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|CHEMISTRY 30 |

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|Notes and problems |

|Unit 1: Electrochemical systems |

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|[pic] |

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|Important formulas |

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|Eo = EoOA - EoRA |

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|ne = I . t and m = I . t . M |

|F F . v |

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|Prepared by Daniel Veraart |

|CIS Abu Dhabi |

Chemistry 30

Day 5

Date: ___________________ Name: ____________________________

Introduction and definitions of REDOX Reactions

This unit examines oxidation and reduction reactions; these reactions are explained as an exchange of electrons from specie to another specie.

In chemistry, species can be elements, ions or compounds.

Oxidation-reduction reactions are also called REDOX reactions. The following are some examples of redox reactions where electrons are exchanged:

• Any combustion: CH4(g) + 2 O2(g) ( CO2(g) + 2 H2O(g)

• Corrosion of metals: 2 Fe(s) + O2(g) + 2 H2O(g) ( 2 Fe(OH)2(s)

• Decomposition of compounds into elements (Electrolysis): NaCl(l) ( Na(s) + Cl2(g)

• Any type of voltaic cells (batteries and dry cells)

• Cellular respiration and photosynthesis

All these reactions involve an exchange of electrons and are classified as REDOX reactions.

The term oxidation originally meant the combining of oxygen with other substances. In combustions or corrosion, oxygen reacts and gains electrons to become O2-.

Example: Fe(s) + O2(g) ( Fe2O3(s) (In iron oxide, the oxygen is O2-.)

CH4(g) + 2 O2(g) ( CO2(g) + 2 H2O(g) (In both the products, the oxygen is O2-.)

The term reduction originally applied to the process of reducing the metallic compounds found in ores to pure metals. For example, iron ore can be changed into pure iron:

Example: Fe2O3(s) + 3 CO(g) ( 2 Fe(s) + 3 CO2(g)

To find if a reaction is a redox reaction, you must find if there was an electron transfer between two species. Lets compare these two reactions:

Reaction 1: BaCl2(aq) + Na2SO4(aq) ( BaSO4(s) + 2 NaCl(aq)

Reaction 2: Mg(s) + 2 HCl(aq) ( H2(g) + MgCl2(aq)

In Reaction 1, there is no transfer of electrons between elements (or ions). None of the elements (or ions) changes their charges. It is not an example of Redox reaction.

Ba2+(aq) ( Ba2+(aq) Cl-(aq) ( Cl-(aq) Na+(aq)( Na+(aq) SO42-(aq) ( SO42-(aq)

In Reaction 2, Mg (neutral) loses electrons and becomes Mg2+ (charge of +2). Also, H+ gains electrons and becomes H2 with no charge.

Mg(s) ( Mg2+(aq) and H+(aq) ( H2(g)

| How to recognize a Redox reaction? |

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|Oxidation-reduction reaction: Reaction that involves a loss and a gain of electrons. |

|If the charge of an element or ion changes during a reaction, it is an example of Redox reaction. |

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|Important Tip: Most redox reactions involve elements. If you see an element in a reaction, the reaction is always a redox reaction. |

|Precipitation and neutralization are not examples of Redox reactions |

Examples:

1. Write the charge of each species (elements or ions) and indicate if the reactions are Redox or not.

A) 3 AgNO3(aq) + FeCl3(aq) ( 3 AgCl(s) +Fe(NO3)3(aq) : ______________

B) Mg(s) +Cu(NO3)2(aq) ( Mg(NO3)2(aq) + Cu(s) : ______________

C) SnCl4(l) + Fe (s) ( SnCl2(s) + FeCl2(s) : ______________

D) H2SO4(aq) + Ba(OH)2(aq) ( BaSO4(s) + 2 HOH(l) : ___________

|What is a half-reaction? |

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|A half reaction represents what is happening to only one reactant in a redox reaction. |

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|Example: For the reaction: Mg(s) + 2 HCl(aq) ( H2(g) + MgCl2(aq) |

|The balanced two half reactions would be: |

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|Mg(s) ( Mg2+(aq) + 2e = Oxidation |

|2H+(aq) + 2e ( H2(g) = Reduction |

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|Oxidation : The Loss of Electrons in a half reaction is called an Oxidation (LEO) |

|Reduction : The Gain of Electrons in a half reaction is called a Reduction (GER) |

Examples

1. Complete the following half-reaction by adding the number of electrons required to balance the half reaction. Indicate if the half reaction is an example of reduction ( R ) or oxidation ( O ).

a) Ba(s) ( Ba2+(aq) b) Al (s) ( Al3+(aq)

c) Fe3+(aq) ( Fe (s) d) F2(g) ( F-(aq)

e) Fe3+(aq) ( Fe2+(s) f) O2(g) ( O-2(aq)

g) H2Se(s) ( Se(aq) + 2 H+(aq) h) 2 H2O(g) ( H2(g) + 2 OH-1(aq)

i) P3-(aq) ( P4(s) j) Cl-(aq) + 2 OH-1(aq) ( OCl-(aq) + H2O(g)

k) PbO2(s) + SO42-(aq) + 4 H+(aq) ( PbSO4(s) + 2 H2O(g)

Reducing agent and oxidizing agent

| Terms used in Redox reactions |

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|1. Reduction half reaction: Reaction involving a gain of electrons. |

|2. Oxidation half reaction: Reaction involving a loss of electrons. |

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|3. Reducing agent: A element or ion that loses electrons to another element or ion |

|4. Oxidizing agent: An element or ion that gains electrons from another element or ion |

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|5. Reduced specie: Element or ion that gained electrons. |

|6. Oxidized specie: Element or ion that lost electrons. |

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|7. Anode: The electrode of a cell where the oxidation half reaction occurs |

|8. Cathode: The electrode of a cell where the reduction half-reaction occurs |

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|To remember these terms, you can use the following memory aid : |

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|GERCOR and LEOARO |

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|GE : Gain Electrons LE : Lost Electrons |

|R : Example of Reduction half-reaction O : Example of Oxidation half-reaction |

|C : Happens at the Cathode of a cell A : Happens at the Anode of a cell |

|O : Done by a n Oxidizing agent R : Done by a Reducing agent |

|R : The substance is Reduced O : The substance is Oxidized |

|Important information to remember |

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|All the metals are examples of Reducing agents because they lose electrons. |

|Non-metals are examples of Oxidizing agents because they gain electrons. |

|Anions (Negatives ions) are examples of Reducing agents because they lose electrons. |

|Most cations (positive ions) are examples of Oxidizing agents because they gain electrons. |

