CHAPTER 19
Electrochemistry
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
19.1. Assign oxidation numbers to the skeleton equation (Step 1).
0 +5 +5 +4
I2 + NO3 - IO3 -
+ NO2
Separate into two incomplete half-reactions (Step 2). Note that iodine is oxidized (increases in
oxidation number), and nitrogen is reduced (decreases in oxidation number).
I2 IO3-
NO3- NO2
Balance each half-reaction separately. The oxidation half-reaction is not balanced in I, so place a
two in front of IO3− (Step 3a). Then add six H2O’s to the left side to balance O atoms (Step 3b),
and add twelve H+ ions to the right side to balance H atoms (Step 3c). Finally, add ten electrons to
the right side to balance the charge (Step 3d). The balanced oxidation half-reaction is
I2 + 6H2O 2IO3− + 12H+ + 10e−
The reduction half-reaction is balanced in N (Step 3a). Add one H2O to the right side to balance
O atoms (Step 3b), and add two H+ ion to the left side to balance H atoms. Finally, add one
electron to the left side to balance the charge (Step 3d).
The balanced reduction half-reaction is
NO3− + 2H+ + e− NO2 + H2O
Multiply the reduction half-reaction by 5 so that, when added, the electrons cancel (Step 4a).
I2 + 6H2O 2IO3− + 12H+ + 10e−
10NO3− + 20H+ + 10e− 10NO2 + 10H2O
I2 + 10NO3− + 20H+ + 6H2O + 10e−
2IO3− + 10NO2 + 12H+ + 10H2O + 10e−
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 756 Chapter 19: Electrochemistry
Simplify the equation by canceling the twelve H+ and six H2O that appear on both sides. The
coefficients do not need to be reduced (Step 4b). The net ionic equation is
I2(s) + 10NO3−(aq) + 8H+(aq) 2IO3−(aq) + 10NO2(g) + 4H2O(l)
19.2. After balancing the equation as though it were in acid solution, you obtain
H2O2 + 2ClO2 2ClO2− + O2 + 2H+
Add two OH− to both sides of the equation (Step 5), and replace the two H+ and two OH− on the
right side with two H2O. No further cancellation is required. The balanced equation for the
reaction in basic solution is
H2O2 + 2ClO2 + 2OH− 2ClO2− + O2 + 2H2O
19.3. Silver ion is reduced at the silver electrode (cathode). The half-reaction is
Ag+(aq) + e− Ag(s)
The nickel electrode (anode) is where oxidation occurs. The half-reaction is
Ni(s) Ni2+(aq) + 2e−
Electron flow in the external circuit is from the nickel electrode (anode) to the silver electrode
(cathode). Positive ions will flow in the solution portion of the circuit opposite to the direction of
the electrons. A sketch of the cell is given below:
e
Ni Ag
(anode) (cathode)
– Salt bridge +
Ni2+ Ag+
Ni Ni2+ + 2e Ag+ + e– Ag
19.4. The notation for the cell is Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s).
19.5. The half-cell reactions are
Cd(s) Cd2+(aq) + 2e−
2H+(aq) + 2e− H2(g)
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 19: Electrochemistry 757
Summing the half-cell reactions gives the overall cell reaction.
Cd(s) + 2H+(aq) Cd2+(aq) + H2(g)
19.6. The half-reactions are
Zn(s) Zn2+(aq) + 2e−
Cu2+(aq) + 2e− Cu(s)
n equals 2, and the maximum work for the reaction as written is
wmax = −nFEcell = −2 9.6485 104 C 1.10 V = −2.122 105 VC = −2.12 105 J
For 6.54 g of zinc metal, the maximum work is
1 mol Zn 2.123 105 J
6.54 g Zn = −2.123 104 = −2.12 104 J
65.39 g Zn 1 mol Zn
19.7. The half-reactions and corresponding electrode potentials are
Ag-(aq) + e - Ag(s) 0.80 V
NO3-(aq) + 4H-(aq) + 4e - NO(g) + 2H2O(l) 0.96 V
The stronger oxidizing agent is the one involved in the half-reaction with the more positive
standard electrode potential, so NO3− is the stronger oxidizing agent.
19.8. In this reaction, Cu2+ is the oxidizing agent on the left side; I2 is the oxidizing reagent on the right
side. The corresponding standard electrode potentials are
Cu2+(aq) + 2e− Cu(s); E = 0.34 V
I2(s) + 2e− 2I−(aq); E = 0.54 V
The stronger oxidizing agent is the one involved in the half-reaction with the more positive
standard electrode potential, so I2 is the stronger oxidizing agent. The reaction is nonspontaneous
as written.
