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⁺ + 2e^1–

o_Cu^2+_s Oxidation half reaction: Cu equilibrium arrow

Cu²⁺ + 2e^1–

2_0__ Oxidation half reaction: Cu equilibrium arrow

Cu²⁺ + 2e^1– e^1–

Red. 1 Ox. 1

Reduction half reaction:

o_Ag^1+_s Reduction half reaction: Ag¹⁺ + 1e^1– equilibrium arrow

1_0__ Reduction half reaction: Ag¹⁺ + 1e^1– equilibrium arrow

Ag e^1–

o_Ag_s Reduction half reaction: Ag¹⁺ + 1e^1– equilibrium arrow

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 ℰ_cathode = +0.80 V.
The anode is the Cu²⁺/Cu couple, so ℰ_anode = +0.34 V.
The cell potential is ℰ = ℰ_cathode – ℰ_anode = 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⁺ + 3e^1–

o_Cr^3+_s Oxidation half reaction: Cr equilibrium arrow

Cr³⁺ + 3e^1–

3_0__ Oxidation half reaction: Cr equilibrium arrow

Cr³⁺ + 3e^1– e^1–

Red. 1 Ox. 1

Reduction half reaction:

o_Cu^2+_s Reduction half reaction: Cu²⁺ + 2e^1– equilibrium arrow

2_0__ Reduction half reaction: Cu²⁺ + 2e^1– equilibrium arrow

Cu e^1–

o_Cu_s Reduction half reaction: Cu²⁺ + 2e^1– equilibrium arrow

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 ℰ_cathode = +0.34 V.
The anode is the Cr³⁺/Cr couple, so ℰ_anode = –0.74 V.
The cell potential is ℰ = ℰ_cathode – _anode = 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⁺ + 2e^1–

o_Fe^2+_s Oxidation half reaction: Fe equilibrium arrow

Fe²⁺ + 2e^1–

2_0__ Oxidation half reaction: Fe equilibrium arrow

Fe²⁺ + 2e^1– e^1–

Red. 1 Ox. 1

Reduction half reaction:

o_Ni^2+_s Reduction half reaction: Ni²⁺ + 2e^1– equilibrium arrow

2_0__ Reduction half reaction: Ni²⁺ + 2e^1– equilibrium arrow

Ni e^1–

o_Ni_s Reduction half reaction: Ni²⁺ + 2e^1– equilibrium arrow

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 ℰ_cathode = –0.23 V.
The anode is the Fe²⁺/Fe couple, so ℰ_anode = –0.44 V.
The cell potential is = ℰ_cathode – ℰ_anode = –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, OH^1–, 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⁺ + e^1–

o_Na^1+_s Oxidation half reaction: Na equilibrium arrow

Na¹⁺ + e^1–

1_0__ Oxidation half reaction: Na equilibrium arrow

Na¹⁺ + e^1– e^1–

Red. 1 Ox. 1

Reduction half reaction:

o_2H_2O_s Reduction half reaction: 2H₂O + 2e^1– equilibrium arrow

H₂ + 2OH^1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."

2_0__ Reduction half reaction: 2H₂O + 2e^1– equilibrium arrow

H₂ + 2OH^1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other." e^1–

o_H_2_s Reduction half reaction: 2H₂O + 2e^1– equilibrium arrow

H₂ + 2OH^1–
Hydroxide ion is neither oxidized or reduced in this reaction, so it is "Other."

o_2OH^1-_s Reduction half reaction: 2H₂O + 2e^1– equilibrium arrow

H₂ + 2OH^1–
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 + 2OH^1–

o_2H_2O_s Net Equation: 2Na + 2H₂O → 2Na¹⁺ + H₂ + 2OH^1–

→

o_2Na^1+_s Net Equation: 2Na + 2H₂O → 2Na¹⁺ + H₂ + 2OH^1–

o_H_2_s Net Equation: 2Na + 2H₂O → 2Na¹⁺ + H₂ + 2OH^1–

o_2OH^1-_s Net Equation: 2Na + 2H₂O → 2Na¹⁺ + H₂ + 2OH^1–

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 ℰ_cathode = –0.83 V.
The anode is the Na¹⁺/Na couple, so ℰ_anode = –2.71 V.
The cell potential is ℰ = ℰ_cathode – ℰ_anode = –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⁺ + 3e^1–

o_Cr^3+_s Oxidation half reaction: Cr equilibrium arrow

Cr³⁺ + 3e^1–

3_0__ Oxidation half reaction: Cr equilibrium arrow

Cr³⁺ + 3e^1– e^1–

Red. 1 Ox. 1

Reduction half reaction:

o_Ag^1+_s Reduction half reaction: Ag¹⁺ + 1e^1– equilibrium arrow

1_0__ Reduction half reaction: Ag¹⁺ + 1e^1– equilibrium arrow

Ag e^1–

o_Ag_s Reduction half reaction: Ag¹⁺ + 1e^1– equilibrium arrow

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 ℰ_cathode = +0.80 V.
The anode is the Cr³⁺/Cr couple, so ℰ_anode = –0.74 V.
The cell potential is ℰ = ℰ_cathode – ℰ_anode = 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⁺ + 2e^1–

o_Cu^2+_s Oxidation half reaction: Cu equilibrium arrow

Cu²⁺ + 2e^1–

2_0__ Oxidation half reaction: Cu equilibrium arrow

Cu²⁺ + 2e^1– e^1–

Red. 1 Ox. 1

Reduction half reaction:

o_NO_3^1-_s Reduction half reaction: NO₃^1– + 4H¹⁺ + 3e^1– equilibrium arrow

NO + 2H₂O

o_4H^1+_s Reduction half reaction: NO₃^1– + 4H¹⁺ + 3e^1– equilibrium arrow

NO + 2H₂O

3_0__ Reduction half reaction: NO₃^1– + 4H¹⁺ + 3e^1– equilibrium arrow

NO + 2H₂O e^1–

o_NO_s Reduction half reaction: NO₃^1– + 4H¹⁺ + 3e^1– equilibrium arrow

NO + 2H₂O

o_2H_2O_s Reduction half reaction: NO₃^1– + 4H¹⁺ + 3e^1– 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 + 2NO₃^1– + 8H¹⁺ → 3Cu²⁺ + 2NO + 4H₂O

o_2NO_3^1-_s Net Equation: 3Cu + 2NO₃^1– + 8H¹⁺ → 3Cu²⁺ + 2NO + 4H₂O

o_8H^1+_s Net Equation: 3Cu + 2NO₃^1– + 8H¹⁺ → 3Cu²⁺ + 2NO + 4H₂O

→ Red. 1 Ox. 2 Other

o_3Cu^2+_s Net Equation: 3Cu + 2NO₃^1– + 8H¹⁺ → 3Cu²⁺ + 2NO + 4H₂O

o_2NO_s Net Equation: 3Cu + 2NO₃^1– + 8H¹⁺ → 3Cu²⁺ + 2NO + 4H₂O

o_4H_2O_s Net Equation: 3Cu + 2NO₃^1– + 8H¹⁺ → 3Cu²⁺ + 2NO + 4H₂O

Ox. 1 Red. 2 Other

The standard cell potential in volts is: 0.62___ Nitric acid is reduced, so ℰ_cathode = +0.96 V.
Copper is oxidized, so ℰ_anode = 0.34 V.
The cell potential is ℰ = ℰ_cathode – ℰ_anode = 0.96 – 0.34 = 0.62 V