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|Exceptions: Some of the metallic ions with multiples charge (like Copper that can be 2+ or 1+) could be either a Reducing agent or an Oxidizing |

|agent. |

Examples

1. For the following redox reactions:

a) Find the charge of each element or ion

b) Write the balanced half reduction and half oxidation reactions.

c) Find the reducing agent and the oxidizing agent. Determine what species are oxidized and reduced.

a) 3 Zn(s) + 2 Au(NO3)3(aq) ( 2 Au(s) + 3 Zn(NO3)2(aq)

The half reactions are: ___________________________ and ____________________________

The oxidizing agent is: ________________ The reducing agent is: _____________

The oxidized specie is: _______________ The reduced specie is: _____________

b) Cl2(g) + 2 NaBr(aq) ( Br2(l) + 2 NaCl(aq)

The half reactions are:_________________________ and ____________________________

The oxidizing agent is: ________________ The reducing agent is: _____________

The oxidized specie is: _______________ The reduced specie is: _____________

c) 2 H2(g) + O2(g) ( 2 H2O(g)

The half reactions are: ___________________________ and ____________________________

The oxidizing agent is: ________________ The reducing agent is: _____________

The oxidized specie is: _______________ The reduced specie is: _____________

|Do Homework #5 |

Lab #1

Testing redox reactions

Date: ___________________ Name: ____________________________

Problem: What are the products of the following redox reactions? Identify the reducing and oxidizing agents.

1. Solid Copper and a Solution of Iron (III) nitrate

2. Solid Aluminum and Hydrochloric acid

3. Solution of Lead (II) nitrate and a Solution of sodium sulphate

4. Solid Zinc and a Solution of copper (II) sulfate

5. Hydrochloric acid and solid Sodium hydroxide (use bromothymol blue)

Procedure:

1. In a small test tube, we will react the reactants listed above.

2. Record all your observations in a table of observations.

3. Complete the table below by filling out the blank areas

4. Dispose of the solutions as directed by your teacher.

Questions:

1. Complete the table of observations.

|Reaction |Observations |Redox or not |Half reactions |Reducing and oxidizing |

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|Cu(s) + Fe(NO3)3(aq) | | | | |

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|Al(s) + HCl(aq) | | | | |

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|Pb(NO3)2(aq) + Na2SO4(aq) | | | | |

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|Zn(s) + CuSO4(aq) | | | | |

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|HCl(aq) + NaOH(aq) | | | | |

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Lab #2

Spontaneity of Redox Reactions

Problem: Which combinations of copper, lead, Iron and zinc metals and their aqueous metal ion solution produce spontaneous reactions.

Procedure:

1. 1/3 fill each of the 5 small test-tubes with the copper solution.

1. Place a small portion of each of the metals into a different tube.

2. Let it sit for 2 minutes, observe and record any changes.

3. Repeat step 1 - 3 for each of the other solutions provided.

4. Dispose of the metal strips and aqueous solutions as directed by the teacher.

Questions:

1. Complete the table of observations.

| |Cu2+(aq) |Pb2+(aq) |Fe3+(aq) |Zn2+(aq) |

|Cu(s) |X | | | |

|Pb(s) | |X | | |

|Fe(s) | | |X | |

|Zn(s) | | | |X |

2. List the metallic ions from the most reactive to least reactive.

___________ - _____________ - ______________- _____________

Most reactive Least

3. List the metals from the most reactive to least reactive.

___________ - _____________ - ______________- _____________

Most reactive Least

Chemistry 30

Day 6

Date: ___________________ Name: ____________________________

The reduction Half-reaction table

Until now in our studies, it has been assumed that all chemical reactions are spontaneous. In Lab #2, you found out that the assumption that all reactions are spontaneous must be unacceptable since many reactions didn’t work. The question that arises is “ How do we know when a chemical reaction will occur?” To answer this question, we have to look at the evidence collected in Lab #2. First, let’s compare the reactivity of the five ions in solution, which are the Oxidizing agents:

Cu2+(aq) Pb2+(aq) Fe3+(aq) Zn2+(aq)

( Decreasing reactivity of the Oxidizing agent (

Second, lets compare the reactivity of the five metals, which are the Reducing agents:

Zn(s) Fe(s) Pb(s) Cu(s)

( Decreasing reactivity of the Reducing agent (

|Important information to remember |

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|Notice the first list is ordered from copper ions to zinc ions and the second list is ordered from zinc to copper. The lists are reversed. |

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|The metal ion with the greatest tendency to gain electrons reacts to produce the metal which is least likely to lose electrons. |

|A strong oxidizing agent becomes a weak reducing agent. |

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|The metal which is most likely to lose electrons reacts to produce the metal ion with the least tendency to gain electrons. |

|A strong reducing agent becomes a weak oxidizing agent. |

The two lists of reactivity can be combined as half reduction equation, as shown below:

Strong Oxidizing Agent Weak Reducing Agent

Cu2+(aq) + 2e- ( Cu(s)

Pb2+(aq) + 2e- ( Pb(s)

Fe3+(aq) + 3e- ( Fe(s)

Weak Oxidizing Agent Zn2+(aq) + 2e- ( Zn(s) Strong Reducing agent

| Important information to remember |

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|This list is called a table of reduction half reactions. The list expands to contain many other half reactions and is included in your data |

|booklet P. 7 |

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|In that list, Fluorine (F2) is the SOA (strongest oxidizing agent). Fluorine is very reactive and will take electrons from almost any |

|substance. |

|In that list, Lithium (Li) is the SRA (strongest reducing agent). Lithium is also very reactive and will give electrons to almost any |

|substance. |

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|Some species can act as either reducing or oxidizing agents (Fe2+, Cr2+, Sn2+, H2O) , |

Evidence from many redox reactions, for which half reactions have been listed in this way, has been used to establish a generalization, called the redox spontaneity rule.

| Redox spontaneity rule |

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|A spontaneous redox reaction occurs only if the oxidizing agent (OA) is LISTED HIGHER than the reducing agent (RA) on the |

|table. |

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|OA |

|( Spontaneous reaction |

|RA |

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|If the oxidizing agent is listed lower than the reducing agent, then the reaction is non-spontaneous. |

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|RA |

|( Non-spontaneous reaction |

|OA |

Examples:

Complete the following table.

|Species |OA or RA |Would react with |Would not react with |

|Cobalt solid | | | |

|Solution of Fe(NO3)3 | | | |

|Solution of AgNO3 | | | |

|Solution of HNO3 | | | |

|Chlorine gas | | | |

|Iron solid | | | |

|Hydrogen gas | | | |

|Hydrogen peroxide | | | |

|Zinc Solid | | | |

Building a reduction Half-reaction table

Using the spontaneity rule, we can generate a reduction table to find the Relative Strengths of oxidizing and reducing agents.