19.9. The reduction half-reactions and standard electrode potentials are
Zn2+(aq) + 2e− Zn(s); EZn = −0.76 V
Cu2+(aq) + 2e− Cu(s); ECu = 0.34 V
Reverse the first half-reaction and its half-cell potential to obtain
Zn(s) Zn2+(aq) + 2e−; −EZn = 0.76 V
Cu2+(aq) + 2e− Cu(s); ECu = 0.34 V
Obtain the cell potential by adding the half-cell potentials.
Ecell = ECu − EZn = 0.34 V + 0.76 V = 1.10 V
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 758 Chapter 19: Electrochemistry
19.10. The half-cell reactions, the corresponding half-cell potentials, and their sums are displayed below:
Sn2+(aq) Sn4+(aq) + 2e− −E = −0.15 V
2Hg2+(aq) + 2e− Hg22+(aq) E = 0.90 V
Sn2+(aq) + 2Hg2+(aq) Sn4+(aq) + Hg22+(aq) Ecell = 0.75 V
Note that each half-reaction involves two electrons; hence n equals 2. Also, Ecell equals 0.75 V,
and the faraday constant, F, is 9.6485 104 C. Therefore,
G = −nFEcell = −2 9.6485 104 C 0.75 V = −1.4472 105 J = −1.4 105 J
Thus, the standard free-energy change is −1.4 102 kJ.
19.11. Write the equation with Gf’s beneath each substance.
Hence,
G = ΣnGf(products) − ΣmGf(reactants)
= [(−454.8) − 65.52] kJ = −520.32 kJ = −5.2032 105 J
Obtain n by splitting the reaction into half-reactions.
Mg(s) Mg2+(aq) + 2e−
Cu2+(aq) + 2e− Cu(s)
Each half-reaction involves two electrons, so n equals 2. Therefore,
G = −nFEcell
−5.2032 105 J = −2 9.6485 104 C Ecell
Rearrange and solve for Ecell. Recall that J = CV.
5.2032 105 J
Ecell = = 2.6963 = 2.696 V
2 9.6485 104 C
19.12. The half-reactions and standard electrode potentials are
Fe(s) Fe2+(aq) + 2e−; −EFe = 0.41 V
Sn4+(aq) + 2e− Sn2+(s); ESn 2+ = 0.15 V
The standard cell potential for the cell is
Ecell = EFe − ESn 2+ = 0.41 V + 0.15 V = 0.56 V
Note that n equals 2. Substitute into the equation relating E and K. Also, K = Kc.
0.0592
0.56 V = log Kc
2
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Electrochemistry
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
19.1. Assign oxidation numbers to the skeleton equation (Step 1).
0 +5 +5 +4
I2 + NO3 - IO3 -
+ NO2
Separate into two incomplete half-reactions (Step 2). Note that iodine is oxidized (increases in
oxidation number), and nitrogen is reduced (decreases in oxidation number).
I2 IO3-
NO3- NO2
Balance each half-reaction separately. The oxidation half-reaction is not balanced in I, so place a
two in front of IO3− (Step 3a). Then add six H2O’s to the left side to balance O atoms (Step 3b),
and add twelve H+ ions to the right side to balance H atoms (Step 3c). Finally, add ten electrons to
the right side to balance the charge (Step 3d). The balanced oxidation half-reaction is
I2 + 6H2O 2IO3− + 12H+ + 10e−
The reduction half-reaction is balanced in N (Step 3a). Add one H2O to the right side to balance
O atoms (Step 3b), and add two H+ ion to the left side to balance H atoms. Finally, add one
electron to the left side to balance the charge (Step 3d).
The balanced reduction half-reaction is
NO3− + 2H+ + e− NO2 + H2O
Multiply the reduction half-reaction by 5 so that, when added, the electrons cancel (Step 4a).
I2 + 6H2O 2IO3− + 12H+ + 10e−
10NO3− + 20H+ + 10e− 10NO2 + 10H2O
I2 + 10NO3− + 20H+ + 6H2O + 10e−
2IO3− + 10NO2 + 12H+ + 10H2O + 10e−
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 756 Chapter 19: Electrochemistry
Simplify the equation by canceling the twelve H+ and six H2O that appear on both sides. The
coefficients do not need to be reduced (Step 4b). The net ionic equation is
I2(s) + 10NO3−(aq) + 8H+(aq) 2IO3−(aq) + 10NO2(g) + 4H2O(l)
19.2. After balancing the equation as though it were in acid solution, you obtain
H2O2 + 2ClO2 2ClO2− + O2 + 2H+
Add two OH− to both sides of the equation (Step 5), and replace the two H+ and two OH− on the
right side with two H2O. No further cancellation is required. The balanced equation for the
reaction in basic solution is
H2O2 + 2ClO2 + 2OH− 2ClO2− + O2 + 2H2O
19.3. Silver ion is reduced at the silver electrode (cathode). The half-reaction is
Ag+(aq) + e− Ag(s)
The nickel electrode (anode) is where oxidation occurs. The half-reaction is
Ni(s) Ni2+(aq) + 2e−
Electron flow in the external circuit is from the nickel electrode (anode) to the silver electrode
(cathode). Positive ions will flow in the solution portion of the circuit opposite to the direction of
the electrons. A sketch of the cell is given below:
e
Ni Ag
(anode) (cathode)
