Assigning Oxidation Numbers and Balancing Redox Equations

+5 -6

1. Ag + NO3 - Ag1+ + NO

0 +5 -2 +1 +2 -2 -4 +4 +2 -2

2. N2H4 + H2O2 N2 + H2O

-2 +1 +1 -1 0 +1 -2 +6 -6 +4 -4

3. CO + Fe2O3 FeO + CO2

+2 -2 +3 -2 +2 -2 +4 -2 +5 -6 +4 -4 +4 -4

4. NO3 - + CO CO2 + NO2

+5 -2 +2 -2 +4 -2 +4 -2 +8 -8

5. H2 + Fe3O4 Fe + H2O

0 +8/3 -2 0 +1 -2 +2 +6 -8 +7 -8 +4 -4

6. H2C2O4 + MnO4 - CO2 + MnO

+1 +3 -2 +7 -2 +4 -2 +2 -2 +5 -6

7. Zn + NO3 - Zn2+ + NO

0 +5 -2 +2 +2 -2 +6 -6

8. C2N2 CN - + CNO –

+3 -3 +2 -3 +4 -3 -2

+4 -4 +3 -4 +3 -4 +5 -6

9. ClO2 + SbO2 - ClO2

- + Sb(OH)6 –

+4 -2 +3 -2 +3 -2 +5 (-1) +12 -14

10. Cr2O7 2- + I - Cr3+ + I2

+6 -2 -1 +3 0 +8 -8 +2 -2

11. Fe3O4 + H2O2 Fe3+ + H2O

+8/3 -2 +1 -1 +3 +1 -2 +7 -8 +4 -4 +5 -6

12. MnO4 - + NH3 MnO2 + NO3

–

+7 -2 -3 +1 +4 -2 +5 -2 +6 -8 +3 -3

13. CN - + CrO4 2- CNO - + Cr(OH)3

+2 -3 +6 -2 +4 -3 -2 +3 (-1) +3 -3 +6 -6 +4 -6 +6 -8 -3 +4 +5 -6 +2 -2

14. NH4NO3 N2O

-3 +1 +5 -2 +1 -2 +3 -4 +7 -8 +5 -6

15. NO2– + MnO4

– NO3– + Mn2+ (in acid solution)

+3 -2 +7 -2 +5 -2 +2 +7 -8 +4 -4

16. I- + MnO4- I2 + MnO2 (in basic solution)

-1 +7 -2 0 +4 -2

+4 -6 +6 -8

17. Cl2 + S2O32- Cl- + SO42- (in acidic solution)

0 +2 -2 -1 +6 -2 -4 +4 +5 -6

18. Br2 Br- + BrO3- (in basic solution)

0 -1 +5 -2

Assigning Oxidation Numbers and Balancing Redox Equations

1. Ag + NO3 - ↔ Ag

1+ + NO

0 +5 -2 +1 +2 -2

ox: 3 (Ag ↔ Ag1+

+ 1 e -)

0 +1

red: NO3 - + 3 e - ↔ NO

+5 +2

3 Ag + NO3 - ↔ 3 Ag

1+ + NO

3 Ag + NO3 - + 4 H + ↔ 3 Ag

1+ + NO + 2 H2O

2. N2H4 + H2O2 ↔ N2 + H2O

-2 +1 +1 -1 0 +1 -2

ox: N2H4 ↔ N2 + 4 e –

-2 0

red: 2 (H2O2 + 2 e - ↔ 2 H2O)

-1 -2 N2H4 + 2 H2O2 ↔ N2 + 4 H2O

3. CO + Fe2O3 ↔ FeO + CO2

+2 -2 +3 -2 +2 -2 +4 -2

ox: CO ↔ CO2 + 2 e –

-2 +4

red: Fe2O3 + 2 e - ↔ 2 FeO

+3 +2 CO + Fe2O3 ↔ 2 FeO + CO2

4. NO3 - + CO ↔ CO2 + NO2

+5 -2 +2 -2 +4 -2 +4 -2

ox: CO ↔ CO2 + 2 e –

+2 +4

red: 2 (NO3 - + 1 e - ↔ NO2)

+5 +4

2 NO3 - + CO ↔ CO2 + 2 NO2

2 NO3 - + CO + 2 H + ↔ CO2 + 2 NO2 + H2O

5. H2 + Fe3O4 ↔ Fe + H2O

0 +8/3 -2 0 +1 -2

ox: 4 (H2 ↔ H2O + 2 e -)

