Chapter 11

[Pages:19]Chapter 4 83. a)

b)

c)

Homework #6

Chapter 11

Electrochemistry

Oxidation ? Reaction Fe + HCl HFeCl4 Fe + 4HCl HFeCl4 Fe + 4HCl HFeCl4 + 3H+ Fe + 4HCl HFeCl4 + 3H+ + 3e-

Reduction ? Reaction H2 2H+ H2 2H+ + 2e- H2

Balanced Reaction 2(Fe + 4HCl HFeCl4 + 3H+ + 3e-) 3(2H+ + 2e- H2) 2Fe(s) + 8HCl(aq) 2HFeCl4(aq) + 3H2(g)

Oxidization ? Reaction I- I33I- I33I- I3- + 2e-

Reduction ? Reaction IO3- I33IO3- I33IO3- I3- + 9H2O 3IO3- + 18H+ I3- + 9H2O 3IO3- + 18H++ 16e- I3- + 9H2O

Balanced Reaction 3IO3- + 18H++ 16e- I3- + 9H2O 8(3I- I3- + 2e-) 3IO3-(aq) + 18H+(aq) + 24I-(aq) 9I3-(aq) + 9H2O(l) Divide trough by 3 IO3-(aq) + 6H+(aq) + 8I-(aq) 3I3-(aq) + 3H2O(l)

Oxidation ? Reaction Cr(NCS)64- Cr3+ + NO3- + CO2 + SO42Cr(NCS)64- Cr3+ + 6NO3- + 6CO2 + 6SO42Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e-

Reduction ? Reaction Ce4+ Ce3+ Ce4+ + e- Ce3+

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Balanced Reaction Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e97(Ce4+ + e- Ce3+) Cr(NCS)64-(aq)+ 54H2O(l) + 97Ce4+(aq) Cr3+(aq) + 6NO3-(aq) + 6CO2(g) + 6SO42(aq) + 108H+(aq) + 97Ce3+(aq)

d) Oxidation ? Reaction CrI3 CrO42- + IO4CrI3 CrO42- + 3IO4CrI3 + 16H2O CrO42- + 3IO4CrI3 + 16H2O CrO42- + 3IO4- + 32H+ CrI3 + 16H2O CrO42- + 3IO4- + 32H+ + 27e-

Reduction ? Reaction Cl2 ClCl2 2ClCl2 + 2e- 2Cl-

Balanced Equation 2(CrI3 + 16H2O CrO42- + 3IO4- + 32H+ + 27e-) 27(Cl2 + 2e- 2Cl-) 2CrI3(s) + 32H2O(l) + 27Cl2(g) 2CrO42-(aq) + 6IO4-(aq) + 64H+(aq) + 54Cl-(aq)

The solution is basic not acidic, add 64 OH- to both sides 2CrI3(s) + 27Cl2(g) + 64OH-(aq) 2CrO42-(aq) + 6IO4-(aq) + 32H2O(l) + 54Cl-(aq)

e) Oxidation ? Reaction Fe(CN)64- Fe(OH)3 + CO32- + NO3Fe(CN)64- Fe(OH)3 + 6CO32- + 6NO3Fe(CN)64- + 39H2O Fe(OH)3 + 6CO32- + 6NO3Fe(CN)64- + 39H2O Fe(OH)3 + 6CO32- + 6NO3- + 75H+ Fe(CN)64- + 39H2O Fe(OH)3 + 6CO32- + 6NO3- + 75H+ + 61e-

Reduction ? Reaction Ce4+ Ce(OH)3 Ce4+ + 3H2O Ce(OH)3 Ce4+ + 3H2O Ce(OH)3 + 3H+ Ce4+ + 3H2O + e- Ce(OH)3 + 3H+

Balance Reaction Fe(CN)64- + 39H2O Fe(OH)3 + 6CO32- + 6NO3- + 75H+ + 61e61(Ce4+ + 3H2O + e- Ce(OH)3 + 3H+) Fe(CN)64-(aq) + 222H2O(l) + 61Ce4+(aq) Fe(OH)3(s) + 6CO32-(aq) + 6NO3-(aq)+ 258H+(aq) + 61Ce(OH)3(s)

The solution is basic not acidic, add 258OH- to both sides Fe(CN)64-(aq) + 258OH-(aq) + 61Ce4+(aq) Fe(OH)3(s) + 6CO32-(aq) + 6NO3-(aq)+ 61Ce(OH)3(s) + 36H2O(l)

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Chapter 11 3. Reactions of Interest

Ni2+ + 2e- Ni

E?=-0.23 V

Cu+ + e- Cu

E?=0.52 V

Cu2+ + 2e- Cu

E?=0.34 V

Zn2+ + 2e- Zn

E?=-0.76 V

In order to plate out Ni you need the Ni reaction to be the reduction ? reaction (cathode). In

addition, you also need the cell to be galvanic (E?>0). The E?cell of the nickel/copper cell is -0.75 V or -0.57 V depending on the ion of copper that is used. Therefore, neither of these cells

would be a galvanic cell, resulting in copper not being an appropriate material. The E?cell of the nickel/zinc cell is 0.53 V. Therefore, the nickel/zink cell would plate out Ni.

