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Exercise 11.10 (Cont.)

Exercise 11.10: Continued

Use the Table of Standard Reduction Potentials to write the net equations for the processes described below and determine their standard cell potentials. Indicate any subscripted characters with an underscore (_) and any superscipted characters with a carat (^). For example, NH_4^1+ for NH41+. Include any coefficients other than 1. Omit any spaces.
Metallic copper is placed in 1 M AgNO3.
Oxidation half reaction:
o_Cu_s Oxidation half reaction: Cu equilibrium arrow Cu2+ + 2e1–
    equilibrium arrow    
o_Cu^2+_s Oxidation half reaction: Cu equilibrium arrow Cu2+ + 2e1–
    +    
2_0__ Oxidation half reaction: Cu equilibrium arrow Cu2+ + 2e1– e1–
Red. 1                                  Ox. 1
Reduction half reaction:
o_Ag^1+_s Reduction half reaction: Ag1+ + 1e1– equilibrium arrow Ag
    +    
1_0__ Reduction half reaction: Ag1+ + 1e1– equilibrium arrow Ag e1–
    equilibrium arrow    
o_Ag_s Reduction half reaction: Ag1+ + 1e1– equilibrium arrow Ag
Ox. 2                                                                                  Red. 2
The number of electrons transferred (or the LCM) is: 2_0__ The lowest common multiple (LCM) of the electrons gained in the reduction and lost in the oxidation is 2.
Write the net equation (if any).
o_Cu_s Net reaction: Cu + 2Ag1+ Cu2+ + 2Ag
    +    
o_2Ag^1+_s Net reaction: Cu + 2Ag1+ Cu2+ + 2Ag
    
o_Cu^2+_s Net reaction: Cu + 2Ag1+ Cu2+ + 2Ag
    +    
o_2Ag_s Net reaction: Cu + 2Ag1+ Cu2+ + 2Ag
Red. 1                          Ox. 2                          Ox. 1                              Red. 2
The standard cell potential in volts is: 0.46___ The cathode is the Ag1+/Ag couple, so ℰcathode = +0.80 V.
The anode is the Cu2+/Cu couple, so ℰanode = +0.34 V.
The cell potential is ℰ = ℰcathode – ℰanode = 0.80 – 0.34 = 0.46 V.
Metallic chromium is placed in 1 M CuSO4.
Oxidation half reaction:
o_Cr_s Oxidation half reaction: Cr equilibrium arrow Cr3+ + 3e1–
    equilibrium arrow    
o_Cr^3+_s Oxidation half reaction: Cr equilibrium arrow Cr3+ + 3e1–
    +    
3_0__ Oxidation half reaction: Cr equilibrium arrow Cr3+ + 3e1– e1–
Red. 1                              Ox. 1
Reduction half reaction:
o_Cu^2+_s Reduction half reaction: Cu2+ + 2e1– equilibrium arrow Cu
    +    
2_0__ Reduction half reaction: Cu2+ + 2e1– equilibrium arrow Cu e1–
    equilibrium arrow    
o_Cu_s Reduction half reaction: Cu2+ + 2e1– equilibrium arrow Cu
Ox. 2                                                                                      Red. 2
The number of electrons transferred (or the LCM) is: 6_0__ The lowest common multiple (LCM) of the electrons gained in the reduction and lost in the oxidation is 6.
Write the net equation (if any).
o_2Cr_s Net reaction: 2Cr + 3Cu2+ 2Cr3+ + 3Cu
    +    
o_3Cu^2+_s Net reaction: 2Cr + 3Cu2+ 2Cr3+ + 3Cu
    
