electrochemistry. transfer of electrons oxidation: electrons are lost (product side of equation),...

Electrochemistry

TRANSFER OF ELECTRONS

Oxidation: Electrons are lost (product side of equation), charge increases. The substance that is oxidized is the reducing agent.

Reduction: Electrons are gained (reactant side of equation), charge decreases. The substance that is reduced is the oxidizing agent.

Oxidation-Reduction (Redox) Reactions

2 Fe2O3(s)4 Fe(s) + 3 O2(g)Rusting of iron:an oxidation of Fe

4 Fe(s) + 3 CO2(g)2 Fe2O3(s) + 3 C(s)Manufacture of iron: a reduction of Fe3+


Oxidation: The loss of one or more electrons by a substance, whether element, compound, or ion.

Reduction: The gain of one or more electrons by a substance, whether element, compound, or ion.


1. An atom in its elemental state has an oxidation number of 0.

Rules for Assigning Oxidation Numbers

Na H2 Br2 S Ne

Oxidation number 0

Oxidation Number (State): A value that indicates whether an atom is neutral, electron-rich, or electron-poor.


Na+

+1

Ca2+

+2

Al3+

+3

Cl–

–1

O2–

–2

2. An atom in a monatomic ion has an oxidation number identical to its charge.


b) Oxygen usually has an oxidation number of –2.

H O1–

–2+1

HH Ca

–1–1 +2

OH O

–1+1 –1

HH O

+1+1 –2

H

+1

3. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion.a) Hydrogen can be either +1 or –1.


H Cl

+1 –1

c) Halogens usually have an oxidation number of -1.3.

ClCl O

+1+1 –2


Cr2O72–

–2+1 x

x = +6

2(x) + 7(-2) = –2 (net charge)

H2SO3

x –2

x = +4

2(+1) + x + 3(–2) = 0 (net charge)

4. The sum of the oxidation numbers is 0 for a neutral compound and is equal to the net charge for a polyatomic ion.


• Causes oxidation• Gains one or more electrons• Undergoes reduction• Oxidation number of atom decreases

• Causes reduction• Loses one or more electrons• Undergoes oxidation• Oxidation number of atom increases

Oxidizing Agent

Reducing Agent

Identifying Redox Reactions

2 Fe23 O2(g)+4 Fe(s) O3 (s)

–20

reduction

oxidation

+30

Oxidizing Agent

Reducing Agent


3 4 Fe(s)C(s) +2 Fe2O3 + 3 (g)(s) O2C

oxidation

+40

0+3

reductionOxidizing Agent

Reducing Agent


In the redox reaction indicated above, which is the oxidizing agent?

a) MnO4–

b) Fe2+

c) H+

d) H2O

e) None of the above

Worked Example 7.9 Identifying Oxidizing and Reducing Agents

a. The elements Ca and H2 have oxidation numbers of 0; Ca2+ is +2 and H+ is +1. Ca is oxidized, because its oxidation number increases from 0 to +2, and H+ is reduced, because its oxidation number decreases from +1 to 0. The reducing agent is the substance that gives away electrons, thereby going to a higher oxidation number, and the oxidizing agent is the substance that accepts electrons, thereby going to a lower oxidation number. In the present case, calcium is the reducing agent and H+ is the oxidizing agent.

Strategy and Solution

Assign oxidation numbers to all atoms, tell in each case which substance is undergoing oxidation and which reduction, and identify the oxidizing and reducing agents.a. b.

Worked Example 7.9 Identifying Oxidizing and Reducing Agents

b. Atoms of the neutral element Cl2 have an oxidation number of 0; the monatomic ions have oxidation numbers equal to their charge:

Fe2+ is oxidized because its oxidation number increases from +2 to +3, and Cl2 is reduced because its oxidation number decreases from 0 to –1. Fe2+ is the reducing agent, and Cl2 is the oxidizing agent.

Continued

The elements that are higher up in the table are more likely to be oxidized.

Thus, any element higher in the activity series will reduce the ion of any element lower in the activity series.