10. E? is the reaction potential at the standard state. The standard state is 1 M concentration of

aqueous solutions or 1 atm pressure of gasses. By convention E? is set to 0 for hydrogen's reduction ? reaction (2H+ + 2e- H2). E is the reaction potential not at the standard state. E is 0 when the cell is at equilibrium.

19. a)

Reactions of interest

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O E? = 1.33 V

Cl2 + 2e- 2Cl-

E? = 1.36 V

3(Cl2 + 2e- 2Cl-)

E? = 1.36 V

2Cr3+ + 7H2O Cr2O72- + 14H+ + 6e-

E? =- 1.33 V

3Cl2(g) + 2Cr3+(aq) + 7H2O(l) 6Cl-(aq) + Cr2O72-(aq) + 14H+(aq) E?=1.36 V-1.33 V=0.03V

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b) Reactions of Interest

Cu2+ + 2e- Cu

E?= 0.34 V

Mg2+ + 2e- Mg

E?=-2.37 V

Cu2+ + 2e- Cu

E?= 0.34 V

Mg Mg2+ + 2e-

E? =2.37 V

Mg(s) + Cu2+(aq) Cu(s) + Mg2+(aq) E? = 0.34 V + 2.37 V = 2.71 V

c)

Reactions of Interest

IO3- + 6H+ + 5e- ?I2 +3H2O E?=1.20 V

Fe3+ + e- Fe2+

E?= 0.77 V

IO3- + 6H+ + 5e- ?I2 +3H2O

E? = 1.20 V

5(Fe2+ Fe3+ + e-)

E?=-0.77 V

IO3-(aq) + 6H+(aq) + 5Fe2+(aq) ?I2(s) +3H2O(l) + 5Fe3+(aq) E?=1.20 V-0.77V= 0.43 V

Note: I2 does not conduct electricity; therefore, the Pt electrode is used d) Reactions of Interest

Ag+ + e- Ag

E?= 0.80 V

Zn2+ + 2e- Zn

E?= -0.76 V

2(Ag+ + e-Ag)

E? = 0.80 V

Zn Zn2+ + 2e-

E? = 0.76 V

2Ag+(aq) + Zn(s) 2Ag(s) + Zn2+(aq) E? = 0.80 V + 0.76 V = 1.56 V

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21. a)

Cl2 + 2e- 2Cl2Br- Br2 + 2eCl2(g) + 2Br-(aq) 2Cl-(aq) + Br2(aq) Note: Br2 is a liquid at room temperature

E?= 1.36 V E?=-1.09 V E?=1.36 V ? 1.09 V= 0.27 V

b)

5(IO3- + 2H+ + 2e- IO3- +H2O)

E?= 1.60 V

2(Mn2+ + 4H2O MnO4- + 8H+ + 5e-)

E?= -1.51 V

5IO3-(aq)+2Mn2+(aq)+3H2O(l)5IO3-(aq)+2MnO4-(aq)+6H+(aq) E?= 1.60V?1.51V=0.09V

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c)

H2O2 +2H+ + 2e- 2H2O

E?= 1.78 V

H2O2 O2 + 2H+ + 2e-

E?= -0.68 V

2H2O2(aq) O2(g) + 2H2O(l) E?= 1.78 V -0.68 V = 1.10 V

d)

2(Fe3+ + 3e- Fe)

3(Mn Mn2+ + 2e-)

2Fe3+(aq) + 3Mn(s) 2Fe(s) + Mn2+(aq)

E?= -0.036V E? = 1.18 V E? = -0.036 V + 1.18 V = 1.14 V

22. a)

For a galvanic cell E? must be positive therefore Cu+ must be oxidized given:

Au3+ + 3e- Au

E? = 1.50V

Cu2+ + e- Cu+

E? = 0.16 V

Au3+ + 3e- Au

E? = 1.50V

3(Cu+ Cu2+ +e-)

E? = -0.16 V

Au3+(aq) + 3Cu+(aq) Au(s) + 3Cu2+(aq)

E? = 1.34 V

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b)

