Worksheet 25 Balancing Redox Reactions in ... - Chemistry 1

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Chemistry 1

Volume 4

Worksheet 25

Balancing Redox Reactions in Basic Solutions

Ion/Electron Method

Part 2


Rules for balancing redox reactions in basic solution:

1. Write the net ionic equation. 2. Write half-reactions. 3. Balance atoms other than hydrogen and oxygen. 4. Balance oxygen by adding H2O. 5. Balance H by adding H+. 6. Balance the charges by adding e-.

a. Add up the total charge on each side. b. Add electrons to the more positive side to balance the charge on both sides.

7. Multiply to balance the electrons gained/lost between the two half-reactions and add reactions together.

8. Cancel common species on both sides of the reaction. 9. Add to both sides of the reaction the same number of OH- as H+. 10. Combine H+ and OH- to form H2O. 11. Cancel any remaining species that appear on both sides of the reaction.

Oxidation number rules: 1. a. For atoms in a neutral atom, molecule, etc., the sum of all oxidation numbers is zero.

b. In an ion, the sum of oxidation numbers is equal to the charge on the ion. 2. In compounds,

a. Group 1A metals always have an oxidation number of +1 b. Group 2A metals always have an oxidation number of +2.

3. In compounds, fluorine (F), has an oxidation number of -1. 4. In most compounds, hydrogen (H) has an oxidation number of +1. 5. In most compounds, oxygen (O) has an oxidation number of -2. 6. In two-element compounds with metals,

a. Group 7A elements have -1 oxidation numbers b. Group 6A elements have -2 oxidation numbers c. Group 5A elements have -3 oxidation numbers

Exceptions:

1. If hydrogen is bonded to a metal, H has a -1 oxidation number. 2. When oxygen is bonded to another oxygen, it has a -1 oxidation number.


1. Balance the redox reaction below that is in basic solution:

NaClO + KCrO2 à K2CrO4 + NaCl



Mn2+ + BiO3- à MnO4

- + Bi3+



SeO2- + Cl2 à Cl- + SeO42-



CN- + MnO4- à CNO- + MnO2



MnO4- + Fe à Mn2+ + Fe2+



ClO3- + Cr(OH)3 à Cl- + CrO4

2-


Answer Key 1. Balance the redox reaction below that is in basic solution:

NaClO + KCrO2 à K2CrO4 + NaCl

Step 1: Write the total ionic equation. Do not worry about balancing elements at this step.

Na+ + ClO- + K+ + CrO2- à K+ + CrO4

2- + Na+ + Cl-

Cancel spectator ions (Na+ and K+) to get the net ionic equation.

ClO- + CrO2- à CrO4

2- + Cl-

Step 2: Break into half-reactions.

CrO2- à CrO4

2-

ClO- à Cl-

Step 3: All elements other than O or H are already balanced.

CrO2- à CrO4

2-

ClO- à Cl-

Step 4: Balance O by adding H2O.

2 H2O + CrO2- à CrO4

2-

ClO- à Cl- + H2O

Step 5: Balance H by adding H+.


2- + 4 H+

2 H+ + ClO- à Cl- + H2O


Step 6: Balance charges by adding e-.


2- + 4 H+ + 3 e-

2 e- + 2 H+ + ClO- à Cl- + H2O

Step 7: Multiply the half-reactions in order to balance electrons and add them together.

2 x [ 2 H2O + CrO2- à CrO4

2- + 4 H+ + 3 e- ]

3 x [ 2 e- + 2 H+ + ClO- à Cl- + H2O ]

6 e- + 6 H+ + 3 ClO- + 4 H2O + 2 CrO2- à 2 CrO4

2- + 8 H+ + 6 e- + 3 Cl- + 3 H2O

Step 8: Cancel common species.