0 +1

red: Fe3O4 + 8 e - ↔ 3 Fe

+8/3 0 4 H2 + Fe3O4 ↔ 3 Fe + 4 H2O

6. H2C2O4 + MnO4 - ↔ CO2 + MnO

+1 +3 -2 +7 -2 +4 -2 +2 -2

ox: 5 (H2C2O4 ↔ 2 CO2 + 2 e -)

+3 +4

red: 2 (MnO4 - + 5 e - ↔ MnO)

+7 +2

5 H2C2O4 + 2 MnO4 - ↔ 10 CO2 + 2 MnO

5 H2C2O4 + 2 MnO4 - + 2 H + ↔ 10 CO2 + 2 MnO + 6 H2O

7. Zn + NO3 - ↔ Zn

2+ + NO

0 +5 -2 +2 +2 -2

ox: 3 (Zn ↔ Zn 2+

+ 2 e -)

0 +2

red: 2 (NO3 - + 3 e - ↔ NO)

+5 +2

3 Zn + 2 NO3 - ↔ 3 Zn2+ + 2 NO

3 Zn + 2 NO3 - + 8 H + ↔ 3 Zn

2+ + 2 NO + 4 H2O

8. C2N2 ↔ CN - + CNO –

+3 -3 +2-3 +4-3-2

ox: C2N2 ↔ 2 CNO - + 2e –

+3 +4

red: C2N2 + 2e - ↔ 2 CN -

+3 +2

2 C2N2 ↔ 2 CN - + 2 CNO -

C2N2 + H2O ↔ CN - + CNO - + 2 H +

9. ClO2 + SbO2 - ↔ ClO2

- + Sb(OH)6

–

+4 -2 +3 -2 +3 -2 +5 (-1)

ox: SbO2 - ↔ Sb(OH)6

- + 2e –

+3

red: 2 (ClO2 + 1 e - ↔ ClO2 -)

+4 +3

2 ClO2 + SbO2 - ↔ 2 ClO2

- + Sb(OH)6

-

2 ClO2 + SbO2 - + 2 OH - + 2 H2O ↔ 2 ClO2

- + Sb(OH)6

-

10. Cr2O7 2-

+ I - ↔ Cr

3+ + I2

+6 -2 -1 +3 0

ox: 3 (2 I - ↔ I2 + 2e -)

-1 0

red: Cr2O7 2-

+ 6 e - ↔ 2 Cr3+

+6 +3

Cr2O7 2-

+ 6I - ↔ 2 Cr

3+ + 3 I2

Cr2O7 2-

+ 6I - + 14 H + ↔ 2 Cr

3+ + 3 I2 + 7 H2O

11. Fe3O4 + H2O2 ↔ Fe3+

+ H2O

+8/3 -2 +1 -1 +3 +1 -2

ox: 2 (Fe3O4 ↔ 3 Fe3+

+ 1 e -)

+8/3 +3

red: H2O2 + 2 e - ↔ 2 H2O

-1 -2

2 Fe3O4 + H2O2 ↔ 6 Fe3+

+ 2 H2O

2 Fe3O4 + H2O2 + 18 H + ↔ 6 Fe3+

+ 10 H2O

12. MnO4 - + NH3 ↔ MnO2 + NO3

–

+7 -2 -3 +1 +4 -2 +5 -2

ox: 3 (NH3 ↔ NO3 -

+ 8e -)

-3 +5

red: 8 ( MnO4 - + 3 e - ↔ MnO2)

+7 +4

8 MnO4 - + 3 NH3 ↔ 8 MnO2 + 3 NO3

- 8 MnO4

- + 3 NH3 + 5 H + ↔ 8 MnO2 + 3 NO3

- + 7 H2O

13. CN - + CrO4

2- ↔ CNO

- + Cr(OH)3

+2 -3 +6 -2 +4-3-2 +3 (-1)

ox: 3 (CN - ↔ CNO

- + 2 e -)

+2 +4

red: 2 (CrO4 2-

+ 3 e - ↔ Cr(OH)3)

+6 +3

3 CN - + 2 CrO4

2- + 5 H2O ↔ 3 CNO

- + 2 Cr(OH)3 + 4 OH –

3 CN - + 2 CrO4

2- + 5 H2O ↔ 3 CNO

- + 2 Cr(OH)3 + 4 OH -

14. NH4NO3 ↔ N2O

-3+1+5-2 +1 -2

ox: 2 NH4 + ↔ N2O + 8e –

-3 +1

red: 2 NO3 - + 8 e - ↔ N2O

+5 +1

2NH4+ + 2 NO3

– ↔ 2 N2O NH4NO3 ↔ N2O + 2 H2O

15. NO2– + MnO4

– ↔ NO3– + Mn2+ (in acid solution)