o_2Cr^3+_s Net reaction: 2Cr + 3Cu2+ 2Cr3+ + 3Cu
    +    
o_3Cu_s Net reaction: 2Cr + 3Cu2+ 2Cr3+ + 3Cu
Red. 1                          Ox. 2                          Ox. 1                              Red. 2
The standard cell potential in volts is: 1.08___ The cathode is the Cu2+/Cu couple, so ℰcathode = +0.34 V.
The anode is the Cr3+/Cr couple, so ℰanode = –0.74 V.
The cell potential is ℰ = ℰcathodeanode = 0.34 – (–0.74) = 1.08 V.
Metallic iron is placed in 1 M NiSO4.
Oxidation half reaction:
o_Fe_s Oxidation half reaction: Fe equilibrium arrow Fe2+ + 2e1–
    equilibrium arrow    
o_Fe^2+_s Oxidation half reaction: Fe equilibrium arrow Fe2+ + 2e1–
    +    
2_0__ Oxidation half reaction: Fe equilibrium arrow Fe2+ + 2e1– e1–
Red. 1                                  Ox. 1
Reduction half reaction:
o_Ni^2+_s Reduction half reaction: Ni2+ + 2e1– equilibrium arrow Ni
    +    
2_0__ Reduction half reaction: Ni2+ + 2e1– equilibrium arrow Ni e1–
    equilibrium arrow    
o_Ni_s Reduction half reaction: Ni2+ + 2e1– equilibrium arrow Ni
Ox. 2                                                                                      Red. 2
The number of electrons transferred (or the LCM) is: 2_0__ The lowest common multiple (LCM) of the electrons gained in the reduction and lost in the oxidation is 2.
Write the net equation (if any).
o_Fe_s Net Equation: Fe + Ni2+ Fe2+ + Ni
    +    
o_Ni^2+_s Net Equation: Fe + Ni2+ Fe2+ + Ni
    
o_Fe^2+_s Net Equation: Fe + Ni2+ Fe2+ + Ni
    +    
o_Ni_s Net Equation: Fe + Ni2+ Fe2+ + Ni
Red. 1                          Ox. 2                              Ox. 1                          Red. 2
The standard cell potential in volts is: 0.21___ The cathode is the Ni2+/Ni couple, so ℰcathode = –0.23 V.
The anode is the Fe2+/Fe couple, so ℰanode = –0.44 V.
The cell potential is = ℰcathode – ℰanode = –0.23 – (–0.44) = 0.21 V.
Metallic sodium is added to water.

Note: in some half reactions, there are substances present that are not involved in the electron transfer but are required for a balanced equations. These substances will be referred to as "other." Other substances are usually H2O, OH1–, or H1+ when they are not involved in the electron transfer. This example contains one "Other" substance.
Oxidation half reaction:
o_Na_s Oxidation half reaction: Na equilibrium arrow Na1+ + e1–
    equilibrium arrow    
o_Na^1+_s Oxidation half reaction: Na equilibrium arrow Na1+ + e1–
    +    
1_0__ Oxidation half reaction: Na equilibrium arrow Na1+ + e1– e1–
Red. 1                                  Ox. 1
Reduction half reaction:
o_2H_2O_s Reduction half reaction: 2H2O + 2e1– equilibrium arrow H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
    +    
2_0__ Reduction half reaction: 2H2O + 2e1– equilibrium arrow H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
e1–
    equilibrium arrow    
o_H_2_s Reduction half reaction: 2H2O + 2e1– equilibrium arrow H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
    +    
o_2OH^1-_s Reduction half reaction: 2H2O + 2e1– equilibrium arrow H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
Ox. 2                                                                          Red. 2                         Other
The number of electrons transferred (or the LCM) is: 2_0__ The lowest common multiple (LCM) of the electrons gained in the reduction and lost in the oxidation is 2.
Write the net equation (if any).
o_2Na_s Net Equation: 2Na + 2H2O 2Na1+ + H2 + 2OH1–
  +  
o_2H_2O_s Net Equation: 2Na + 2H2O 2Na1+ + H2 + 2OH1–
    
o_2Na^1+_s Net Equation: 2Na + 2H2O 2Na1+ + H2 + 2OH1–
  +  
o_H_2_s Net Equation: 2Na + 2H2O 2Na1+ + H2 + 2OH1–
  +  
o_2OH^1-_s Net Equation: 2Na + 2H2O 2Na1+ + H2 + 2OH1–
Red. 1                      Ox. 2                          Ox. 1                      Red. 2                     Other
The standard cell potential in volts is: 1.88___ The cathode is the H2O/H2 couple, so ℰcathode = –0.83 V.
The anode is the Na1+/Na couple, so ℰanode = –2.71 V.
The cell potential is ℰ = ℰcathode – ℰanode = –0.83 – (–2.71) = 1.88 V.
Metallic chromium is placed in 1 M AgNO3.
Oxidation half reaction:
o__s Oxidation half reaction: Cr equilibrium arrow Cr3+ + 3e1–
    equilibrium arrow    
o_Cr^3+_s Oxidation half reaction: Cr equilibrium arrow Cr3+ + 3e1–
    +    
3_0__ Oxidation half reaction: Cr equilibrium arrow Cr3+ + 3e1– e1–
Red. 1                                  Ox. 1
Reduction half reaction:
o_Ag^1+_s Reduction half reaction: Ag1+ + 1e1– equilibrium arrow Ag
    +    
1_0__ Reduction half reaction: Ag1+ + 1e1– equilibrium arrow Ag e1–
    equilibrium arrow    
o_Ag_s Reduction half reaction: Ag1+ + 1e1– equilibrium arrow Ag
Ox. 2                                                                                      Red. 2
The number of electrons transferred (or the LCM) is: 3_0__ The lowest common multiple (LCM) of the electrons gained in the reduction and lost in the oxidation is 3.
Write the net equation (if any).
o_Cr_s Net reaction: Cr + 3Ag1+ Cr3+ + 3Ag
    +    
o_3Ag^1+_s Net reaction: Cr + 3Ag1+ Cr3+ + 3Ag
    