24. a) b) c) d) e)

25. a) b)

For a galvanic cell E? must be positive therefore Cd must be oxidized given:

VO2+ + 2H+ + e- VO2+ + H2O E? = 1.00V

Cd2+ +2e- Cd

E? = -0.40 V

2(VO2+ + 2H+ + e- VO2+ + H2O)

E? = 1.00V

Cd Cd2+ +2e-

E? = 0.40 V

2VO2+(aq) +4H+(aq) + Cd(s) 2VO2+(aq) +2H2O(l) + Cd2+(aq) E? = 1.40 V

Cu2+ + 2e- Cu

E? = 0.34 V

In order for the cell to be a galvanic cell Cu2+ must be reduced, therefore, SCE is

oxidized and at the anode.

E?cell = 0.34 V + -0.242 V = 0.10 V

Fe3+ + e- Fe2+

E? = 0.77 V

In order for the cell to be a galvanic cell, Fe3+ must be reduced, therefore, SCE is oxidized

and at the anode.

E?cell = 0.77 V + -0.242 V = 0.53 V

AgCl + e- Ag + Cl-

E? = 0.22 V

In order for the cell to be a galvanic cell, Ag must be oxidized, therefore, SCE is reduced and at the cathode.

E?cell = 0.242 V ? 0.22 V = 0.02 V

Al3+ + 3e- Al E? = -1.66 V

In order for the cell to be a galvanic cell, Al must be oxidized, therefore, SCE is reduced and at the cathode.

E?cell = 0.242 V + 1.66 V = 1.90 V

Ni2+ + 2e- Ni E? = -0.23 V

In order for the cell to be a galvanic, cell Ni must be oxidized, therefore, SCE is reduced and at the cathode.

E?cell = 0.242 V + 0.23 V = 0.47 V

Cu Cu2+ +2e-

E?= -0.34 V

2H+ + 2e- H2

E?= 0.0 V

No H+ cannot oxidize Cu

2I- I2 +2e-

E?=-0.54 V

Fe3+ + e- Fe2+

E?= 0.77 V

Yes Fe3+ is capable of oxidizing I- if it is going to Fe2+

2I- I2 +2e-

E?=-0.54 V

Fe3+ + 3e- Fe

E?= -0.04 V

No Fe3+ cannot oxidize I- if it is going to Fe

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c)

Ag+ + e- Ag

E?=0.80 V

H2 2H+ +2e-

E?= 0.0 V

Yes H2 is capable of reducing Ag+

d)

Cr3+ + e- Cr2+

E? = -0.50 V

Fe2+ Fe3+ + e-

E?=-0.77 V

No Fe2+ is not capable of reducing Cr3 + to Cr2+

26. The oxidizing agent is the species that is reduced. Therefore, the best oxidizing reagent is the

species that has the largest E? value for the reduction ? reaction. K+ < H2O < Cd2+ < I2 < AuCl4 - < IO3-

27. The reducing agent is the species that is oxidized. Therefore, the best reducing agent is the

species that has the smallest E? value for the reduction ? reaction. F- < H2O < I2 < Cu+ < H- < K

28. Choices Cl2 + 2e- 2ClAg+ + e- Ag

E? = 1.36 V E? = 0.80 V

Pb2+ + 2e- Pb

E? = -0.13 V

Zn2+ + 2e- Zn

E? = -0.76 V

Na+ + e- Na

E? = -2.71 V

Underlined are possible answers

a) The oxidizing agent is the species that is being reduced or species in the reduction

reaction. Therefore Cl-, Ag, Pb, and Zn can be eliminated because if they were on the

reactants side of the reaction the reaction would be an oxidation reaction, therefore,

they can only be reducing agents. Out of the remaining species the best oxidizing agent

is the one with the largest E? value in the reduction ? reactions.

Ag+ b) The reducing agent is a species that is being oxidized or the species in the oxidizing

reaction. Therefore Ag+, Zn2+, and Na+ can be eliminated because if you flip the reaction to make it an oxidation reaction they are on the product side of the reaction and not the reactants, therefore, they can only be oxidizing agents. Out of the remaining species the

best oxidizing agent is the one with the most negative E? value in the reduction ?

reactions (the E? value will become positive when the reaction is flipped).

Zn

c)

SO42- + 4H+ + 2e- H2SO3 + H2O

E?= 0.20 V

In order for SO42- to oxidize a species the species has to be in an oxidation reaction;

therefore, Ag+, Zn2+, and Na+ can be eliminated because they are on the reactant side of

reduction reactions. Of the remaining species the E? value of the reduction ? reaction

must be smaller than 0.20 V. Pb and Zn

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