+3 -2 +7 -2 +5 -2 +2

ox: 5 (NO2– ↔ NO3

– + 2 e -)

+3 +5

red: 2 (MnO4– + 5 e - ↔ Mn2+)

+7 +2

5 NO2- + 2 MnO4

- + 6 H+ ↔ 5 NO3- + 2 Mn2+ + 3 H2O

5 NO2- + 2 MnO4

- + 6 H+ ↔ 5 NO3- + 2 Mn2+ + 3 H2O

16. I- + MnO4- ↔ I2

+ MnO2 (in basic solution)

-1 +7 -2 0 +4 -2

ox: 3 (2 I– ↔ I2– + 2 e -)

-1 0

red: 2 (MnO4– + 3 e - ↔ MnO2)

+7 +4

6 I- + 2 MnO4- ↔ 3 I2

+ 2 MnO2

6 I- + 2 MnO4- + 4 H2O ↔ 3 I2

+ 2 MnO2 + 8 OH-

17. Cl2 + S2O32- ↔ Cl- + SO42- (in acidic solution)

0 +2 -2 -1 +6 -2

ox: S2O32- ↔ 2 SO42- + 8 e -

+2 +6

red: 4 (Cl2 + 2 e - 2 Cl-)

0 -1

4 Cl2 + S2O32- + 5 H2O ↔ 8 Cl- + 2 SO42- + 10 H+

4 Cl2 + S2O32- + 5 H2O ↔ 8 Cl- + 2 SO42- + 10 H+

18. Br2 Br- + BrO3- (in basic solution)

0 -1 +5 -2

ox: Br2 ↔ 2 BrO3- + 10 e -

0 +5

red: 5 (Br2 + 2 e - 2 Br-)

0 -1

6 Br2 ↔ 10 Br- + 2 BrO3-

3 Br2 + 6 OH- ↔ 5 Br- + BrO3- + 3 H2O

X √

X √

X

X

√

Predicting REDOX Reactions

Building a REDOX Table

1. The following reactions were performed. Construct a table of relative strengths of oxidizing and

reducing agents written as reductions and with the SOA to WOA.

Zn + Co2+ Zn2+ + Co

Mg2+ + Zn no rxn

Half rxns: Final Order:

Zn2+ + 2e¯ ↔ Zn SOA Co2+ + 2e¯ ↔ Co

Co2+ + 2e¯ ↔ Co Zn2+ + 2e¯ ↔ Zn

√

Mg2+ + 2e¯ ↔ Mg Mg2+ + 2e¯ ↔ Mg SRA

2. In a school laboratory four metals were combined with each of four solutions. Construct a table of

relative strengths of oxidizing and reducing agents written as reductions and with the SOA to WOA.

Be + Cd2+ Be2+ + Cd

Cd + 2 H+ Cd2+ + H2

Ca2+ + Be no rxn

Cu + 2 H+ no rxn

Half rxns: Final Order:

Be2+ + 2e¯ ↔ Be SOA Cu2+ + 2e¯ ↔ Cu

Cd2+ + 2e¯ ↔ Cd 2 H+ + 2e¯ ↔ H2

√

2 H+ + 2e¯ ↔ H2 Cd2+ + 2e¯ ↔ Cd

Ca2+ + 2e¯ ↔ Ca Be2+ + 2e¯ ↔ Be

Cu2+ + 2e¯ ↔ Cu Ca2+ + 2e¯ ↔ Ca SRA

3. Write and rank the two half reaction equations for each of the following reactions: (a) Co + Cu(NO3)2 Cu + Co(NO3)2

Cu2+ + 2e¯ ↔ Cu

Co2+ + 2e¯ ↔ Co

√

√

(b) Cd + Zn(NO3)2 Zn + Cd(NO3)2

Zn2+ + 2e¯ ↔ Zn

Cd2+ + 2e¯ ↔ Cd

(c) Br2 + 2KI ↔ I2 + 2 KBr

Br2 + 2e¯ ↔ 2 Br¯

I2 + 2e¯ ↔ 2 I¯

4. Prepare a REDOX table of half-reactions showing the relative strengths of oxidizing and reducing agent

for the following:

OA

RA

Al3+ Tl+ Ga2+ In3+

Al X √ √ √

Tl X X X X

Ga X √ X √

In X √ X X

WOA Rank 4th SOA Rank 1st Rank 3rd Rank 2nd

SOA Tl+ + e¯ Tl

In3+ + 3e¯ In

Ga2+ + 2e¯ Ga

Al3+ + 3e¯ Al SRA

Prediction REDOX Reaction in Solution

1. List all the entities initially present in the following mixtures and identify all possible oxidizing and

reducing agents. Write the resulting REDOX reaction (or no rxn). (a) A lead strip is placed in a copper (II) sulfate solution.

(Cu+ +0.15)

OA (Cu +0.34) (not in H+) (H2 -0.83)

Pb Cu2+ SOA SO4- + Pb SRA H2O

RA (Pb2+ +0.13) (S2O8

2- -2.01) (PbSO4 +0.36) (O2 -1.23)

ox: Pb + SO4- ↔ PbSO4 + 2 e-

red: Cu2+ + 2e- ↔ Cu

Pb + SO4-

+ Cu2+ ↔ PbSO4 + Cu

(b) A potassium dichromate solution is added to an acidic iron (II) nitrate solution.

OA (K-2.92) (with H+, Cr3+ +1.23) (Fe -0.45) (NO2 +0.80) (H2 -0.83)

K+ Cr2O72- SOA Fe2+ SRA H+, NO3

- H2O

RA (Fe3+ -0.77) (O2 -1.23) 6x ox: 6 Fe2+ ↔ 6 Fe3+ + 6 e-

red: Cr2O72- + 14 H+ + 6 e- ↔ 2 Cr3+ + 7 H2O

Cr2O72- + 6 Fe2+ + 14 H+ ↔ 2 Cr3+ + 6 Fe3+ + 7 H2O

(c) An aqueous chlorine solution is added to a phosphorous acid solution.

OA (Cl-+1.36) (H2 -0.83)

Cl2 SOA H+, PO33- H2O SRA

RA (O2 -1.23) ox: 2 H2O ↔ O2 + 4 H+ + 4 e-

2x red: 2 Cl2 + 4 e- ↔ 4 Cl-

2 Cl2 + 2 H2O ↔ 4 Cl- + O2 + 4 H+

(d) A potassium permanganate solution is mixed with an acidified tin (II) chloride solution.

OA (K -2.92) (Mn2+ +1.51) (Sn -0.14) (H2 -0.83) K+ MnO4

-, H+ SOA Sn2+ SRA Cl- H2O RA (Sn4+ -0.15) (Cl2 -1.36) (O2 -1.23) 5x ox: 5 Sn2+ ↔ 5 Sn4+ + 10 e-

2x red: 2 MnO4- + 10 e- ↔ 2 Mn2+

2 MnO4- + 5 Sn2+ + 16 H+ ↔ 2 Mn2+ + 5 Sn4+ + 8 H2O

Electrochemical (Galvanic or Voltaic) Cells Worksheet

1. a) Determine the anode, cathode and calculate the standard cell potential produced by a galvanic cell

consisting of a Ni electrode in contact with a solution of Ni2+ ions and a Ag electrode in contact with a

solution of Ag1+ ions.

Ni2+ + 2e- ↔ Ni E° = -0.26 V (lesser flip)

Ag+ + e- ↔ Ag(s) E° = +0.80 V

ANODE: Ni ↔ Ni2+ + 2e- E° = +0.26 V

CATHODE: 2 Ag+ + 2e- ↔ 2 Ag E° = +0.80 V

E° = +1.06 V b) Write the shorthand cell notation.

Ni (s) | Ni2+ (aq) || Ag1+ (aq) | Ag (s)

2. a) Determine the anode, cathode and calculate the voltage produced by a galvanic cell consisting of an

Fe electrode in contact with a solution of Fe2+ ions and a Al electrode in contact with a solution of

Al3+ ions.

Fe2+ + 2e- ↔ Fe E° = -0.44 V

Al3+ + 3e- ↔ Al E° = -1.66 V (lesser flip)

ANODE: 2 Al ↔ 2 Al3+ + 6e- E° = +1.66 V

CATHODE: 3 Fe2+ + 6e- ↔ 3 Fe E° = -0.44 V

E° = +1.22 V b) Write the shorthand cell notation.

