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 Cu2+ + 2e1–
o_Cu^2+_s
Oxidation half reaction: Cu Cu2+ + 2e1–
+
2_0__
Oxidation half reaction: Cu Cu2+ + 2e1–
e1–
Red. 1 Ox. 1
Reduction half reaction:
o_Ag^1+_s
Reduction half reaction: Ag1+ + 1e1– Ag
+
1_0__
Reduction half reaction: Ag1+ + 1e1– Ag
e1–
o_Ag_s
Reduction half reaction: Ag1+ + 1e1– 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).
Ox. 1 Red. 2
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
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.
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 Cr3+ + 3e1–
o_Cr^3+_s
Oxidation half reaction: Cr Cr3+ + 3e1–
+
3_0__
Oxidation half reaction: Cr Cr3+ + 3e1–
e1–
Red. 1 Ox. 1
Reduction half reaction:
o_Cu^2+_s
Reduction half reaction: Cu2+ + 2e1– Cu
+
2_0__
Reduction half reaction: Cu2+ + 2e1– Cu
e1–
o_Cu_s
Reduction half reaction: Cu2+ + 2e1– 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).
Ox. 1 Red. 2
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
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 = ℰcathode –anode = 0.34 – (–0.74) = 1.08 V.
The anode is the Cr3+/Cr couple, so ℰanode = –0.74 V.
The cell potential is ℰ = ℰcathode –
Metallic iron is placed in 1 M NiSO4.
Oxidation half reaction:
o_Fe_s
Oxidation half reaction: Fe Fe2+ + 2e1–
o_Fe^2+_s
Oxidation half reaction: Fe Fe2+ + 2e1–
+
2_0__
Oxidation half reaction: Fe Fe2+ + 2e1–
e1–
Red. 1 Ox. 1
Reduction half reaction:
o_Ni^2+_s
Reduction half reaction: Ni2+ + 2e1– Ni
+
2_0__
Reduction half reaction: Ni2+ + 2e1– Ni
e1–
o_Ni_s
Reduction half reaction: Ni2+ + 2e1– 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.
The anode is the Fe2+/Fe couple, so ℰanode = –0.44 V.
The cell potential is
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.
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 Na1+ + e1–
o_Na^1+_s
Oxidation half reaction: Na Na1+ + e1–
+
1_0__
Oxidation half reaction: Na Na1+ + e1–
e1–
Red. 1 Ox. 1
Reduction half reaction:
Red. 2 Other
o_2H_2O_s
Reduction half reaction: 2H2O + 2e1– H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
+ Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
2_0__
Reduction half reaction: 2H2O + 2e1– H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other." e1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other." e1–
o_H_2_s
Reduction half reaction: 2H2O + 2e1– H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
+ Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
o_2OH^1-_s
Reduction half reaction: 2H2O + 2e1– H2 + 2OH1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."
Ox. 2 Hydroxide ion is neither oxidized or reduced in this reaction, so it is "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).
Other
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
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.
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 Cr3+ + 3e1–
o_Cr^3+_s
Oxidation half reaction: Cr Cr3+ + 3e1–
+
3_0__
Oxidation half reaction: Cr Cr3+ + 3e1–
e1–
Red. 1 Ox. 1
Reduction half reaction:
o_Ag^1+_s
Reduction half reaction: Ag1+ + 1e1– Ag
+
1_0__
Reduction half reaction: Ag1+ + 1e1– Ag
e1–
o_Ag_s
Reduction half reaction: Ag1+ + 1e1– 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).
Ox. 1 Red. 2
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
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.
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 Cu2+ + 2e1–
o_Cu^2+_s
Oxidation half reaction: Cu Cu2+ + 2e1–
+
2_0__
Oxidation half reaction: Cu Cu2+ + 2e1–
e1–
Red. 1 Ox. 1
Reduction half reaction:
Other
o_NO_3^1-_s
Reduction half reaction: NO31– + 4H1+ + 3e1– NO + 2H2O
+
o_4H^1+_s
Reduction half reaction: NO31– + 4H1+ + 3e1– NO + 2H2O
+
3_0__
Reduction half reaction: NO31– + 4H1+ + 3e1– NO + 2H2O
e1–
o_NO_s
Reduction half reaction: NO31– + 4H1+ + 3e1– NO + 2H2O
+
o_2H_2O_s
Reduction half reaction: NO31– + 4H1+ + 3e1– NO + 2H2O
Ox. 2 Other 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).
Ox. 2 Other
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
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
Copper is oxidized, so ℰanode = 0.34 V.
The cell potential is ℰ = ℰcathode – ℰanode = 0.96 – 0.34 = 0.62 V