Example #1

Three reactions were performed to find the Relative Strengths of different oxidizing and reducing agents. Construct a reduction Half-reaction table and find the Relative Strengths of the OA and RA.

Reaction #1: 3 Co2+(aq) + 2 In(s) ( 2 In3+(aq) + 3 Co(s)

Reaction #2: Cu2+(aq) + Co(s) ( Co2+(aq) + Cu(s)

Reaction #3: Cu2+(aq) + Pd(s) ( No evidence of reaction

Example #2

What is the Relative Strengths of oxidizing agents among A2, B2, C2, and D2? Each substance is added to solutions of the other three ions. Use the following table of observation to find the Relative Strengths of these oxidizing agents.

|Reactions of different oxidizing agents with solutions of ions |

| A(aq) B(aq) C(aq) D(aq) |

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|A+(aq) --- ( ( --- |

|B+(aq) --- --- ( --- |

|C+(aq) --- --- --- --- |

|D+(aq) ( ( ( --- |

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|( indicates an evidence of reaction |

Example #3

Given the following spontaneous reactions,

Y(s) + X+2(aq) ( Y+2(aq) + X(s)

Q(s) + X+2(aq) ( Q+2(aq) + X(s)

Z(s) + Y+2(aq) ( Z+2(aq) + Y(s)

Q(s) + Z+2(aq) ( Q+2(aq) + Z(s)

Construct a reduction Half-reaction table and find the Relative Strengths of the OA and RA.

|Do Homework #6 |

Lab #3

Identification of a metal using the Redox spontaneity rule

Problem: How can we identify a metal as lead, silver, zinc, cobalt or iron using the rule of redox spontaneity?

Design a procedure of how you could identify the metal given by your teacher.

Questions:

1. Write a step by step procedure.

2. Make a table of observations.

3. Identify the metal. Explain your answer.

Chemistry 30

Day 7

Date: ___________________ Name: ____________________________

Predicting Redox reactions

The following steps will be used to help us writing a balanced net ionic equation for a Redox reaction.

|Steps to write a net ionic Redox equation |

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|Step 1 : List all the chemical entities present in the mixture. Label all possible oxidizing and reducing agent. Identify the strongest oxidizing|

|agent (SOA) and strongest reducing agent (SRA). |

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|Important notes: Water is always present in aqueous solutions. |

|Acidic solutions contains H+ ions. |

|Some Oxidizing agent or Reducing agent are combinations; for example, MnO4- (aq) and H+(aq) |

|H2O(l), Fe2+(aq), Sn2+(aq) and Cr2+(aq) may act as either OA or RA. |

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|Step 2 : Using the Reduction table, write both half reactions (Oxidation and Reduction). |

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|You will have to reverse the Oxidation reaction because the half reaction given |

|in the data booklet are all Reduction half reaction. |

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|Step 3 : Multiply the half reactions by the smallest whole number to make the gain of electrons equal the loss of electrons. |

|Add the half reactions together. Write the net ionic equation. |

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|Step 4 : Determine whether or not the reaction is spontaneous (Using the spontaneity rule). |

Example #1

Predict the most likely net ionic redox reaction to occur when a solution of iron (II) nitrate is titrated with an acidic solution of potassium dichromate.

Step 1:

Step 2:

Step 3-4:

Example #2

Predict the most likely net ionic redox reaction to occur when a strip of silver is placed in a mixture of nitric acid and aqueous sodium iodide.

Step 1:

Step 2:

Step 3-4:

Example #3

Predict the most likely net ionic redox reaction to occur when sulfuric acid and aqueous solutions of potassium permanganate and tin (II) chloride are mixed together.

Step 1:

Step 2:

Step 3-4:

Redox stoichiometry

Redox titration is the progressive measured addition of an oxidizing agent to a reducing agent (or vice-versa) until the endpoint is reached (change of color).

The endpoint occurs when the reaction is completed and therefore stoichiometric calculation can be done. If one of the reacting species undergoes a color change, the endpoint can easily be detected.

|Important information to remember |

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|Two commonly used oxidizing agents in redox titration are: |

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|1. Acidic permanganate ions |

|MnO4-(aq) + 8 H+(aq) + 5e- ( Mn2+(aq) + 4 H2O(l) |

|(Purple) (Very light pink) |

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|2. Acidic dichromate ions. |

|Cr2O72-(aq) + 14 H+(aq) + 6e- ( 2 Cr3+(aq) + 7 H2O(l) |

|(orange) (Green) |

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|These substances are often used in redox stoichiometry because they are very strong oxidizing agents and they undergo a visible |

|color change. |

|Stoichiometric calculation |

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|The endpoint is the point where a 100% reaction has occurred and a stoichiometric calculation can be made with the volume of oxidizing agent used|

|(equivalence point). The steps for stoichiometric calculation are: |

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|Chem 20 steps: |

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|Step 1: Write the balanced net ionic equation for the redox reaction. |

|Step 2: Calculate the number of moles of the oxidizing agent or the reducing agent using the formulas: n = C . V or n = m/M |

|Step 3: Find the molar ratio between the Oxidizing agent and the Reducing agent |

|Step 4: Answer the question by using the formula C = n/V or m = n . M |

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|New Lazy Chem 30 step: |

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|Step 1: Use the following “Homemade” formula created by Mr. Veraart: |

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|R x CO x VO = O x CR x VR |

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|Where: R = Coefficient of the RA |

|O = Coefficient of the OA. |

|CO = Concentration of the OA |

|CR = Concentration of the RA |

|VO = Volume of the OA |

|VR = Volume of the RA |

Example #1:

During an experiment, a student titrates 15.0 mL of an acidic chromium (II) solution with a 0.100 mol/L dichromate solution.

These are the results:

Trial Initial volume of buret Final volume of buret

1 0.0 mL 21.0 mL

2 21.0 mL 45.0 mL

3 0.0 mL 24.2 mL

Find the concentration of the chromium (II) solution.