The Activity Series of the Elements

2 Ag+(aq) + Cu(s)2 Ag(s) + Cu2+(g)

Cu2+(aq) + 2 Ag(s)Cu(s) + 2 Ag+(g)

Which one of these reactions will occur?

The Activity Series of the Elements

a)

b)

c)

d)

e)

Silver cation will oxidize copper, becoming solid silver, and liberating copper (II) ions. Looking at the diagram to the left, which one of the four reactions below will occur?

Silver cation will oxidize copper, becoming solid silver, and liberating copper (II) ions. Looking at the diagram to the left, which one of the four reactions below will occur?

a)

b)

c)

d)

e)

Worked Example 7.10 Predicting the Products of a Redox Reaction

Look at Table 7.5 to find the relative reactivities of the elements.

Solution

a. Zinc is above mercury in the activity series, so this reaction will occur.b. Copper is below hydrogen in the activity series, so this reaction will not occur.

Strategy

Predict whether the following redox reactions will occur:a. b.

Balancing Redox Reactions by the Half-Reaction Method

Cr3+(aq) + IO3–(aq)I–(aq) + Cr2O7

2–(aq)

Balance the following net ionic equation in acidic solution:


IO3–(aq)I–(aq)

Cr3+(aq)Cr2O72–(aq)

• Write the two unbalanced half-reactions.


2 Cr3+(aq)Cr2O72–(aq)

IO3–(aq)I–(aq)

• Balance both half-reactions for all atoms except O and H.


2 Cr3+(aq) + 7 H2O(l) 14 H+(aq) + Cr2O72–(aq)

IO3–(aq) + 6 H+(aq) 3 H2O(l) + I–(aq)

• Balance each half-reaction for O by adding H2O, and then balance for H by adding H+.


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

IO3–(aq) + 6 H+(aq) + 6 e–3 H2O(l) + I–(aq)

• Balance each half-reaction for charge by adding electrons to the side with greater positive charge.


IO3–(aq) + 6 H+(aq) + 6 e– 3 H2O(l) + I–(aq)

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

oxidation:

reduction:

• Multiply each half-reaction by a factor to make the electron count the same in both half-reactions.


oxidation:

reduction:

IO3–(aq) + 2 Cr3+(aq) + 4 H2O(l)

8 H+(aq) + I–(aq) + Cr2O72–(aq)

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

IO3–(aq) + 6 H+(aq) + 6 e– 3 H2O(l) + I–(aq)

• Add the two balanced half-reactions together and cancel species that appear on both sides of the equation.


Example #1

1. MnO4-(aq) + Fe2+

(aq) Mn2+(aq) + Fe3+

(aq)

2. H+(aq) +Cr2O7

2-(aq) + C2H5OH(l)

Cr3+(aq) + CO2(g) + H2O(l)

MnO2(s) + BrO3–(aq)MnO4

–(aq) + Br–(aq)

Balance the following net ionic equation in basic solution:


BrO3–(aq)Br–(aq)

MnO2(s)MnO4–(aq)

• Write the two unbalanced half-reactions.


• Balance both half-reactions for all atoms except O and H.

BrO3–(aq)Br–(aq)

MnO2(s)MnO4–(aq)


BrO3–(aq) + 6 H+(aq)3 H2O(l) + Br–(aq)

MnO2(s) + 2 H2O(l)4 H+(aq) + MnO4–(aq)

• Balance each half-reaction for O by adding H2O, and then balance for H by adding H+.


MnO2(s) + 2 H2O(l)3 e– + 4 H+(aq) + MnO4–(aq)

• Balance each half-reaction for charge by adding electrons to the side with greater positive charge.

BrO3–(aq) + 6 H+(aq) + 6 e–3 H2O(l) + Br–(aq)


2

• Multiply each half-reaction by a factor to make the electron count the same in both half-reactions.

