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Chemistry 3202 Unit 4: Electrochemistry

Introductory Concepts

Electrochemistry involves the study of all reactions in which a transfer of electrons takes place. Some reactions produce electricity for our use (e.g. batteries) Some reactions require electricity (e.g. electrolysis of water) Some reactions occur where no electricity is produced or used (e.g. 2Na + Cl2 → 2NaCl)

Oxidation & Reduction Reactions p.712

Reactions in which electrons are transferred from one substance to another Also known as redox reactions Common examples:

o reactions in cellular respiration o the rusting of an apple o combustion reactions o chemical reactions that occur in batteries

Examples:

1. A Zn strip is placed in a solution of copper(II)sulfate:

Nonionic Equation:

Total Ionic Equation:

Net Ionic Equation:

In terms of electron transfer, Cu2+ gains 2 electrons and Zn(s) loses 2 electrons to Cu2+ . There are 2 half reactions that together form the overall redox equation

Half reactions: a reaction that involves the gain or loss of electrons, with the electrons written in the equation

Zn(s) → Zn2+ (aq) + 2 e- (Zn loses electrons, Oxidation half reaction) Cu2+ (aq) + 2 e- → Cu(s) (Cu gains electrons, Reduction half reaction) _________________

Overall: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) (The reaction should be balanced for the number of atoms/ions and charge)

Oxidation:

the loss of one or more electrons The substance being oxidized (losing electrons) is the reducing agent

Reduction:

the gain of one or more electrons The substance being reduced (gaining electrons) is the oxidizing agent

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In this example: oxidizing agent is Cu2+ (it is being reduced by Zn; it gains 2 electrons from Zn) reducing agent is Zn (it is being oxidized by Cu2+; it loses 2 electrons to Cu2+)

To help you remember: LEO the lion says “GER” (Loses Electrons: Oxidation, Gains Electrons: Reduction. OIL RIG (Oxidation Involves Loss of e-, Reduction Involves Gain of e-) 2. A strip of copper is placed in a solution of silver nitrate:

3. Sodium chloride can be produced from sodium metal and chlorine gas. 4. Zinc metal displaces hydrogen from aqueous solution of acids according to the total ionic

equation:

Zn(s) + 2H+ (aq) → Zn2+ (aq) + H2(g)

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Practice: For each of the following:

a. write the net ionic equation b. write the oxidation and reduction half-reactions c. state what is oxidized and reduced d. identify the reducing and oxidizing agent

1. A copper strip is placed in a solution of zinc nitrate 2. A copper strip is placed in a solution of lead (II) nitrate 3. A lead strip is placed in a solution of silver nitrate 4. A silver strip is placed in a solution of lead (II) nitrate 5. An iron strip is placed in a copper(II)sulfate solution. Iron(III) ions are produced. 6. Potassium chloride is decomposed into its elements. 7. Page 715: #1 - 4

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OXIDATION NUMBERS The oxidation number of an element in a compound is defined as the charge an atom has, or appears to have, when the electrons are counted, according to the following set of rules. Guidelines:

1. Free elements have the oxidation # of 0 eg. Zn = 0 , O2= 0 , I2= 0 2. For monatomic (simple) ions, the oxidation # is equal to the charge of the ion.

e.g. Cl- = -1, Zn2+= +2 3. The oxidation # of hydrogen in most compounds is +1

e.g. CH4 or H2S, oxidation # for H= +1;

(exception: metal hydrides eg: NaH, sodium hydride. Na is +1 and H is -1)

4. The oxidation # of oxygen in most compounds is -2.

e.g. Li2O or KNO3, oxidation # for O is -2

(exceptions: peroxides eg. H2O2, H is +1 and O2 is -1; In OF2, O is +2) 5. The sum of all of the oxidation numbers in a neutral compound must total zero. e.g. Li2O 2(+1) + (-2) = 0

Note: Known oxidation numbers can be used to find oxidation #s of unknown elements in a compound: e.g. H3PO4 neutral compound = 0 H is known, O is known, P is not

6. The sum of all oxidation numbers in a polyatomic ion must equal the charge on the ion

e.g. MnO4- charge is -1 O is known

7. In covalent compounds that do not include hydrogen or oxygen, the more electronegative element is assigned the negative charge it would have in an ionic compound. e.g. CS2 Oxidation # for S is -2