Example #2:

In a titration experiment all the Br-(aq) ions in an acidic solution were oxidized to Br2(aq) by an acidic 0.0200 mol/L KMnO4(aq) solution.

The volume of Br-(aq) solution used was 25.0 mL and the volume of KMnO4(aq) solution required was 15.0 mL. Calculate the concentration of Br-(aq) ions in the solution.

|Do Homework #7 |

Lab #4

Find the concentration of a reducing agent using acidified KMnO4

Problem: What is the molar concentration of Fe2+(aq) in an acidic sample solution?

Experimental design:

10.0 mL of an acidic solution of Fe2+(aq) is titrated using a standard solution of KMnO4(aq) . The molar concentration is obtained by stoechiometric calculations and the volume of KMnO4(aq used at the equivalence point.

Questions:

1. Write a step by step procedure of the titration with a diagram of the set-up.

2. Make a table of observation.

3. Determine the molar concentration of the iron (II) solution.

4. Calculate the percent of error.

5. Why do we need an acidified solution of Fe2+(aq)

Chemistry 30

Day 8

Date: ___________________ Name: ____________________________

Predicting Redox reactions by constructing Half-Reactions

In the previous section you were able to predict a net ionic redox reaction using half reaction provided in the Reduction table. What if half reactions are not provided in the table? In this case, you have to write your own half-reactions using the following rules:

| Rules for writing half reactions |

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|1. Split the unbalanced equation into two half reactions. |

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|2. Balance each half reaction separately: - Balance all atoms except oxygen and hydrogen |

|- Balance O atom by adding H2O |

|- Balance H atoms by adding H+ |

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|3. Balance the electron exchange. |

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|4. Add the two half reactions for a net reaction, check atoms and charges. |

Example#1: Balance the following reaction: Zn(s) + VO3-(aq) ( Zn2+(aq) + VO2+(aq)

Example#2: Write the following reaction: Fe(s) + NO3-(aq) ( Fe3+(aq) + NH4+(aq)

Oxidation number

|The oxidation numbers |

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|An oxidation number is a positive or negative number corresponding to the oxidation state assigned to an atom. |

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|These numbers provide information about the number of electrons that have been shifted towards or away from an atom during a redox reaction. |

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|Formal charge : the hypothetical charge that would result if all bonding electrons are shared equally |

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|Oxidation state (or oxidation number) : the hypothetical charge that would result if all bonding electrons are assigned to the more |

|electronegative atom in the bond |

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|(# of VE in free atom) – (# of VE in bonded atom) |

Common Oxidation number

|Atom or Ion |Oxidation number |Examples |

|All atoms in elements |0 |Na , Cl2 , H2, Mg, … |

|Hydrogen in most compounds |+1 |H in HCl is +1 |

| | |H in C6H6O6 is +1 |

|Hydrogen in hydrides |-1 |H in LiH is –1 |

|Oxygen in most compounds |-2 |O in H2O is –2 |

| | |O in C6H6O6 is –2 |

|Oxygen in peroxides |-1 |O in H2O2 is –1 |

|Alkali metals |+1 |Na+ is +1 |

|Earth alkali metals |+2 |Mg+2 is +2 |

|Oxidation numbers of other elements |

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|The oxidation number of any other elements in a molecule or complex ions are assigned so that the sum of the oxidation numbers |

|of all atoms is equal to the net charge of the molecule or ion. |

Examples : Assign oxidation numbers to all species in the following :

1. CH4 2. MnO4- 3. N2 4. C4H10 5. H2O2

6. C6H6O6 7. CO 8. CO2 9. Na2CO3 10. CH3OH

|Important information to remember |

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|During oxidation, the oxidation number of the reducing agent increases. |

|The oxidation number of an atom increases when it is oxidized (loses electrons). |

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|During reduction, the oxidation number of the oxidizing agent decreases. |

|The oxidation number of an atom decreases when it is reduced (gains electrons). |

Examples : Assign oxidation numbers to all species in the following reactions and state which species increases its oxidation number and which species decreases its oxidation number.

1. 2 Sn(S) + O2(g) ( 2 SnO(S)

Oxidation number increases: _________ Oxidation number decreases: _________

2. 2 Cl-(l) + 2 H2O(l) ( Cl2(g) + H2(g) + 2 OH-(aq)

Oxidation number increases: _________ Oxidation number decreases: _________

3. MnO-(aq) + 8H+(aq) + 5Fe2+(aq) ( Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

Oxidation number increases: _________ Oxidation number decreases: _________

Disproportionation

|What is a disproportionation reaction? |

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|A disproportionation is the reaction of a single substance with itself to produce two different substances. The same substance is oxidized and |

|reduced. Some molecules of this substance oxidize some other molecules, which become reduced. |

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|The substances that can do a disproportionation reaction are substances that can act either as an OA or as a RA. |

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|Example #1: 3 OCl-(aq) ( 2 Cl-(aq) + ClO3-(aq) |

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|In this example, some Cl+1 become Cl+5 and some become Cl-1 |

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|Example #2: 3 Fe2+(aq) ( Fe(s) + 2 Fe3+(aq) |

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|In this example, some Fe+2 become Fe0 and some become Fe+3 |

| Important redox reactions |

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|Photosynthesis: Plants are the only living organisms capable of using energy from the Sun in a series of endothermic reactions called |

|photosynthesis. |

|The photosynthesis reaction is a redox reaction in which the carbon in CO2 is the oxidizing agent and the oxygen ion in the water is the |

|reducing agent. |

|6 CO2(g) + 6 H2O(l) ( C6H12O6(aq) + 6 O2(g) |

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|Cellular respiration: Both animals and plants breakdown glucose to produce their energy in a series of exothermic reactions called cellular |

|respiration. |

|The cellular respiration reaction is a redox reaction in which the carbon in C6H12O6 is the reducing agent and the oxygen gas is the |

|oxidizing agent. |

|C6H12O6(aq) + 6 O2(g) ( 6 CO2(g) + 6 H2O(l) |

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|Combustion of hydrocarbons: Hydrocarbons (Methane, propane,…) burn to produce energy. |

|Combustion is a redox reaction in which the carbon in the hydrocarbon is the reducing agent and the oxygen gas is the oxidizing agent. |

|CH4(g) + 2 O2(g) ( CO2(g) + 2 H2O(l) |

|Do Homework #8 |

Chemistry 30

Day 9

Date: ___________________ Name: ____________________________

Introduction and definitions of cells and batteries

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|Important definitions to remember |

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|Electrochemical cell: A cell that converts chemical energy into electrical energy (or vice-versa) by a redox reaction. |

|Voltaic cell: A cell that spontaneously produces electricity by redox reactions. |

|Electrolytic cell: A cell in which electrolysis occurs by redox reactions. |

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|Voltaic Cell Electrolytic Cell |

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|Battery: A battery is a group of two or more voltaic cells connected to each other to produce an electric current |

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|Electrodes: Solid electrical conductor (usually a metal or carbon rod) in a cell where the connections are made; the site of oxidation or |

|reduction. |

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|Cathode: The positive electrode in a cell. |

|Anode: The negative electrode in a cell. |

|Electrolyte : A solute that forms a solution that conducts electricity. |

Design of a voltaic cell

The design of a cell “plays a trick” on OA and RA, resulting in electrons passing through an external circuit rather than directly from one substance to another.

