chemistry 40s electrochemistry€¦ · chemists have designed a system to keep track of electrons...

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Chemistry 40S Electrochemistry

(This unit has been adapted from https://bblearn.merlin.mb.ca)

Name:

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Lesson 1: Oxidation and Reduction Reactions

Goals: • Relate the role of oxygen to the process of rusting and burning. • Define oxidation and reduction. • Determine oxidation numbers for atoms in simple compounds and ions. • Identify the oxidizing agent, the reducing agent, the substance reduced

and the substance oxidized, given a balanced chemical equation. Defining Oxidation and Reduction

Oxidation-reduction reactions

For example, a reaction of the burning of magnesium: 2Mg (s) + O2 (g) → 2MgO (s) In this reaction magnesium begins as a neutral atom and loses two electrons to become a Mg2+ ion in MgO. Magnesium is oxidized. Oxygen begins as a neutral atom and gains the two electrons from magnesium to become an O2– ion in MgO. Oxygen is reduced.

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Oxidation Numbers Not all redox reactions are as simple as the ones on the previous page, so chemists have designed a system to keep track of electrons in a chemical reaction. They have assigned oxidation numbers to all atoms and ions. The oxidation number represents the charge the atom would have if every bond were ionic. Not every bond is ionic, but chemists assume they are for this system. The oxidation number is not always the actual charge, but it is very helpful to follow electrons in redox reactions. The rules for assigning oxidation numbers are as follows: • Oxidation numbers are always assigned PER ATOM. • The oxidation numbers of all uncombined elements is zero. For

example: O2, H2, etc… • The oxidation number of monatomic ions equals the charge of that ion. • In compounds, the oxidation number for alkali metals. For example, Li,

Na, K, etc… is always +1. • In compounds, the oxidation number for the alkaline earth metals. For

example, Be, Mg, Ca, etc… is always +2. • In compounds, the oxidation number of aluminum is always +3. • In compounds, the oxidation number of fluorine is -1. • In compounds, the oxidation number of hydrogen is +1. An exception is

in metal hydrides, such as NaH or MgH2, when hydrogen is -1. • In compounds, the oxidation number of oxygen is -2. An exception is in

peroxides, such as H2O2 or Na2O2, when its oxidation number is -1. • For any neutral compound, the sum of the oxidation numbers for each

atom must be zero. • For a polyatomic ion, the sum of the oxidation numbers for each atom

must be the charge of that ion. There is a difference between ion charge and oxidation numbers. For example, the magnesium ion, Mg2+, has an ion charge of 2+ and an oxidation number of +2.

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Example 1: Assign oxidation numbers to each atom in SO2. Example 2: Assign oxidation numbers for each atom in K2Cr2O7. Example 3: Assign oxidation numbers for each atom in Fe(NO3)3.

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Example 4: Is the reaction SO2 + H2O → H2SO3 a redox reaction? Example 5: Is the reaction Cu (s) + 2AgNO3 (aq) → Cu(NO3)2 (aq) + 2Ag (s) a redox reaction?

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Practice: Oxidation and Reduction Reactions

1. Assign oxidation numbers for each element in the following compounds.

a) Na2Cr2O7 b) KNO3 c) FeCl2

d) H2C2O4 e) HClO3 f) KMnO4

g) MnSO4 h) H2SO4 i) H2

k) NH4NO3 l) NO2 l) C2H5OH

2. Identify which of the following are redox reactions.

a) 2NO2 → N2O4

b) 2 Mg + O2 → 2MgO

c) Mg+ + 2Ag+ + 2NO3– → Mg2+ + 2NO3

– + 2Ag

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Lesson 2: Oxidizing and Reducing Agents

Goals: • Identify the substance oxidized and the substance reduced in a redox

reaction. • Identify the oxidizing and reducing agents in a redox reaction. • Determine the number of electrons transferred in a redox reaction. Defining Oxidizing and Reducing Agents

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Example 1: Nitric acid reacts with hydrogen sulfide according to the balanced equation below. Identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent for the burning of propane.

2HNO3 (aq) + 3H2S (g) → 2NO (g) + 3S (s) + 4H2O (l)

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Electrons Transferred Redox reaction is defined as a chemical reaction in which electrons are transferred from one atom to another. We can use a balanced equation to determine how many electrons have been transferred. To determine the number of electrons transferred in a redox reaction… Step 1: Assign oxidation numbers Step 2: Determine atoms gaining and losing reactions Step 3: Multiply the number electrons lost or gained by the number of atoms gaining or losing the electrons Example 2: For the redox reaction Cu + 2AgNO3 → Cu(NO3)2 + 2Ag. How many electrons are transferred?