6 e- + 6 H+ + 3 ClO- + 1 4 H2O + 2 CrO2- à 2 CrO4

2- + 2 8 H+ + 6 e- + 3 Cl- + 3 H2O

3 ClO- + H2O + 2 CrO2- à 2 CrO4

2- + 2 H+ + 3 Cl-

Step 9: Add as many OH- to both sides of the equation as there are total H+.

2 OH- + 3 ClO- + H2O + 2 CrO2- à 2 CrO4

2- + 2 H+ + 3 Cl- + 2 OH-

Step 10: Combine H+ and OH- into H2O.

2 OH- + 3 ClO- + H2O + 2 CrO2- à 2 CrO4

2- + 3 Cl- + 2 H2O


2 OH- + 3 ClO- + H2O + 2 CrO2- à 2 CrO4

2- + 3 Cl- + 2 H2O


Step 12: Make sure both atoms and charges are balanced.

2 OH- + 3 ClO- + 2 CrO2- à 2 CrO4

2- + 3 Cl- + H2O

Reactants Products O 9 O 9 H 2 H 2 Cl 3 Cl 3 Cr 2 Cr 2 Charge -7 Charge -7 Everything is balanced! Since there are OH- left over, this confirms it was carried out in basic solution.

Correct answer:

2 OH- + 3 ClO- + 2 CrO2- à 2 CrO4

2- + 3 Cl- + H2O



Mn2+ + BiO3- à MnO4

- + Bi3+

Step 1: The net ionic equation is already written.


Mn2+ à MnO4-

BiO3- à Bi3+


Mn2+ à MnO4-

BiO3- à Bi3+


4 H2O + Mn2+ à MnO4-

BiO3- à Bi3+ + 3 H2O


4 H2O + Mn2+ à MnO4- + 8 H+

6 H+ + BiO3- à Bi3+ + 3 H2O


4 H2O + Mn2+ à MnO4- + 8 H+ + 5 e-

2 e- + 6 H+ + BiO3- à Bi3+ + 3 H2O



2 x [ 4 H2O + Mn2+ à MnO4- + 8 H+ + 5 e- ]

5 x [ 2 e- + 6 H+ + BiO3- à Bi3+ + 3 H2O ]

10 e- + 30 H+ + 5 BiO3- + 8 H2O + 2 Mn2+ à 2 MnO4

- + 16 H+ + 10 e- + 5 Bi3+ + 15 H2O


10 e- + 14 30 H+ + 5 BiO3- + 8 H2O + 2 Mn2+ à 2 MnO4

- + 16 H+ + 10 e- + 5 Bi3+ + 7 15 H2O

14 H+ + 5 BiO3- + 2 Mn2+ à 2 MnO4

- + 5 Bi3+ + 7 H2O

Step 9: Add as many OH- as there are H+ on both sides.

14 OH- + 14 H+ + 5 BiO3- + 2 Mn2+ à 2 MnO4

- + 5 Bi3+ + 7 H2O + 14 OH-

Step 10: Combine OH- and H+ to form H2O.

14 H2O + 5 BiO3- + 2 Mn2+ à 2 MnO4

- + 5 Bi3+ + 7 H2O + 14 OH-

Step 11: Cancel common species (H2O).

7 14 H2O + 5 BiO3- + 2 Mn2+ à 2 MnO4

- + 5 Bi3+ + 7 H2O + 14 OH-



7 H2O + 5 BiO3- + 2 Mn2+ à 2 MnO4

- + 5 Bi3+ + 14 OH-

Reactants Products H 14 H 14 O 22 O 22 Bi 5 Bi 5 Mn 2 Mn 2 Charge -1 Charge -1 Everything is balanced! Since there are OH- left over, this confirms it was carried out in basic solution.

Correct answer:

7 H2O + 5 BiO3- + 2 Mn2+ à 2 MnO4

- + 5 Bi3+ + 14 OH-



SeO2- + Cl2 à Cl- + SeO42-



SeO2- à SeO42-

Cl2 à Cl-

Step 3: Balance all elements other than O or H.