Since the RA and OA in a cell would react together, we have to separate them into two different compartments so they don’t touch each other.

|Important parts of a cell |

|Two electrodes. Each electrode is in contact with an electrolyte, but the electrolytes surrounding each electrode are separated by a porous |

|boundary. |

|Two common examples of porous boundaries used in laboratories are a salt bridge and a porous cup. |

|A salt bridge (ex. NaCl) A porous cup |

|[pic] |

|The role of the porous boundary is to separate the electrolytes while still permitting ions to move between the two solutions. The migration of ions|

|maintains the overall electrical neutrality of the cell. |

|A cell is always split into two parts connected by a porous boundary. Each part, called a half-cell, consists of one electrode and one electrolyte.|

| |

|In almost all cells, one of the electrodes is the RA and one of the ELECTRODES is the OA. |

Example of a Copper-zinc cell

In a copper-zinc cell, copper metal in a solution of copper ions is in one half cell and zinc metal in a solution of zinc ions is in another half cell. It can be represented as follows:

Cu(s) | Cu(NO3) 2(aq) || Zn(s) | Zn(NO3) 2(aq)

[pic]

As the Copper-Zinc cell operates, this is what happens:

Observations on the right side: The zinc electrode decreases in mass.

Reaction happening : Zn(s) ( Zn2+(aq) + 2e (Oxidation)

This half reaction produces an excess of positively charged Zn2+(aq) in the solution surrounding the zinc electrode.

The electrode where oxidation occurs is called the anode.

Observation of the left side: - The copper electrode increases in mass (more copper deposit)

- The intensity of the blue color of the electrolyte decreases.

Reaction happening : Cu2+(aq) + 2e ( Cu(s) (Reduction)

This half reaction removes positively charged Cu2+(aq) ions from the solution leaving an excess of negative charge.

The electrode where reduction occurs is called the cathode.

The Silver-Copper Cell

Ag(s) | AgNO3(aq) || Cu(s) | Cu(NO3) 2(aq)

[pic]

Observations on the right side: - The copper electrode decreases in mass

- The intensity of the blue color of the electrolyte increases.

- Oxidation occurs: Cu(s) ( Cu2+(aq) + 2e

Observation of the left side: - The silver electrode increases in mass (more silver deposit)

- Reduction occurs: Ag+(aq) + e ( Ag(s)

Observation in the U-tube: - A blue color slowly moves up the U-tube from the copper half-cell.

( Meaning that the Cu2+(aq) are moving

Other observations: - Electrons move from the copper electrode to the silver electrode and record a voltage of 1.56 V

- We can’t see it but the NO3-(aq) is also moving from the cathode to the anode. The Na+(aq) is also moving towards the cathode.

|Summary on Voltaic cells |

| |

|A voltaic cell consists of two half-cells separated by a porous boundary with solid electrodes connected by an external circuit. |

| |

|The porous boundary helps the flow of ions through the solution, maintaining the overall electrical neutrality of the cell. |

| |

| |

|The cathode is the positive electrode and the location of the reaction of the strongest oxidizing agent present. This is where the reduction |

|happens. |

| |

|GERCOR |

| |

|The anode is the negative electrode and the location of the reaction of the strongest reducing agent present. This is where the oxidation happens. |

| |

|LEOARO |

| |

|Electrons travel in the external circuit from the anode to the cathode. |

| |

|Internally, anions move toward the anode and cations move toward the cathode as the cell operates. |

| |

|The migration of ions is essential to the operation of the cell since the accumulation of ionic charges around the electrode would oppose the |

|movement of the electrons. |

| |

|If the OA is very strong or the RA is a solution or a gas, we must use an inert electrode. |

| |

|An inert electrode is made of solid conductor that will not react in the cell or interfere with the cell reaction. |

|A carbon (graphite) rod or platinum metal foil is two commonly used inert electrode (Figure 16.6 in the book). |

| |

|Do Homework #9 |

Project on commercial cell

Objective: Each group will research the components of a commercial cell and present the information gathered on a poster.

Parts of the research:

1. Name of the cell

2. Diagram of the cell

3. Half reactions and net reaction

4. Identify the RA and the OA

5. Voltage

6. Observations at the anode and cathode

7. Ion flow

8. Give one application of the cell

9. List at least two characteristics of the cell.

9. References

Examples of cells:

- Car battery (Lead-Acid)

- Common dry cell (Zinc-MnO2)

- Rechargeable cell (Cadmium)

- Fuel cells (Aluminum-Air fuel cell)…

- Molicel (Lithium ion cell)

- ….

|A very special cell: The fuel cell |

| |

|A fuel cell produces electricity by the reaction of a fuel that is continually supplied to keep the cell operating. |

|The first fuel cell was the hydrogen-oxygen fuel cell. |

|A aluminum-air fuel cell is used in electric cars and also the Ballard fuel cell is used in buses (Book P.661). |

| |

|In principle, a fuel cell operates like a normal cell. Unlike a normal cell, a fuel cell does not run down or require recharging. |

| |

|It will produce energy in the form of electricity and heat as long as fuel is supplied. |

| |

|The hydrogen-oxygen fuel cell consists of two electrodes sandwiched around an electrolyte (See picture on the next page). |

| |

|Oxygen passes over one electrode and hydrogen over the other, generating electricity, water and heat. |

The Fuel Cell

[pic]

Hydrogen fuel is fed into the "anode" of the fuel cell. Oxygen (or air) enters the fuel cell through the cathode. Encouraged by a catalyst, the hydrogen atom splits into a proton and an electron, which take different paths to the cathode. The proton passes through the electrolyte. The electrons create a separate current that can be utilized before they return to the cathode, to be reunited with the hydrogen and oxygen in a molecule of water.