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Using Oxidation Numbers to Identify Redox Reactions

Since redox reactions involve real or apparent electron transfers, the species involved will undergo changes in oxidation number the species that is oxidized will have an increase in oxidation number the species that is reduced will have a decrease in oxidation number

If there is no change in oxidation number, the reaction is not a redox reaction

Examples: 1. NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl (aq)

2. Cu(s) + 2AgNO3 (aq) → 2Ag(s) + Cu(NO3)2 (aq) 3. ZnO (s) + C(s) → CO(g) + Zn(s) 4. Zn(s) + I2(s) → ZnI2(s) 5. 2 Al(s) + 3 Br2 (l) → Al2Br6(s)

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Identifying a Redox Reaction by Reaction Type

Any reaction that involves an element as a reactant or product is definitely redox. All composition, decomposition, single replacement and combustion are redox reactions. Double replacement reactions are never redox reactions!

Balancing Simple Redox Equations A. In neutral solution Steps:

1. Write the half reactions (including number of electrons) 2. Balance each half reaction for number of atoms/ions 3. Balance each half reaction for charge:

Multiply one or both of the half reactions by a value that will make the number of electrons lost during oxidation equal the number of electrons gained during reduction

4. Add the two half reactions to give the overall equation

Examples: 1. Balance the following redox equation: Ag+ (aq) + Cu(s) → Cu2+ (aq) + Ag(s)

a) Half reactions: b) Balance the atoms/ions.

c) Balance for charge (# of electrons lost = # of electrons gained)

d) Add the two half reactions:

2. Balance the following redox reaction: Au3+ (aq) + Zn(s) → Au(s) + Zn2+(aq)

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Balancing Simple Redox Equations B. In acidic solution General Steps:

1. Write the half reactions (do not include electrons) 2. Balance all atoms except Oxygen and Hydrogen 3. Balance Oxygen atoms by adding H2O to the side that needs O. 4. Balance hydrogen by adding H+ to the side that needs H 5. Balance the charge by adding electrons to the side with more positive charge 6. Make electrons gained equal to electrons lost 7. Add the two half-reactions to get the overall equation.

Examples: 1. Balance the following equation for the reaction of Cr2O72- with Fe2+ in an acidic solution.

Cr2O72- + Fe2+ → Cr3+ + Fe3+ Step 1: Divide into half-reactions

Step 2: Balance atoms other than H & O Step 3: Balance oxygen atoms by adding H2O to the side that needs O.

Step 4: Balance hydrogen by adding H+ to the side that needs H Step 5: Balance the charge by adding electrons Step 6: Make electrons gained equal to the electrons lost Step 7: Add the two half-reactions to get the overall equation

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2. Balance the following equation in acidic solution: MnO4-(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)

3. Balance the following equation in acidic solution:

MnO4-(aq) + H2SO3(aq) → SO42-(aq) + Mn2+(aq)

4. Balance the following equation in acidic solution: Zn(s) + NO3-(aq) → Zn2+(aq)+ NO(g)

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Balancing Simple Redox Equations C. In basic solution The easiest way to balance these equations is to treat the solution is acidic and then complete 3 more steps General Steps:

1. Write the half reactions (do not include electrons) 2. Balance all atoms except Oxygen and Hydrogen 3. Balance Oxygen atoms by adding H2O to the side that needs O. 4. Balance hydrogen by adding by adding H+ to the side that needs H 5. Balance the charge by adding electrons to the side with more positive charge 6. Make electrons gained equal to the electrons lost. 7. Add the two half-reactions to get the overall equation. 8. Add the same # of BOH as there are H+ to both sides of the equation 9. Combine OH- and H+ to form H2O 10. Cancel any H2O that you can and get overall equation

Examples: 1. Balance the following equation

SO32- (aq) + MnO4- (aq) → SO34- (aq) + MnO2(aq) Steps 1-7: Step 8: Add the same # of BOH as there are H+ to both sides of the equation Step 9: Combine OH- and H+ to form H2O Step 10: Cancel any H2O that you can and get overall equation

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2. Balance the following redox equation in basic solution: ClO-(aq) + CrO2-(aq) → Cl-(aq) + CrO42-(aq)