Al (s) | Al3+ (aq) || Fe2+ (aq) | Fe (s)

3. a) Determine the anode, cathode and calculate standard cell potential produced by a galvanic cell

consisting of a C electrode in contact with an acidic solution of ClO4- ions and a Cu electrode in contact

with a solution of Cu2+ ions. Which is anode and which is the cathode?

ClO4- + 8H+ + 8e- ↔ Cl- + 4H2O E° = +1.39 V

Cu2+ + 2e- ↔ Cu E° = +0.34 V (lesser flip)

ANODE: 4 Cu ↔ 4 Cu2+ + 8e- E° = -0.34 V

CATHODE: ClO4- + 8H+ + 8e- ↔ Cl- + 4H2O E° = +1.39 V

E° = +1.05 V b) Write the shorthand cell notation.

Cu (s) | Cu2+ (aq) || ClO4- , H+ (aq) | C (s)

4. An electrochemical cell is constructed using electrodes based on the following half reactions:

Pb2+ + 2e- ↔ Pb Au3+ + 3e- ↔ Au a) Which is the anode and which is the cathode in this cell?

ANODE: Pb CATHODE: Au b) What is the standard cell potential?

ANODE: 3 Pb ↔ 3 Pb2+ + 6e- E° = +0.13 V

CATHODE: 2 Au3+ + 6e- ↔ 2 Au E° = +1.50 V

E° = +1.63 V

5. Use complete half-reactions and potentials to predict whether the following reactions are spontaneous or

non-spontaneous in aqueous solutions. If the cell is spontaneous, write the cell shorthand notation.

a) Ca2+(aq) + 2 I-(aq) Ca(s) + I2(aq)

ANODE: 2 I- ↔ I2 + 2e- E° = - 0.54 V

CATHODE: Ca2+ + 2e- ↔ Ca E° = - 2.87 V

E° = - 3.41 V

E° is negative, therefore the cell is non-spontaneous.

b) 2 H2S(g) + O2(g) 2 H2O(l) + 2 S(s)

ANODE: 2 H2S ↔ 2 S + 4H+ + 4e- E° = - 0.14 V

CATHODE: O2 + 4H+ + 4e- ↔ 2 H2O(l) E° = +1.23 V

E° = +1.09 V

E° is positive, therefore the cell is spontaneous.

Pt (s) | H2S (g) ; S (g) || O2 (g) , H+ (aq) | Pt (s)

c) SO2(g) + MnO2(s) Mn2+(aq) + SO4

2-(aq)

ANODE: SO2 + 2 H2O ↔ SO42- + 4H+ + 2e- E° = - 0.18 V

CATHODE: MnO2 + 4H+ + 2e- ↔ Mn2+ + 2 H2O E° = +1.22 V

E° = +1.04 V

E° is positive, therefore the cell is spontaneous.

Pt (s) | SO2(g) ; SO42-

(aq) || MnO2(s) , H+

(aq) ; Mn2+(aq) | C (s)

d) 2 H+(aq) + 2 Br-(aq) H2(g) + Br2(aq)

ANODE: 2 Br- ↔ Br2 + 2e- E° = -1.07 V

CATHODE: 2 H+ + 2e- ↔ H2 E° = 0.00 V

E° = -1.07 V

E° is negative, therefore the cell is non-spontaneous.

e) Ce4+(aq) + Fe2+

(aq) Ce3+(aq) + Fe3+

(aq)

ANODE: Fe2+ ↔ Fe3+ + e- E° = - 0.77 V

CATHODE: Ce4+ + e- ↔ Ce3+ E° = +1.44 V

E° = +0.67 V

E° is negative, therefore the cell is spontaneous.

C (s) | Fe2+ (aq) ; Fe3+(aq) || Ce4+ (aq) ; Ce3+(aq) | C (s)

f) Cr2+(aq) + Cu2+

(aq) Cr3+(aq) + Cu+

(aq)

ANODE: Cr2+ ↔ Cr3+ + e- E° = +0.41 V

CATHODE: Cu2+ + e- ↔ Cu+ E° = +0.15 V

E° = +0.56 V

E° is positive, therefore the cell is spontaneous.

C (s) | Cr2+ (aq) ; Cr3+(aq) || Cu2+ (aq) ; Cu+(aq) | C (s)

Electrolytic Cells Worksheet

1. a) Give the cathode, anode and overall equations including cell potentials to conclude what happens to

the pH of the solution near the cathode and anode during the electrolysis of KNO3? Consider all

possible reactions.

