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Ch. 20 Kimball Spring 2019
Discussion Worksheet - Chapter 20
Electrochemistry
Dr Cabirac Lecture
1. Determine the oxidation numbers for all the elements in each of the following:
A. C2H4C: -2 H: +1
B. MgPtCl6Mg: +2 Pt: +4 Cl: -1
2. For each of the following reactions, identify
0 +1 +5 -2 +1 -2 +1 +4 -2 +1 -2
A. 2 S(s) + 2 HNO3 (aq) + H2O(l) → 2 H2SO3 (aq) + N 2O(g)
i) oxidation numbers for all the elements in the reactants and products see above
ii) element that is oxidized S: 0 → +4
iii) element that is reduced N: +5 → +1
iv) the total number of electrons that are transferred 8 e-
+5 -2 -2 +1 +1 0 0 +1 -2
B. 4 BrO3-(aq) + 5N2H4 (g) + 4 H+(aq) → 2 Br2 (l) + 5 N2 (g) + 12 H2O(l)
i) oxidation numbers for all the elements in the reactants and products see above
ii) the oxidizing agent Br is reduced, so is oxidizing agent
iii) the reducing agent N is oxidized, so is reducing agent
iv) the total number of electrons that are transferred 20 e-
3. Complete and balance the following half reactions and indicate whether it is an oxidation
or a reduction:
A. TiO2 (s) → Ti2+(aq) (acidic solution)
TiO2 (s) + 4H+(aq) + 2e- → Ti2+(aq) + 2H2O(l)
B. N2 (g) → NH3 (g) (basic solution)
N2 (g) + 6H2O(l) + 2e- → 2NH3 (g) + 6OH-(aq)
C. Mn2+(aq) → MnO2 (s) (basic solution)
Mn2+(aq) + 4OH-(aq) → MnO2 (s) + 2H2O(l) + 2e-
D. H2SO3 (aq) → SO42- (aq) (acidic solution)
H2SO3 (aq) + H2O(l) → SO42-(aq) + 4H+(aq) + 2e-
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Ch. 20 Kimball Spring 2019
4. Complete and balance the following redox equations:
A. H2O2 (aq) + Fe2+(aq) → Fe3+(aq) + H2O(l) (acidic solution)
H2O2 (aq) + 2H+(aq) + 2e- → 2H2O(l)
2( Fe2+(aq) → Fe3+(aq) + e- )
H2O2 (aq) + 2Fe2+(aq) + 2H+(aq) → 2H2O(l) + 2Fe3+(aq)
B. NO2- (aq) + Al(s) → NH4+ (aq) + AlO2- (aq) (basic solution)
NO2- (aq) + 6H2O(l) + 6e- → 4NH4+(aq) + 8OH-(aq)
2( Al(s) + 4OH-(aq) →AlO2-(aq) + 2H2O(l) + 3e- )
NO2- (aq) + 2Al(s) + 2H2O(l) → NH+(aq) + 2AlO2-(aq)
5. Consider the following voltaic cell:
The cell potentials are Mg2+(aq) + 2e- → Mg(s) -2.37 V
Ag+(aq) + e- → Ag(s) +0.80 V
A. Which is the cathode? Ag
B. Which is the anode? Mg
C. Which electrode gains mass? Ag
D. Which way do electrons flow through the wire? Mg → Ag
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Ch. 20 Kimball Spring 2019
6. Calculate the standard emf for each of the following reactions and determine whether the
reaction would take place spontaneously in aqueous solutions at 25°C.
A. 2Br-(aq) + Sn2+(aq) → Br2 (l) + Sn(s)
E°cell = Ered (Sn2+/Sn) - Ered (Br2/Br-) = -0.44 V – 1.07 V = -1.21 V nonspontaneous
B. Br2 (l) + 2I-(aq) → 2Br-(aq) + I2 (s)
E°cell = Ered (Br2/Br-) - Ered (I2/I-) = 1.07 V – 0.54 V = +0.52 V spontaneous
C. Ca(s) | Ca2+ (1M) ‖ Cd2+ (1M) | Cd(s)
E°cell = Ered (Cd2+/Cd) - Ered (Ca2+/Cd) = -0.40 V – (-2.87 V) = +2.47 V spontaneous
D. Cu+(aq) + Fe3+(aq) → Cu2+(aq) + Fe2+(aq)
E°cell = Ered (Fe3+/Fe2+) - Ered (Cu2+/Cu+) = 0.77 V – 0.15 V = +0.62 V spontaneous
Consider the following reduction potentials for questions 7, 8:
Al3+(aq) + 3e- → Al(s) Eº = -1.66 V
Co3+(aq) + e- → Co2+(s) Eº = +1.84 V
Cu+(aq) + e- → Cu(s) Eº = +0.52 V
Mn2+(aq) + 2e- → Mn(s) Eº = -1.18 V
7. A. What is the strongest reducing agent? Al
B. What is strongest oxidizing agent? Co3+
8. Which of the following combinations will undergo a spontaneous reaction?
A. Cu+ with Al yes
B. Co2+ with Cu no
C. Mn with Co3+ yes
D. Cu with Al3+ no
E. Mn2+ with Co2+ no
9. Consider the electrochemical reaction Sn2+ + X → Sn + X2+. Given that Eºcell = 0.14V,
what is the Eº for the X2+/X half-reaction?
E°cell = Ered(cathode) – Ered(anode)
0.14 V = -0.14 – Ered(X2+/X)
Ered(X2+/X) = -0.28 V
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Ch. 20 Kimball Spring 2019
10. In an acidic solution, copper (I) ion is oxidized to copper (II) ion by nitrate ion.
A. Write a balanced equation.
3Cu+(aq) + NO3-(aq) + 4H+(aq) → NO(g) + 3Cu2+(aq) + 2H2O(l)
B. Calculate the standard emf.
E°cell = 0.81 V
C. Calculate ∆G° at 298K.
ΔG° = -nFE° = -234 kJ/mol
D. Calculate the equilibrium constant K at 298K.
ΔG° = -RTlnK
K = 1.3 x 1041
11. Under standard-state conditions, what spontaneous reaction will occur in aqueous
solution among the ions Ce4+, Ce3+, Fe3+, Fe2+? Calculate Eº, ∆G°, and Kc for the
reaction. Fe3+ + e- → Fe2+ E = +0.77V
Ce4+ + e- → Ce3+ E = + 1.61
Fe2+ + Ce4+ → Fe3+ + Ce3+ E°cell = 1.61 – 0.77 = 0.84 V
ΔG° = -81 kJ/mol
K = 1.6 x 1014
12. Calculate Eº, E, and ∆G for the following cell reactions:
A. Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s)
[Mg2+] = 0.045 M, [Sn2+] = 0.035 M
E°cell = +2.23 V
𝐸°𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙 −0.0257
𝑛𝑙𝑛𝑄 = 2.23 −0.0257
2𝑙𝑛0.045
0.035 = +𝟐.𝟐𝟑 𝑽
ΔG=-nFE = -431 kJ/mol
B. 4 Fe2+(aq) + O2(g) + 4 H+(aq) → 4 Fe3+(aq) + 2 H2O(l)
[Fe2+] = 1.3 M , [Fe3+] = 0.010 M, PO2=0.50 atm, pH = 3.5 (cathode half-cell)
E°cell = +0.46 V
𝐸°𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙 −0.0257
𝑛𝑙𝑛𝑄 = 2.23 −0.0257
4𝑙𝑛 (0.010)4
(1.3)4(3.16𝑥10−4)4(0.50)= +𝟎.𝟑𝟕 𝑽
ΔG=-nFE = -143 kJ/mol
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Ch. 20 Kimball Spring 2019
13. Consider the electrolysis of molten barium chloride, BaCl2.
A. Write the half reactions and identify the anode and the cathode reactions.
Ba2+(aq) + 2e- → Ba(s) cathode
2Cl-(aq) → Cl2 (g) + 2e- anode
B. How many grams of barium metal can be produced by supplying 0.50 A for 30
min?
30 𝑚𝑖𝑛(0.5 𝐶
1 𝑠𝑒𝑐)(1 𝑚𝑜𝑙 𝑒−1
96,500 𝐶 )(1 𝑚𝑜𝑙 𝐵𝑎2+
2 𝑚𝑜𝑙 𝑒−)(137.3 𝑔
1𝑚𝑜𝑙 ) = 𝟎.𝟔𝟒 𝒈 𝑩𝒂
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