SeO2- à SeO42-

Cl2 à 2 Cl-


3 H2O + SeO2- à SeO42-

Cl2 à 2 Cl-


3 H2O + SeO2- à SeO42- + 6 H+

Cl2 à 2 Cl-


3 H2O + SeO2- à SeO42- + 6 H++ 6 e-

2 e- + Cl2 à 2 Cl-



3 H2O + SeO2- à SeO42- + 6 H++ 6 e-

3 x [ 2 e- + Cl2 à 2 Cl- ]

6 e- + 3 Cl2 + 3 H2O + SeO2- à SeO42- + 6 H++ 6 e- + 6 Cl-


6 e- + 3 Cl2 + 3 H2O + SeO2- à SeO42- + 6 H+ + 6 e- + 6 Cl-

3 Cl2 + 3 H2O + SeO2- à SeO42- + 6 H+ + 6 Cl-


6 OH- + 3 Cl2 + 3 H2O + SeO2- à SeO42- + 6 H+ + 6 Cl- + 6 OH-


6 OH- + 3 Cl2 + 3 H2O + SeO2- à SeO42- + 6 Cl- + 6 H2O


6 OH- + 3 Cl2 + 3 H2O + SeO2- à SeO42- + 6 Cl- + 3 6 H2O



6 OH- + 3 Cl2 + SeO2- à SeO42- + 6 Cl- + 3 H2O

Reactants Products O 7 O 7 H 6 H 6 Cl 6 Cl 6 Se 1 Se 1 Charge -8 Charge -8 Everything is balanced! Since there are OH- left over, this confirms it was carried out in basic solution.

Correct answer:

6 OH- + 3 Cl2 + SeO2- à SeO42- + 6 Cl- + 3 H2O



CN- + MnO4- à CNO- + MnO2



CN- à CNO-

MnO4- à MnO2


CN- à CNO-

MnO4- à MnO2


H2O + CN- à CNO-

MnO4- à MnO2 + 2 H2O


H2O + CN- à CNO- + 2 H+

4 H+ + MnO4- à MnO2 + 2 H2O


H2O + CN- à CNO- + 2 H+ + 2 e-

3 e- + 4 H+ + MnO4- à MnO2 + 2 H2O



3 x [ H2O + CN- à CNO- + 2 H+ + 2 e- ]

2 x [ 3 e- + 4 H+ + MnO4- à MnO2 + 2 H2O ]

6 e- + 8 H+ + 2 MnO4- + 3 H2O + 3 CN- à 3 CNO- + 6 H+ + 6 e- + 2 MnO2 + 4 H2O


6 e- + 2 8 H+ + 2 MnO4- + 3 H2O + 3 CN- à 3 CNO- + 6 H+ + 6 e- + 2 MnO2 + 1 4 H2O

2 H+ + 2 MnO4- + 3 CN- à 3 CNO- + 2 MnO2 + H2O


2 OH- + 2 H+ + 2 MnO4- + 3 CN- à 3 CNO- + 2 MnO2 + H2O + 2 OH-


2 H2O + 2 MnO4- + 3 CN- à 3 CNO- + 2 MnO2 + H2O + 2 OH-


1 2 H2O + 2 MnO4- + 3 CN- à 3 CNO- + 2 MnO2 + H2O + 2 OH-



H2O + 2 MnO4- + 3 CN- à 3 CNO- + 2 MnO2 + 2 OH-

Reactants Products H 2 H 2 O 9 O 9 Mn 2 Mn 2 C 3 C 3 N 3 N 3 Charge -5 Charge -5 Everything is balanced! Since there are OH- left over, this confirms it was carried out in basic solution.