OA K (-2.92) H2 (-0.83) √

K+ NO3- SRA H2O SOA

RA O2 (-1.23) √

ox: 2 H2O ↔ O2 + 4H+ + 4 e- E°ox = - 1.23 V

2x red: 2 H2O + 2 e- ↔ 2H2 + 2OH- E°red = - 0.83 V

6 H2O ↔ O2 + 2H2 + 4H+ + 4 OH- E°cell = - 2.06 V

2 H2O ↔ O2 + 2H2 E°cell = - 2.06 V

at the anode pH , at the cathode pH

b) Write the shorthand cell notation.

C(s) │ │ C(s)

or Pt(s) │ KNO3(aq) │ or Pt(s)

2. Given the following molten systems, predict the products at each electrode. Assume inert electrodes and

sufficient voltage to cause a reaction to take place. Consider all possible rxns.

a) FeBr2

OA Fe (-0.45) √

SRA Fe2+ SOA Br-

RA Fe3+ (-0.77) √ Br2 (-1.07)

Fe3+ is produced at the anode, Fe at the cathode.

b) NiCl2

OA Ni (-0.26) √

Ni2+ SOA Cl- SRA

RA Cl2 (-1.36) √ Cl2 is produced at the anode, Ni at the cathode.

c) Na2SO4

OA Na (-2.71) √

SOA Na+ SO42-

SRA

RA S2O82-

(-2.01) √

S2O42-

is produced at the anode, Na at the cathode.

3. Given the following 1.00 M solutions at 25°C predict the anode and cathode half cell reactions. What is

the minimum voltage required for each cell to operate? a) LiMnO4

OA Li (-3.00) MnO2 (+0.60) √ H2 (-0.83)

Li+ MnO4- SOA H2O SRA

RA O2 (-1.23) √ E°cell = + 0.60 -1.23 V = -0.63 V ; 0.63 V are needed

b) CrI3

OA Cr (-0.76), Cr2+ (-0.41) √ H2 (-0.83)

Cr3+ SOA I- SRA H2O SRA

RA I2 (-0.54) √ O2 (-1.23) E°cell = -0.41 -0.54 V = -0.95 V ; 0.95 V are needed

c) Sn(NO3)2

OA Sn (-0.14) √ H2 (-0.83)

SRA Sn2+ SOA NO3- H2O SRA

RA Sn4+ (-0.15) √ O2 (-1.23) E°cell = -0.14 -0.15 V = -0.29 V ; 0.29 V are needed

d) Ag2SO4

OA Ag (+0.80) √ SO32-

(-0.93) H2 (-0.83)

Ag+ SOA SO42-

H2O SRA

RA S2O82-

(-2.01) O2 (-1.23) √ E°cell = + 0.80 -1.23 V = -0.43 V ; 0.43 V are needed

Stoichiometry and Free Energy Worksheet

1. How many coulombs, q, are required to deposit 0.587 g of Ni from a solution of Ni2+?

Ni2+ + 2e¯ Ni

m = 0.587 g

M = 58.69 g/mol

C 10 x 1.93 C 1930

e mol

C 10 x 9.65 x

Ni mol 1

e mol 2 x

g 58.69

Ni mol 1 x g 0.587 q

F x n q then F

q n

3

-

4-

e-e -

2. Three electrolysis cells are connected in series. They contain, respectively, solutions of copper (II)

nitrate, silver nitrate, and chromium (III) sulfate. If 1.00 g of copper is electrochemically deposited in the

first cell, calculate the mass of silver and chromium deposited in the other cells.

1) Cu2+ + 2e¯ Cu

m = 1.00 g

M = 63.55 g/mol

mol 0.0315 Cu mol 1

e mol 2 x

g 63.55

Cu mol 1 x g 1.00 )Cu(n

-

-e

2) Ag+ + e¯ Ag

n = 0.315 mol m = ?

M = 107.87 g/mol

g 3.40 Ag mol 1

g 107.87 x

e mol 1

Ag mol 1 x e mol 0.0315 m

-

-Ag

3) Cr3+ + 3e¯ Cr

n = 0.315 mol m = ?

M = 52.00 g/mol

g 0.546 Cr mol 1

g 52.00 x

-e mol 3

Cr mol 1 x -e mol 0.0315 Crm

3. A constant current of 3.7 milliampere is passed through molten sodium chloride for 9.0 minutes. The

sodium produced is allowed to react with water (500 mL). What is the pH of the resulting solution?