o_Cr^3+_s Net reaction: Cr + 3Ag1+ Cr3+ + 3Ag
    +    
o_3Ag_s Net reaction: Cr + 3Ag1+ Cr3+ + 3Ag Net reaction: Cr + 3Ag1+ Cr3+ + 3Ag
Red. 1                          Ox. 2                          Ox. 1                              Red. 2
The standard cell potential in volts is: 1.54___ The cathode is the Ag1+/Ag couple, so ℰcathode = +0.80 V.
The anode is the Cr3+/Cr couple, so ℰanode = –0.74 V.
The cell potential is ℰ = ℰcathode – ℰanode = 0.80 – (–0.74) = 1.54 V.
Metallic copper is placed in 1 M nitric acid.
Oxidation half reaction:
o_Cu_s Oxidation half reaction: Cu equilibrium arrow Cu2+ + 2e1–
    equilibrium arrow    
o_Cu^2+_s Oxidation half reaction: Cu equilibrium arrow Cu2+ + 2e1–
    +    
2_0__ Oxidation half reaction: Cu equilibrium arrow Cu2+ + 2e1– e1–
Red. 1                                  Ox. 1
Reduction half reaction:
o_NO_3^1-_s Reduction half reaction: NO31– + 4H1+ + 3e1– equilibrium arrow NO + 2H2O
  +  
o_4H^1+_s Reduction half reaction: NO31– + 4H1+ + 3e1– equilibrium arrow NO + 2H2O
  +  
3_0__ Reduction half reaction: NO31– + 4H1+ + 3e1– equilibrium arrow NO + 2H2O e1–
  equilibrium arrow  
o_NO_s Reduction half reaction: NO31– + 4H1+ + 3e1– equilibrium arrow NO + 2H2O
  +  
o_2H_2O_s Reduction half reaction: NO31– + 4H1+ + 3e1– equilibrium arrow NO + 2H2O
Ox. 2                        Other                                                             Red. 2                    Other
The number of electrons transferred (or the LCM) is: 6_0__ The lowest common multiple (LCM) of the electrons gained in the reduction and lost in the oxidation is 6.
Write the net equation (if any).
o_3Cu_s Net Equation: 3Cu + 2NO31– + 8H1+ 3Cu2+ + 2NO + 4H2O
    +    
o_2NO_3^1-_s Net Equation: 3Cu + 2NO31– + 8H1+ 3Cu2+ + 2NO + 4H2O
    +    
o_8H^1+_s Net Equation: 3Cu + 2NO31– + 8H1+ 3Cu2+ + 2NO + 4H2O
           Red. 1                              Ox. 2                                  Other
o_3Cu^2+_s Net Equation: 3Cu + 2NO31– + 8H1+ 3Cu2+ + 2NO + 4H2O
    +    
o_2NO_s Net Equation: 3Cu + 2NO31– + 8H1+ 3Cu2+ + 2NO + 4H2O
    +    
o_4H_2O_s Net Equation: 3Cu + 2NO31– + 8H1+ 3Cu2+ + 2NO + 4H2O
Ox. 1                                Red. 2                                Other
The standard cell potential in volts is: 0.62___ Nitric acid is reduced, so ℰcathode = +0.96 V.
Copper is oxidized, so ℰanode = 0.34 V.
The cell potential is ℰ = ℰcathode – ℰanode = 0.96 – 0.34 = 0.62 V