MnO2(s) + 2 H2O(l)3 e– + 4 H+(aq) + MnO4–(aq)



2 MnO2(s) + H2O(l) + BrO3–(aq)

2 H+(aq) + 2 MnO4–(aq) + Br–(aq)


2 MnO2(s) + 4 H2O(l)6 e– + 8 H+(aq) + 2 MnO4–(aq)

• Add the two balanced half-reactions together and cancel species that appear on both sides of the equation.


2 MnO2(s) + H2O(l) + BrO3–(aq) + 2 OH–(aq)

2 OH–(aq) + 2 H+(aq) + 2 MnO4–(aq) + Br–(aq)

2 H2O

2 MnO2(s) + BrO3–(aq) + 2 OH–(aq)

H2O(l) + 2 MnO4–(aq) + Br–(aq)

• Since the reaction occurs in a basic solution, “neutralize” the excess H+ by adding OH– and cancel any water (if possible).


Worked Example Balancing an Equation for a Reaction in Base

Follow the steps outlined in Figure 7.4.

Solution

Steps 1 and 2. The unbalanced net ionic equation shows that chromium is oxidized (from +3 to +6) and chlorine is reduced (from +1 to –1). Thus, we can write the following half-reactions:

Step 3. The half-reactions are already balanced for atoms other than O and H.

Strategy

Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with chromite ion [Cr(OH)4

–] in basic solution to yield chromate ion (CrO42–) and chloride ion. The net ionic equation is

Balance the equation using the half-reaction method.

Figure 7.4Using the half-reaction method to balance redox equations for reactions in acidic solution.


Step 4. Balance both half-reactions for O by adding H2O to the sides with less O, and then balance both for H by adding H+ to the sides with less H:

Step 5. Balance both half-reactions for charge by adding electrons to the sides with the greater positive charge:

Next, multiply the half-reactions by factors that make the electron count in each the same. The oxidation half-reaction must be multiplied by 2, and the reduction half-reaction must be multiplied by 3 to give 6 e – in both:

Step 6. Add the balanced half-reactions:

Continued


Now, cancel the species that appear on both sides of the equation:

Finally, since we know that the reaction takes place in basic solution, we must add 2 OH– ions to both sides of the equation to neutralize the 2 H+ ions on the right giving 2 additional H2O. The final net ionic equation, balanced for both atoms and charge, is

Continued

Example #2

1. CN-(aq) + MnO4

-(aq) CNO-

(aq) + MnO2 (s)

2. S2-(aq) + MnO4

-(aq) S(s) + MnO2 (s)

Galvanic Cells

Electrochemistry: The area of chemistry concerned with the interconversion of chemical and electrical energy.

In a Galvanic (Voltaic) Cell: A spontaneous chemical reaction generates an electric current.

In an Electrolytic Cell: An electric current drives a nonspontaneous reaction.

How to draw a Galvanic CellThe oxidation reaction occurs at the anode. The reduction

reaction occurs at the cathode.

You will be give the unbalanced net ionic reaction or a list of the substances present (line notation). From the information given you need to decide what half reactions occur in each beaker. Draw each beaker with its substances present. The electrons leave the anode and travel to the cathode. Remember to draw a salt bridge. If the substance reduced produces a gas you will have a platinum electrode and an inverted test tube to catch the gas. Use platinum as electrode for redox involving ions; otherwise, use the metal given.

Example #3 Draw the galvanic cell for the reaction:

1. Cu(s) + Ag+(aq) Ag(s) + Cu2+

(aq)

2. Zn(s) + 2H+(aq) Zn2+

(aq) + H2(g)

3. Pt(s) |Fe2+(aq), Fe3+

(aq) |Ag+(aq), Ag(s)|

Cell Potentials and Free-Energy Changes for Cell Reactions

Electromotive Force (emf): The force (or electrical potential) that pushes the negatively charged electrons away from the anode (– electrode) and pulls them toward the cathode (+ electrode).

It is also called the cell potential (o) or the cell voltage.


1 J = 1 C × 1 V

voltSI unit of electric potential

jouleSI unit of energy

coulombElectric charge

1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second.