Na+ + e¯ Na

I = 3.7 mA n = ?

t = 9.0 min = 540 s

mol 10 x 2.1

e mol 1

Na mol 1 x

C 10 x 9.65

e mol x s 540 x

s

C 10 x 3.7 n

5-

-4

--3

Na

2 Na + 2 H2O 2 NaOH + H2

n = 2.1 x 10-5 mol C = ?

V = 0.500L

9.62

)10 x (4.2 log 14 pH

M 10 x 4.2

L 0.500

1 x

Na mol 2

NaOH mol 2 x mol 10 x 2.1 C

5-

5-

5-NaOH

4. Given these half-reactions and their standard reduction potentials.

2 ClO4- + 16 H+ + 14 e- Cl2 + 8 H2O Eo (ClO4

-) = + 1.47 V

S2O82- + 2 e- 2 SO4

2- Eo (S2O82-) = + 2.01 V

Calculate:

(a) Complete the REDOX reaction and calculate the Eocell.

an (ox): Cl2 + 8 H2O 2 ClO4- + 16 H+ + 14 e- Eo = - 1.47 V

7 x cat (red): 7 S2O82- + 14 e- 14 SO4

2- Eo = + 2.01 V

7 S2O82- + Cl2 + 8 H2O 14 SO4

2- + 2 ClO4- + 16 H+ Eo

cell = +0.54 V (b) If Ca(NO3)2 (aq) is added and 2.59 g of CaSO4 is produced, calculate the pH of a 30.0 mL solution.

Ca(NO3)2 (aq) + SO42-

(aq) CaSO4 (s)

+ 2 NO3- (aq)

m = 2.59

M = 136.14 g/mol

n = 2.59 g x 1mol

136.14g = 0.0190 mol

7 S2O82- + Cl2 + 8 H2O 14 SO4

2- + 2 ClO4- + 16 H+

0.140

)(0.724 log - pH

M 0.724

L 0.0300

1 x

SO mol 14

H mol 16 x SO mol 0.0190 ][H

-24

-24

5. The system 2 AgI + Sn Sn2+ + 2 Ag + I- has a current of 8.46A run through it for 1.25

minutes. Calculate the mass of silver produced.

Ag+ + e- Ag

I = 8.46A m = ?

t = 75.0 s M = 107.90 g/mol

mAg = -1

4 -1

8.46 C s x 75.0 s 107.90 g x

1 mol Ag9.65 x 10 C mol Ag

= 0.709 g

6. Calculate the current needed to produce 5.0 mL of chlorine gas after 100. seconds at for the following

reaction, if:.

NiO2 + 2 Cl- + 4 H+ Cl2 + Ni2+ + 2 H2O

A 0.39

s 100.

e molC 10 x 9.65 x

Cl mol 1

e mol 2 x

K 298 x KCl 1-molLkPa 8.314

L 0.0050 x kPa 100. I

RT

PV n and

t

Fn I ;

F

It n

-1-4

2

-

1-2

-e-e

Review Questions for SCH 4U Electrochemistry Test

1. Balance the following REDOX reaction in acidic solution

Zn + NO3 Zn

2+ + NH4

+

0 +5 -2 +2 -3 +1

ox: 4 (Zn Zn2+

+ 2 e -)

0 +2

red: NO3 + 8 e - NH4

+

+5 -3

4 Zn + NO3 4 Zn

2+ + NH4

+

4 Zn + NO3 + 10 H+ 4 Zn

2+ + NH4

+ + 3 H2O

2. Balance the following REDOX reaction in acidic solution

MnO4 + C2O4

2 CO2 + MnO2

+7 -2 +3 -2 +4 -2 +4 -2

ox: 3 (C2O42 2 CO2 + 2 e -)

+3 +4

red: 2 (MnO4 + 3 e - MnO2)

+7 +4

2 MnO4 + 3 C2O4

2 2 MnO2 + 6 CO2

2 MnO4 + 3 C2O4

2 + 8 H+ 2 MnO2 + 6 CO2 + 4 H2O

3. Given the following reactions, generate a standard reduction potential table:

W2+ + Z Z2+ + W

X2+ + W W2+ + X

X2+ + Y no rxn

SOA Y2+ + 2e- Y

X2+ + 2e- X

W2+ + 2e- W

Z2+ + 2e- Z SRA

4. Describe and explain what will happen if carbon electrodes are placed in a FeCl2 solution.

Give ALL possible half reactions.