Correct answer:

H2O + 2 MnO4- + 3 CN- à 3 CNO- + 2 MnO2 + 2 OH-



MnO4- + Fe à Mn2+ + Fe2+



MnO4- à Mn2+

Fe à Fe2+


MnO4- à Mn2+

Fe à Fe2+


MnO4- à Mn2++ 4 H2O

Fe à Fe2+


8 H+ + MnO4- à Mn2+ + 4 H2O

Fe à Fe2+


5 e- + 8 H+ + MnO4- à Mn2+ + 4 H2O

Fe à Fe2+ + 2 e-



2 x [ 5 e- + 8 H+ + MnO4- à Mn2+ + 4 H2O ]

5 x [ Fe à Fe2+ + 2 e- ]

5 Fe + 10 e- + 16 H+ + 2 MnO4- à 2 Mn2+ + 8 H2O + 5 Fe2+ + 10 e-


5 Fe + 10 e- + 16 H+ + 2 MnO4- à 2 Mn2+ + 8 H2O + 5 Fe2+ + 10 e-

5 Fe + 16 H+ + 2 MnO4- à 2 Mn2+ + 8 H2O + 5 Fe2+


16 OH- + 5 Fe + 16 H+ + 2 MnO4- à 2 Mn2+ + 8 H2O + 5 Fe2+ + 16 OH-


16 H2O + 5 Fe + 2 MnO4- à 2 Mn2+ + 8 H2O + 5 Fe2+ + 16 OH-


8 16 H2O + 5 Fe + 2 MnO4- à 2 Mn2+ + 8 H2O + 5 Fe2+ + 16 OH-



8 H2O + 5 Fe + 2 MnO4- à 2 Mn2+ + 5 Fe2+ + 16 OH-

Reactants Products H 16 H 16 O 16 O 16 Fe 5 Fe 5 Mn 2 Mn 2 Charge -2 Charge -2 Everything is balanced! Since there are OH- left over, this confirms it was carried out in basic solution.

Correct answer:

8 H2O + 5 Fe + 2 MnO4- à 2 Mn2+ + 5 Fe2+ + 16 OH-



ClO3- + Cr(OH)3 à Cl- + CrO4

2-



ClO3- à Cl-

Cr(OH)3 à CrO42-


ClO3- à Cl-

Cr(OH)3 à CrO42-


ClO3- à Cl- + 3 H2O

H2O + Cr(OH)3 à CrO42-


6 H+ + ClO3- à Cl- + 3 H2O

H2O + Cr(OH)3 à CrO42- + 5 H+


6 e- + 6 H+ + ClO3- à Cl- + 3 H2O

H2O + Cr(OH)3 à CrO42- + 5 H+ + 3 e-



6 e- + 6 H+ + ClO3- à Cl- + 3 H2O

2 x [ H2O + Cr(OH)3 à CrO42- + 5 H+ + 3 e- ]

2 H2O + 2 Cr(OH)3 + 6 e- + 6 H+ + ClO3- à Cl- + 3 H2O + 2 CrO4

2- + 10 H+ + 6 e-


2 H2O + 2 Cr(OH)3 + 6 e- + 6 H+ + ClO3- à Cl- + 1 3 H2O + 2 CrO4

2- + 4 10 H+ + 6 e-

2 Cr(OH)3 + ClO3- à Cl- + H2O + 2 CrO4

2- + 4 H+


4 OH- + 2 Cr(OH)3 + ClO3- à Cl- + H2O + 2 CrO4

2- + 4 H+ + 4 OH-


4 OH- + 2 Cr(OH)3 + ClO3- à Cl- + H2O + 2 CrO4

2- + 4 H2O


4 OH- + 2 Cr(OH)3 + ClO3- à Cl- + 2 CrO4

2- + 5 H2O




2- + 5 H2O

Reactants Products O 13 O 13 H 10 H 10 Cr 2 Cr 2 Cl 1 Cl 1 Charge -5 Charge -5 Everything is balanced! Since there are OH- left over, this confirms it was carried out in basic solution.

Correct answer:


2- + 5 H2O

Worksheet 25 Balancing Redox Reactions in ... - Chemistry 1

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