3. Balance the following redox equation in basic solution:

Bi(OH)3 (s) + SnO22-(aq) → Bi(s) + SnO32-(aq) 4. Balance the following equation in basic solution:

MnO2 (aq) + ClO3-(aq) → MnO4-(aq) + Cl-(aq)

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ELECTROCHEMISTRY WORKSHEET #2 1. Balance the following reactions in acidic solution.

a. BrO3- (aq) + SO2 (g) → SO42-(aq) + Br2 (l)

b. I(s) + NO3- (aq) → IO3-(aq) + NO (g)

c. C3H7OH (aq) + Cr2O72- (aq) →C2H5COOH (aq) + Cr3+(aq)

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2. Balance the following reactions in basic solution.

a. Ni(s) + NO3-(aq) → NH4+(aq) + Ni2+(aq)

b. P4 (s) + IO3- (aq) → H2PO42-(aq) + I-(aq)

c. Cr(OH)63-(aq) + BrO-(aq) → CrO42-(aq) + Br-(aq)

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Redox Stoichiometry (Titrations) · As with any titration, a redox titration uses a known concentration of an oxidizing agent

to find an unknown concentration of a reducing agent, or vice-versa. · Practical applications of this include finding the iron content of water or the Vitamin C

content in foods. General Steps:

1. Balance the redox equation 2. Organize data for the given and required substances 3. Find the moles of the given substance

If given volume and concentration, use n=cv If given mass, find molar mass, then use n= m/M

4. Find the moles of the required substances, using mole ratio 5. Convert moles to concentration (C=n/v) or volume (v=n/C)

or mass (m=nM) of required substance Example 1: The following example involves the titration of a solution of sodium oxalate, Na2C2O4(aq) with a solution of potassium permanganate, KMnO4(aq). Important Notes: · The permanganate ion, MnO4-, is often used as an oxidizing agent because of its deep

purple color, which allows to act as its own indicator. · It gets reduced to pale pink Mn2+ by the oxalate ions. · Once all oxalate ions have been oxidized the next drop of permanganate ions turns the

solution a faint purple color, signaling the endpoint of the titration. Find the volume of 0.200 M solution of MnO4- which will react with 50.0 mL of 0.100 M solution of C2O42-.

MnO4- + C2O42- Mn2+ + CO2

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Example 2: A 23.30 ml sample of KMnO4 solution is decolorized by 0.1111g of oxalic acid (H2C2O4). The products are Mn2+ and CO2 gas. Calculate the concentration of KMnO4 using the following equation:

KMnO4 (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) Example 3: The oxidation of Fe2+ ions by permanganate in acidic solution is accompanied by the formation of Fe3+ ions and Mn2+ ions. How many moles of FeSO4 would be oxidized by 100.0 ml of 0.02118 mol/L KMnO4 solution.

Fe2+ + MnO4- Fe3+ + Mn2+

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ELECTROCHEMISTRY WORKSHEET #3 1. In a redox titration 12.50 mL of 0.0800 mol/L K2Cr2O7 (aq) was used in acidic solution to

oxidize Sn2+ (aq) ions to Sn4+ (aq) ions. The volume of K2Cr2O7 (aq) used was just sufficient to oxidize all the Sn2+ (aq) in 10.0 mL of the solution. Calculate the concentration of the Sn2+(aq) ions in the solution according to the following unbalanced equation.

(Ans: 0.300 mol/L) Cr2O72-(aq) + Sn2+(aq) Sn4+(aq) + Cr3+(aq)

2. The copper (II) ions in a solution can be converted to copper metal by trickling the

solution over scrap iron. The reaction produced iron (II) ions from scrap iron. If the process produces 25.00 L of solution containing 0.00200 mol/L of Fe2+(aq) ions, what mass of copper is produced?

(Ans: 3.18g) Cu2+(aq) + Fe(s) Fe2+(aq) + Cu(s)

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3. What volume of 0.0500 mol/L KmnO4 (aq) is needed to oxidize all the Br- (aq) ions in 25.0 mL of an acidic 0.200 mol/L NaBr(aq) solution according to the following unbalanced equation.

(Ans: v = 20.0mL) MnO4-(aq) + Br-(aq) Br2 (aq) + Mn2+(aq)