Since there is no hydrocarbon combustion, emissions of carbon dioxide and carbon monoxide are nil.

Oxidation ½ reaction: _______________________ Reduction ½ reaction: _______________________

Overall reaction with potential: __________________________________________

Chemistry 30

Day 10

Date: ___________________ Name: ____________________________

The Standard Cell Potentials = Voltage ((Eo values)

|Standard Cell and Standard Cell Potential ((Eo or E°net) |

| |

|Definitions: A standard cell is a voltaic cell in which each half-cell contains all entities at SATP conditions, with a concentration of 1.0 |

|mol/L for the electrolytes. |

|A standard cell potential or voltage ((Eo) is the electric potential difference of a standard cell. It’s the energy difference between the cathode |

|and the anode. |

| |

| |

|Standard Reduction Potential (Eor) |

|A standard reduction potential is the ability of a half-cell to attract electrons. |

| |

|Important Formula to find the Potential (Voltage) of a cell |

|The standard cell potential is the difference between the reduction potentials of the two half-cells. |

| |

| |

|(Eonet = Eor(OA) - Eor(RA) |

| |

It is impossible to determine the reduction potential of a single half cell. A voltmeter can only measure a potential difference.

In order to assign values for standard reduction potentials, the standard hydrogen half-cell is internationally used as the reference half-cell.

A half cell such as this, is chosen as reference and arbitrarily assigned an electrode potential of exactly zero volts.

|The Standard Reference Hydrogen Half-Cell |

| |

|The standard reference Hydrogen half cell consists of an inert platinum electrode immersed in a 1.00 mol/L solution of H+(aq) ions, with hydrogen |

|gas at SATP bubbling over the electrode. |

| |

|Standard reduction potentials for all other half-cells are measured relative to that of the standard hydrogen half-cell defined as 0.00 V. |

| |

|2H+(aq) + 2e H2(g) Eor = 0.00 V |

|[pic] |

The hydrogen half-cell is then paired with each substance on the reduction table in a voltaic cell at SATP. The recorded voltage represents the Eor of the substance.

[pic]

Examples

1. Calculate the standard cell potential (Eonet) of the following cells.

a) Copper-hydrogen: ___________________________________________________

b) Aluminum-hydrogen: _________________________________________________

c) Copper-Zinc: _______________________________________________________

d) Dichromate-Lead: ____________________________________________________

Other types of problems

Problem #1: What if another half-cell was assigned an E0 value of 0.00 V instead of the hydrogen half-cell?

Example : If the Cu2+(aq) / Cu(s) reduction half-reaction was assigned as the reference cell, then what would be reduction potential of the Ni2+(aq) / Ni(s) half-reaction on that table?

Problem #2: What if cells are connected in series, how will the voltage change?

Example : If a copper-silver and a copper-zinc standard cells are connected in series. What would be the overall potential?

Problem #3: How can you determine the identity of an unknown half-cell from the cell potential involving a know half-cell?

Example : Determine the reduction potential and the identity of the unknown X2+(aq) / X(s) if

2 Ag+(aq) + X(s) ( 2 Ag(s) + X2+(aq) (Eo = 1.08V

Example : The voltage of an electrochemical cell is +0.20 V . If one of the half reactions is the reduction of Cu2+(aq) , then what is the other half reaction?

Problem #4: How can you determine the oxidation potential of a half-reaction?

Example : What is the standard oxidation potential for the oxidation of Lead?

| |

|Do Homework #10 |

Lab #6

Testing Voltaic Cells

Problem: What are the voltages of different cells constructed from various combinations?

Experimental design:

Using different electrodes and electrolytes, each group will build 2 cells (using a porous cup and a U-tube) and record the voltage then compare with the theoretical voltage.

Questions:

1. Draw a picture of your 2 cells and identify all the important components and flow of ions and electrons.

2. Make a table of observation for your recorded voltage and the theoretical voltage. Calculate the % error between the recorded voltage and the theoretical voltage for each cell.

3. For both cells, write the ½ reactions and observations that you should see happening (over time) at the anode and the cathode.

Lab #7

Concentration of the electrolytes and net potential

Problem: Is the concentration of the electrolyte affects the potential of a cell?

Questions:

1. Write a short experimental procedure that would help you answer the question?

2. Make a table of observation.

List the different variables of the experiment.

Chemistry 30

Day 11

Date: ___________________ Name: ____________________________

Corrosion and its problems

|Corrosion (Oxidation) of metals |

| |

|For most metal, as they oxidize (corrode), an oxide forms and adheres tightly to the surface of the metal. This prevents further corrosion by sealing any|

|exposed surfaces. |

| |

|Magnesium, chromium, tin, lead, aluminum and many more metals form such a thin almost transparent layer. |

| |

|For magnesium, the process can be simply represented by two half reactions: |

| |

|Oxidation : Mg(s) ( Mg2+(s) + 2e |

|Reduction : O2(g) + 4e ( 2 O2-(g) |

|Overall : 2 Mg(s) + O2(g) ( Mg2+(s) + 2 O2-(g) (or MgO(s)) |

| |

|The MgO2(s), that forms on the surface on the magnesium protects further corrosion of the magnesium. |

|Most of these oxides are transparent and shinny. The copper (II) oxide is more of a greenish color and the lead oxide is almost black. |

|Corrosion of iron or steel |

| |

|Unfortunately iron does not corrode that way. For example, as iron corrodes, the iron oxide that forms on the surface does not adhere very well and |

|flakes off, exposing new iron to be corroded. |

|Corrosion of iron (or steel) costs the public millions of dollars annually. The corrosion of iron requires the presence of both oxygen and water. |

| |

|The process can be simply represented by two half reactions: |

| |

|Oxidation : Fe(s) ( Fe2+(aq) + 2e |

|Reduction : O2(g) + H2O(l) + 4e ( 4 OH-(aq) |

|Overall : 2 Fe(s) + O2(g) + H2O(l) ( Fe(OH)2(s) |

| |

|The Fe(OH)2(s), that forms is further oxidised to form rust, a mixture of hydrated oxides Fe2O3 xH2O(s) . |

|The presence of salt or acidity speeds up the process. Ships rust faster in seawater and cars rust faster where salt is used on roads. |

|Steel is an alloy of iron, it is made with iron and carbon combined together to make it stronger. The iron in the steel rusts the same way. |