Species in Solution: Fe2+ Cl1- H2O Possible Reduction Half Reactions (OA):

Fe2+ + 2e- Fe Eo = - 0.44 V SOA

2 H2O + 2e- O2 + 4H+ + 4e- Eo = - 0.83 V

Possible Oxidation Half Reactions (RA):

2 Cl1- Cl2 + 2e- Eo = - 1.36 V

Fe2+ Fe3+ + e- Eo

= - 0.77 V SRA

2 H2O O2 + 4H+ + 4e- Eo = - 1.23 V

Full Reaction:

3 Fe2+ Fe + 2 Fe3+ Eocell = -1.21 V

5. Use the redox spontaneity rule to predict whether the following mixtures will be spontaneous or not.

(a) Nickel metal in a solution of silver ions

OA Ag (+0.80) H2 (-0.83)

Ni SRA Ag+ SOA H2O

RA Ni2+ (+0.26) O2 (-1.23)

Eocell = +0.80 V + 0.26 V = +1.06 V

spontaneous

(b) Chlorine gas bubbled into a bromide ion solution

OA Cl- (+1.36) H2 (-0.83)

Cl2 SOA Br- SRA H2O

RA Br2 (-1.07) O2 (-1.23)

Eocell = + 1.36 V - 1.07 V = + 0.29 V

spontaneous

(c) Copper metal in nitric acid

OA NO2 (+0.80) H2 (-0.83)

Cu SRA NO3-, H+

SRA H2O

RA Cu+ (-0.52) (-1.07) O2 (-1.23)

Cu2+ (-0.34)

Eocell = + 0.80 V – 0.34 V = + 0.46 V

spontaneous

6. Three electrolysis cells are connected in series. They contain, respectively, solutions of zinc nitrate,

aluminum nitrate and silver nitrate. If 1.00 g of silver is deposited in the third cell what mass of

aluminum and zinc were deposited in the other cells.

1) Ag+ + e¯ Ag

m = 1.00 g

M = 107.87 g/mol

mol 3-10 x 9.27 Ag mol 1

-e mol 1 x

g 107.87

Ag mol 1 x g 1.00 -en

2) Zn2+ + 2e¯ Zn

n = 9.27 x 10-3 mol m = ?

M = 65.39 g/mol

g 0.303 Znmol 1

g 65.39 x

-e mol 2

Znmol 1 x mol 3-10 x 9.27 Znm

3) Al3+ + 3e¯ Al

n = 9.27 x 10-3 mol m = ?

M = 26.98 g/mol

g 0.0834 Znmol 1

g 26.98 x

-e mol 3

Al mol 1 x mol 3-10 x 9.27 Alm

7. For the cell:

Ag (s) | Ag1+ (aq) || Zn2+ (aq) | Zn (s) a) List all possible half-reactions that will occur at the cathode, including their cell potentials.

Zn2+ + 2e- Zn Eo = - 0.76 V SOA

2 H2O + 2 e- 2 OH1- + H2 Eo = - 0.83 V

b) List the possible half-reactions that will occur at the anode, including their cell potentials.

Ag Ag1+ + e- Eo = - 0.80 V SRA

2 H2O O2 + 4H+ + 4e- Eo = - 1.23 V

c) Give the full balanced REDOX reaction with the value for the cell’s Eo

Zn2+ + Ag Zn + Ag1+ Eocell = -0.80 V – (+0.76V) = -1.56 V

d) Draw a fully labeled diagram of the electrolytic cell.

or

8. For the cell:

Ag (s) | S2- (aq) || HCl (aq) | Pt (s)

a) List all the possible anode reactions with their Eo values.

2 Ag + S2- Ag2S Eo = +0.69 V SRA

2 H2O O2 + 4H+ + 4e- Eo = - 1.23 V

b) List all the possible cathode reactions with their Eo values.

2 H2O + 2 e- 2 OH1- + H2 Eo = - 0.83 V

2 H+ + 2e- H2 Eo = - 0.00 V SOA

c) Give the most probable reaction for the electrochemical cell and the value for the cell’s Eo

2 Ag + S2- + 2 H+ Ag2S + H2 Eocell

= +0.69 V

d) Draw a fully labeled diagram of the cell.

e) As this reaction proceeds, what will happen to the Eo value?

Voltage will decrease because concentration of reactants is decreasing over time.

f) What would happen if HCl(aq) was added to the cathodic half-cell?

If HCl was added, then [H+] would increase, shifting the half reaction to the products.