined with either substance. May 2, Spring 1998
5 PS15.6. Select a suitable species for each of the following (a) an oxidizing agent able to convert Sn to Sn 2+, but not Cu to Cu 2+. Sn Sn e Cu Cu e So the species which is capable of oxidizing Sn, but not oxidizing Cu, must have reduction E which is between v and v. Some examples include Pb 2+, H + and Sn 4+. 2e + Pb 2+ Pb e + 2H + H e + Sn 4+ Sn (b) a reducing agent capable of converting H + to H 2, but not Zn 2+ to Zn. 2e + 2H + H e + Zn 2+ Zn So the species which is capable of reducing H +, but not reducing Zn 2+, must have reduction E which is between 0 v and v. Some examples include Pb, Sn, Ni and Cr. Pb Pb e Sn Sn e Ni Ni e Cr Cr e (c) an metal capable of reacting with HNO 3 but not HCl; it displaces Ag + (aq) put not Cu 2+ (aq). 3e + 4H + + NO 3 NO + 2H 2 O e + 2H + H e + Cu 2+ Cu e + Ag + Ag So the species which is capable of reducing Ag +, but not reducing Cu 2+, must have reduction E which is between and v. The only example from our redox table is Hg. May 2, Spring 1998
6 PS15.7. Calculate E and determine which of the following reactions will occur in the forward direction under standard conditions? Balance the equations in acid solution. (a) Pb(s) + Cl 2 (aq) Pb 2+ (aq) + Cl (g) Pb Pb e Cl 2 + 2e 2Cl Pb + Cl 2 Pb Cl volts The reaction is spontaneous because the overall E is positive. (b) Mn 2+ (aq) + Cr 2 O 7 2 (aq) MnO4 (aq) + Cr 3+ (aq) 6[4H 2 O(l) + Mn 2+ (aq) MnO 4 (aq) + 8H + (aq) + 5e ] [6e + Cr 2 O 7 2 (aq) + 14H + (aq) 2Cr 3+ (aq) + 7H2 O(l)] Mn Cr 2 O H + 6MnO Cr H 2 O volt The reaction is nonspontaneous because the overall E is negative. (c) Ag(s) + NO 3 (aq) Ag + (aq) + NO(g) 3[Ag Ag + + 1e ] e + NO 3 + 4H + NO + 2H 2 O Ag(s) + NO 3 (aq) + 4H + 3Ag + (aq) + NO(g) + 2H 2 O(l) The reaction will occur spontaneously. PS15.8. Use standard reduction potentials to predict the spontaneous reaction, if any, that occurs between the following. If no spontaneous reaction occurs, write NR. (a) Ca 2+ (aq) + Mg(s) Mg Mg e e + Ca 2+ Ca Ca 2+ (aq) + Mg(s) NR The reaction is nonspontaneous because the E is negative! (b) H + (aq) + Cu 2+ (aq) 2e + Cu 2+ Cu e + 2H + H Both reactions are reductions half-reactions. Need one oxidation half-reaction and one reduction half-reaction! No reaction. May 2, Spring 1998
7 PS15.8. (Continued) (c) Cl 2 (g) + Ag(s) Cl 2 + 2e 2Cl Ag Ag + + 1e Cl 2 + 2Ag 2Cl + 2Ag V (d) Mn 2+ (aq) + Cr 2 O 7 2 (aq) + H + (aq) 3[Mn 2+ (aq) + 2H 2 O(l) 2e + 4H + (aq) + MnO 2 (s)] e + Cr 2 O 7 2 (aq) + 14H + (aq) 2Cr 3+ (aq) + 7H2 O(l) Mn 2+ (aq) + Cr 2 O 7 2 (aq) + 2H + (aq) 2Cr 3+ (aq) + H2 O(l) + 3MnO 2 (s) (e) Sn 4+ (aq) + Sn(s) 2e + Sn 4+ Sn Sn Sn e Sn 4+ (aq) + Sn(s) 2Sn 2+ (aq) PS15.9. Calculate G and K for the spontaneous reactions in problem PS15.8. (a) Ca 2+ (aq) + Mg(s) (b) H + (aq) + Cu 2+ (aq) (c) Cl 2 (g) + Ag(s) May 2, Spring 1998
8 PS15.9. (Continued) (d) Mn 2+ (aq) + Cr 2 O 7 2 (aq) + H + (aq) (e) Sn 4+ (aq) + Sn(s) PS Calculate the solubility product constant for the iron(ii) hydroxide, given the following information. Fe(OH) 2 (s) + 2e Fe(s) + 2OH (aq) E = volts Fe 2+ (aq) + 2e Fe(s) E = volts May 2, Spring 1998
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