Corrosion of iron

[pic]

|Corrosion Prevention |

| |

|Corrosion protection can be divided into two categories: |

| |

|Protective coating like paint or other coatings (like tin in cans) are a simple way or corrosion prevention. It creates a barrier against|

|the oxidizing agents like oxygen and water. |

| |

|Cathodic protection (forcing the metals to become a cathode). |

|Paint and coatings |

| |

|Paint and other similar coatings (tape, oil, chromium plating, etc.) are a simple method of corrosion prevention. This method works well as long as |

|the coating remains intact. |

| |

|Unfortunately, a scratch or a chip in the surface can easily expose a small surface of iron and corrosion begins. |

| |

|Tin, chromium, and zinc are often used as metallic coatings. These metals rapidly oxidize in air to form an oxide compound. |

| |

|Ex. : 2 Sn(S) + O2(g) ( 2 SnO(S) |

| |

|The tin oxide that forms adheres tightly to the surface of the metal. This prevents further corrosion by effectively sealing any exposed surfaces. |

|Cathodic protection |

| |

|The process of corrosion prevention called cathodic protection is used when protective coating can’t be used. For example, we can’t paint a ship or |

|underground tanks. |

| |

|During the cathodic protection of iron for example, the iron is forced to become the cathode (instead of the anode where it usually rusts); this is |

|done by supplying the iron with a flow of electrons. |

| |

| |

|This supply of electrons can be done two different ways: |

| |

|1. By connecting the iron with a continuous electric current. |

| |

|This method of corrosion prevention requires a constant electric power supply and is used for pipelines and culverts. |

|When the iron is supplied continuously with the electrons, it becomes like a cathode and the iron is now prevented from corroding. |

|The electrons coming from a source of electricity will be gained by the oxidizing agent; the OA will not take electrons from the iron = No corrosion. |

| |

| |

|Cathodic protection using an impressed current |

|[pic] |

| |

|2. By using a sacrificial anode. |

| |

|( A sacrificial anode is a metal that is more easily oxidised than Iron (a stronger Reducing agent than Iron) and is connected to the iron object to |

|be protected. |

|( Zinc and magnesium are the most common sacrificial anodes used because they are strong RA and there oxides are non-toxic. |

|( The metal is slowly consumed or sacrificed at the anode, forcing the iron object to be the cathode of the cell. |

|( The practice of zinc plating (galvanizing) iron objects is an example of this method. |

|( Sacrificial zinc or magnesium is also connected to the exposed underwater metal surfaces of ships to prevent the corrosion of the iron in the steel.|

| |

| |

| |

Examples of questions

Question #1

Talon Tapes of Edmonton manufactures plastic tape containing small pieces of magnesium. This tape is completely wrapped around iron pipes that will be buried underground. Explain in chemical terms the purpose(s) of each component of the tape. Your response should include relevant half-reactions.

Question #2

A student leaves a copper penny and a gold coin outside for a month to find out which one corrode the most. After the month has passed, the student finds out that the gold coin hasn’t changed at all while the penny has turn green.

Explain why the gold hasn’t changed while he copper has. Your explanation should include chemical equations.

|Do Homework #11 |

Chemistry 30

Day 12

Date: ___________________ Name: ____________________________

What is electrolysis? What is an electrolytic cell?

|Important definitions to remember |

| |

|Electrolysis: The process of supplying electrical energy to force a non-spontaneous redox reaction to occur. Electrolysis is mostly used to produce elements |

|from compounds. |

| |

| |

|Electrolytic cell: This type of cell consists: |

| |

|- Two electrodes (usually inert) |

|- One electrolyte (usually the electrolyte acts as the oxidizing agent) |

|- An external power source. |

| |

| |

|[pic] |

|In this particular cell, a solution of NaCl(aq) is changed into Chlorine gas, Hydrogen gas and Sodium hydroxide aqueous using an external source of |

|electricity. |

| |

|Electrolytic cell |

|Reactants + Electrical energy ( Products |

Comparaison of voltaic and Electrolytic Cells

| |Voltaic Cell |Electrolytic Cell |

|Spontaneity |Spontaneous reaction |Non-spontaneous reaction |

|Cell potential (Eo |Positif |Negatif |

| |SOA undergoes reduction |SOA undergoes reduction |

|Cathode |Positive electrode |Negative electrode |

| |SRA undergoes oxidation |SRA undergoes oxidation |

|Anode |Negative electrode |Positive electrode |

|Electron movement |Anode ( Cathode |Anode ( Cathode |

|Ion movement |Anions ( Anode |Anions ( Anode |

| |Cations ( Cathode |Cations ( Cathode |

The use of electrolysis

|Three types of electrolysis |

| |

|1. The electrolysis of an aqueous solution. |

| |

|( In this type of electrolysis, the water is often the Reducing agent. |

|( The cations in the electrolyte are the oxidizing agent. The cations will often change into an element. |

| |

|2. The electrolysis of a molten ionic compound (The compounds has to be pure and liquid). |

| |

|( In this type of electrolysis, the compound is melted so water is not present. |

|( The cation are changes to metals at the cathode |

|( The anions are changes to non metals at the anode. |

| |

|3. The electrolysis of Brine = The Chlor-Alkali Cell |

| |

|( In this type of electrolysis, a high voltage is used to decompose a saturated solution |

|of salt (NaCl(aq)) into Chlorine gas, Hydrogen gas and Sodium hydroxide aqueous. |

|( The water acts as the oxidizing agent. |

|( The chloride ion acts as the reducing agent. |

Electrolysis #1: The electrolysis of an aqueous solution

Example: Electrolysis of CuSO4(aq)

|Explanation of the electrolysis |

| |

|In the CuSO4(aq) cell, the Oxidizing agent is the Cu2+(aq) and the Reducing agent is the H2O(l) . |

| |

|The reaction at the anode is: 2 H2O(l) ( O2(gl + 4 H+(aq) + 2e |

|The reaction at the cathode is: Cu+2(aq) + 2e ( Cu(s) |

|The net reaction is: Cu+2(aq) + 2 H2O(l) ( Cu(s) + O2(gl + 4 H+(aq) |

| |

|The standard cell potential is: (Eo = Eo(OA) - Eo(RA) = +0.34V – (+1,23V) = -0.89 V |

| |

|Observations: A deposit of copper should appear at the cathode. |

|Bubbles of gas (Oxygen) should appear at the anode. |

|The pH should decrease at the anode. |

|The minimum voltage required is 0.89 V. |

Electrolysis #2: The electrolysis of molten sodium chloride = Downs Cell

Example: In the Downs cell, the salt (NaCl(l)) is heated to its melting point and then put into the Downs Cell where a electric current is applied.