– Salt bridge +
Ni2+ Ag+
Ni Ni2+ + 2e Ag+ + e– Ag
19.4. The notation for the cell is Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s).
19.5. The half-cell reactions are
Cd(s) Cd2+(aq) + 2e−
2H+(aq) + 2e− H2(g)
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 19: Electrochemistry 757
Summing the half-cell reactions gives the overall cell reaction.
Cd(s) + 2H+(aq) Cd2+(aq) + H2(g)
19.6. The half-reactions are
Zn(s) Zn2+(aq) + 2e−
Cu2+(aq) + 2e− Cu(s)
n equals 2, and the maximum work for the reaction as written is
wmax = −nFEcell = −2 9.6485 104 C 1.10 V = −2.122 105 VC = −2.12 105 J
For 6.54 g of zinc metal, the maximum work is
1 mol Zn 2.123 105 J
6.54 g Zn = −2.123 104 = −2.12 104 J
65.39 g Zn 1 mol Zn
19.7. The half-reactions and corresponding electrode potentials are
Ag-(aq) + e - Ag(s) 0.80 V
NO3-(aq) + 4H-(aq) + 4e - NO(g) + 2H2O(l) 0.96 V
The stronger oxidizing agent is the one involved in the half-reaction with the more positive
standard electrode potential, so NO3− is the stronger oxidizing agent.
19.8. In this reaction, Cu2+ is the oxidizing agent on the left side; I2 is the oxidizing reagent on the right
side. The corresponding standard electrode potentials are
Cu2+(aq) + 2e− Cu(s); E = 0.34 V
I2(s) + 2e− 2I−(aq); E = 0.54 V
The stronger oxidizing agent is the one involved in the half-reaction with the more positive
standard electrode potential, so I2 is the stronger oxidizing agent. The reaction is nonspontaneous
as written.
19.9. The reduction half-reactions and standard electrode potentials are
Zn2+(aq) + 2e− Zn(s); EZn = −0.76 V
Cu2+(aq) + 2e− Cu(s); ECu = 0.34 V
Reverse the first half-reaction and its half-cell potential to obtain
Zn(s) Zn2+(aq) + 2e−; −EZn = 0.76 V
Cu2+(aq) + 2e− Cu(s); ECu = 0.34 V
Obtain the cell potential by adding the half-cell potentials.
Ecell = ECu − EZn = 0.34 V + 0.76 V = 1.10 V
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 758 Chapter 19: Electrochemistry
19.10. The half-cell reactions, the corresponding half-cell potentials, and their sums are displayed below:
Sn2+(aq) Sn4+(aq) + 2e− −E = −0.15 V
2Hg2+(aq) + 2e− Hg22+(aq) E = 0.90 V
Sn2+(aq) + 2Hg2+(aq) Sn4+(aq) + Hg22+(aq) Ecell = 0.75 V
Note that each half-reaction involves two electrons; hence n equals 2. Also, Ecell equals 0.75 V,
and the faraday constant, F, is 9.6485 104 C. Therefore,
G = −nFEcell = −2 9.6485 104 C 0.75 V = −1.4472 105 J = −1.4 105 J
Thus, the standard free-energy change is −1.4 102 kJ.
19.11. Write the equation with Gf’s beneath each substance.
Hence,
G = ΣnGf(products) − ΣmGf(reactants)
= [(−454.8) − 65.52] kJ = −520.32 kJ = −5.2032 105 J
Obtain n by splitting the reaction into half-reactions.
Mg(s) Mg2+(aq) + 2e−
Cu2+(aq) + 2e− Cu(s)
Each half-reaction involves two electrons, so n equals 2. Therefore,
G = −nFEcell
−5.2032 105 J = −2 9.6485 104 C Ecell
Rearrange and solve for Ecell. Recall that J = CV.
5.2032 105 J
Ecell = = 2.6963 = 2.696 V
2 9.6485 104 C
19.12. The half-reactions and standard electrode potentials are
Fe(s) Fe2+(aq) + 2e−; −EFe = 0.41 V
Sn4+(aq) + 2e− Sn2+(s); ESn 2+ = 0.15 V
The standard cell potential for the cell is
Ecell = EFe − ESn 2+ = 0.41 V + 0.15 V = 0.56 V
Note that n equals 2. Substitute into the equation relating E and K. Also, K = Kc.
0.0592
0.56 V = log Kc
2
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