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Example 3: How many electrons are transferred in the reaction below? 2HNO3 (aq) + 3H2S (g) → 2NO (g) + 3S (s) + 4H2O (l)

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Practice: Oxidizing and Reducing Agents

Which of the following equations represents a redox reaction? For the redox reactions, identify:

• element oxidized • element reduced • oxidizing agent • reducing agent • number of electrons transferred

1. H2 + I2 → 2HI

2. 2Ag + Mg(NO3)2 → 2AgNO3 + Mg

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3. I2 + 2Na2S2O3 → Na2S4O6 + 2NaI

4. MgO + SO3 → MgSO4

5. 2KMnO4 + 8H2SO4 + 10KBr → 5Br2 + 6K2SO4 + 2MnSO4 + 8H2O

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6. 3Ag2S + 8HNO3 → 6AgNO3 + 2NO + 3S + 4H2O

7. 2C2H6 + 3O2 → 4CO2 + 6H2O

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Lesson 3: Balancing Equations Using Redox Methods

Goals: • Balance redox equations using the oxidation number method. • Balance redox equations in acidic and basic solutions using the half

reaction method.

The Oxidation Method Example 1: Use the oxidation method to balance the equation below.

K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3

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Example 2: Balance the following equation using the oxidation method. P + HNO3 + H2O → NO + H3PO4

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Why We Balance Electrons Lost and Gained First? Example 3. Balance the reaction of Cu + Cl2 → Cu+ + Cl-. Half-Reaction

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Balancing Redox Reactions by Half Reaction Method

Steps for Balancing Redox Reactions in Acidic Solutions: There are about 8 steps to this method. Eventually, you may wish to combine some of theses steps to speed up the balancing process, but never skip steps. The order of the steps is also important.

Step 1: Assign oxidation numbers. Step 2: Identify the substances oxidized and reduce then write the oxidation and reduction half-reactions. Step 3: Balance all elements except hydrogen and oxygen. Step 4: Add electrons lost and gained. Step 5: Balance oxygen atoms by using H2O. Step 6: Balance hydrogen atoms using H+ ions. Step 7: Balance the number of electrons lost and gained. Step 8: Add the two half-reactions.

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Example 4: Balance the following reaction in an acidic solution. Cr2O7

2- (aq) + SO32- (aq) → Cr3+ (aq) + SO4

2- (aq)

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Example 5: Balance the following equation in an acidic solution. MnO4

- + I-→ MnO2 + I2

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Steps for Balancing Redox Reactions in Acidic Solutions: The first part of balancing redox reactions in basic solutions follows the same steps as that for acidic solutions. Step 1: Assign oxidation numbers. Step 2: Identify the substances oxidized and reduce then write the oxidation and reduction half-reactions. Step 3: Balance all elements except hydrogen and oxygen. Step 4: Add electrons lost and gained. Step 5: Balance oxygen atoms by using H2O. Step 6: Balance hydrogen atoms using H+ ions. Since the reaction occurs in a base, we must add hydroxide ions. Step 7: Before adding the two half-reactions, you add the same number of hydroxide ions as hydrogen ions to BOTH sides of the equation. Step 8: We eliminate the hydrogen ions by forming water

(H+ + OH-→ H2O).

Then continue as we did with acid. Step 9: Balance the number of electrons lost and gained. Step 10: Add the two half-reactions.

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Example 6: Balance the following reaction in a basic solution. MnO4

- + C2O42- → CO2 + MnO2

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Example 7: Balance the following reaction in a basic solution. N2O + ClO-

→ NO2- + Cl-

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Practice: Balancing Equations Using Redox Methods Complete the questions on loose-leaf.

1. Balance the following using the Oxidation Number Method.

a) HNO3 + C2H6O + K2Cr2O7 → KNO3 + C2H4O + H2O + Cr(NO3)3

b) Al + CuSO4→ Al2(SO4)3 + Cu

c) Cu + HNO3→ Cu(NO3)2 + NO2 + H2O

d) KMnO4 + KCl + H2SO4→ MnSO4 + K2SO4 + Cl2 + H2O 2. Use the Half-Reaction Method to balance the following redox reactions

in acidic solutions.

a) S2- + Cr2O72- → S + Cr3+

b) Br- + SO4

2- → Br2 + SO2

c) Zn + NO3-→ Zn2+ + N2O

d) C2H5OH + NO3

-→ HC2H3O2 + NO2

3. Use the Half-Reaction Method to balance the following redox reactions

in a basic solution.

a) AsO43- + NO2

-→ AsO2

- + NO3-

b) ClO3

- + N2H4→ NO + Cl-

c) Zn + BrO4-→ Zn(OH)4

2- + Br-

d) ClO3- + BH4

-→ Cl- + H2BO3

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