[pic]

|Explanation of the electrolysis |

| |

|In this cell, the Oxidizing agent is the Na+(l) and the Reducing agent is the Cl-(l) . |

| |

|The reaction at the anode is: 2 Cl-(l) ( Cl2(g) + 2e |

|The reaction at the cathode is: 2 ( Na+(l) + e ( Na(l) ) |

| |

|The net reaction is: 2 Cl-(l) + 2 Na+(l) ( Cl2(g) + 2 Na(l) |

| |

| |

|The standard cell potential is: (Eo = Eo(OA) - Eo(RA) = -2.71 – (+1,36V) = -4.07 V |

| |

| |

|Observations: Sodium liquid is produced at the cathode. |

|Bubbles of gas (Chlorine) should appear at the anode. |

|The minimum voltage required is 4.07 V. |

| |

| |

|Note: Electrolysis of molten ionic compounds is not a simple technology. The high temperature to melt the compounds causes problems for cell |

|components and increases the cost of production. |

|No water is present in this cell. |

Electrolysis #3: The electrolysis of brine = Chlor-Alkali Cell

Example: Electrolysis of Brine = Saturated NaCl(aq)

To overcome the cost of melting sodium chloride, a third design can be used. The Chlor-Alkali Cell.

This technology is used near Edmonton, Alberta. In the Chlor-Alkali cells, a fairly high voltage and very high current are applied to a saturated sodium chloride solution.

The use of high voltage and current overpower the reduction of water permitting the chloride ions to be reduced instead of the water.

[pic]

|Explanation of the electrolysis |

| |

|In this cell, the Oxidizing agent is the H2O(l) but you have to remember that the Reducing agent is not water but Cl-(aq) . This is only the case |

|with the Chlor-Alkali Cells due to the high voltage and high current. |

| |

|The reaction at the anode is: 2 Cl-(aq) ( Cl2(g) + 2e |

|The reaction at the cathode is: 2 H2O(l) + 2e ( H2(g) + 2 OH-(aq) |

| |

|The net reaction is: 2 Cl-(l) + 2 H2O(l) ( Cl2(g) + H2(g) + 2 OH-(aq) |

| |

| |

|Observations: Bubbles of gas (Hydrogen) is produced at the cathode. |

|Bubbles of gas (Chlorine) is produced at the anode. |

|The OH-(aq) produced at the cathode combines with Na+(aq) to produce NaOH(aq). |

| |

|Notes: The chlorine gas is used to produce bleaches, plastics, pesticides, solvents,… |

|The NaOH(aq) is concentrated by evaporation and sold as caustic soda. |

Commercial uses of electrolysis

|Three major uses of electrolysis |

| |

|1. The production of elements from ionic compounds. |

| |

|Most elements in nature are combined with oxygen or other elements in a stable ionic compound. |

|If the compound is soluble in water, the electrolysis of the aqueous solution is used. |

|If the compound has a low solubility or is a weak RA or OA, the molten compound has to be used. |

| |

|2. The refining of metals. |

| |

|Electrolysis is used to refine all the metals |

| |

|3. Electroplating |

| |

|Several metals, such as silver, gold, copper, and chromium, are valuable because of their beauty and their resistance to corrosion. |

|The coating of an object with these metals is called electroplating. |

|Plating of a metal happens at the cathode of an electrolytic cell. |

| |

|Do Homework #12 |

Chemistry 30

Day 13

Date: ___________________ Name: ____________________________

Stoichiometry of cell reactions

|The Faraday constant |

| |

|In an electrolytic cell, the quantity of electricity (charge in Coulomb) that passes through a cell determines the mass of substances that react at |

|the electrodes. |

| |

|The charge (in Coulomb) is determined by multiplying the current (in amp) by the time (in sec): |

| |

|q = I.t |

| |

|Michael Faraday was the first to discover the relationship between electricity and electrochemical changes. He came up with a constant that |

|corresponds to the charge of 1.0 mol of electrons. |

| |

|Faraday Constant (F) = 9.65 x 104 C/mol |

| |

|The formula to find the number of moles of electrons |

| |

|The Faraday Constant can be used as a conversion factor in converting electric charge to an amount of moles of electrons transferred in a cell. |

| |

|ne = I . t_____ |

|F |

Example: Convert a current of 3.50 A for 1.00 hour into an amount in moles of electrons.

|Stoichiometry of cells |

| |

|Since the mass of an element produced at an electrode depends on the number of moles of electrons transferred, a half reaction is necessary in order|

|to do stoichiometric calculations. |

| |

|The four steps of stoichiometry in cells: |

| |

|Step #1: Write the half reaction happening at the electrode. |

| |

|Step #2: Find the number of moles of electrons using: ne = (I. t) / F |

| |

|Step #3: Molar ratio between the number of electrons and the substance. |

| |

|Step #4: Find the mass of the substance. |

| |

|Simplified formula: |

| |

|Step #1: Use the following “Homemade” formula created by Mr. Dan: |

| |

|m = ne . M or m = I . t . M |

|e F . e |

| |

|Where: m = mass of the substance (g) I = Current (A) |

|T = Time (s) F = Faraday’s Constant 9.65 x 104 C/mol |

|e = Number of electrons in the half reaction of the substance |

Example:

1. If 20.0 A of current flows for 1.00 h through an electrolytic cell containing molten aluminum oxide. What mass of aluminium will be deposit at the cathode?

2. During the electrolysis of KI(l), if the mass of the element formed at the anode is 18.1 g, what would be the mass of element formed at the cathode?

|Do Homework #13 |

-----------------------

Read the following page: P. 606 to 611

Look at the following: Figure 15.3–15.4–15.5-15.6-15.7

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[pic]

Read the following page: P. 568 to 572

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Read the following page: P. 573 to 578

Read the following page: P. 596 - 597

[pic]

[pic]

Read the following page: P. 581 to 582

Read the following page: P. 585 to 588

[pic]

[pic]

Read the following page: P. 612 to 615

[pic]

Read the following page: P. 622 to 626

[pic]

Read the following page: P. 627 to 631

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Read the following page: P. 634 to 637

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Read the following page: P. 639 to 640

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Read the following page: P. 641 to 648

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Read the following page: P. 649 to 650

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Read the following page: P. 652 to 656

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