∆G° = –nFo

cell potentialfree-energy change

number of moles of electrons transferred in the reaction

faraday or Faraday’s constantthe electric charge on 1 mol of electrons

96,500 C/mol e–

∆G = –nF or

Calculating electromotive force,

emf or cell

From table of standard reduction potentials, write the half reactions according to the way the reaction is written.

Flip the sign of o for the species being oxidized.

Add the two values.

The reaction with 2H+(aq) and H2(g) is known at the standard

hydrogen electrode (SHE). The voltages for all half

reactions are based on this reaction. o for SHE is 0.

Example #4

Determine the electromotive force, for the following reactions.

1. Ag(s) + Cu2+(aq) Cu(s) + Ag+

(aq)

2. Zn(s) + 2H+(aq) Zn2+

(aq) + H2(g)

cell and spontaneity

If cell is positive the reaction is spontaneous as it is written (runs forward).

If cell is negative it is not spontaneous as it is written (runs backward).

If cell is zero, it is at equilibrium.

Look at the previous example, are these reactions spontaneous?

To Decide which way a galvanic

cell will run from o

If you are given the substances in each cell without any indication which is the oxidation and which is the reduction you will look up the reactions and the numbers

in the o table. Decide which reaction you can reverse and still have a positive number after you add the two values.

Example #5

Draw the galvanic cell based on the following half-reactions under standard conditions.

Ag+ +e- Ag

Fe3+ + e- Fe2+

Example #6

Determine the spontaneous reaction that will occur from the following half reactions.

1. Sn2+ + 2e- Sn

I2 + 2e- 2I-

2. Au3+ + 3e- Au

Cu2+ + 2e- Cu

Example #7

Using o values, predict whether 1M HNO3 will dissolve gold to form a 1M Au3+ solution.

cell and G

Go = -nF o

n: number of moles of electrons

F: Faradays constant, 96500C/mole

Joule = V C (Volt-Coulomb)

Named for Michael Faraday.

Example #8

Using o values calculate Go for the reaction:

Fe(s) + Cu2+(aq) Cu(s) + Fe2+

(aq)

Example #9Write the balanced oxidation reduction reaction for

the following:

Zn + 2H+ H2 + Zn2+

A. Calculate Go for the reaction using the Gof

(kJ/mol) values below:

Gof Zn H+ H2 Zn2+

0 0 0 -146.68

B. Calculate o using Go in part a.

C. Compare to o from Table.

Effects of concentration on emf.

R = 8.314 J/K mol F = 96500 C/mol

Go = -nF o and Go = -RTlnK

Example #10

Calculate the emf for the reaction:

2Al(s) + 3Mn2+(aq) 2Al3+

(aq) + 3Mn(s)

Calculate the equilibrium constant for this reaction.

Predict whether the reaction will be spontaneous at 30oC when:

A. [Al3+] = 2M, [Mn2+] = 1M

B. [Al3+] = 2M, [Mn2+] = 3M

Example #11: Putting Everything Together

A galvanic cell is composed of aqueous Cr2O72-, Cr3+, Fe3+ and Fe2+ ions. At

25°C it runs at a pH of 2.75. a. Write the balanced spontaneous reaction for the galvanic cell.b. Draw a diagram of the galvanic cell. Identify the anode and the cathode. Indicate the direction of electron flow.c. Determine the εo for this reaction. d. Calculate ΔGo and K values for this reaction.e. If the concentrations of all the species present are: [Cr2O7

2-]= 0.025M [Cr3+]= 0.45M [Fe3+]= 2.3M [Fe2+]= 1.75MCalculate the ε value. Is the reaction spontaneous under these conditions?

Example #11: Putting Everything Together

A galvanic cell is composed of aqueous Cr2O72-, Cr3+, Fe3+ and Fe2+ ions. At

25°C it runs at a pH of 2.75. Acidic balancea. Write the balanced spontaneous reaction for the galvanic cell.14H+ + Cr2O7

2- + 6Fe2+ 2Cr3+ + 6Fe3+ + 7H2Ob. Draw a diagram of the galvanic cell. Identify the anode and the cathode. Indicate the direction of electron flow.*c. Determine the εo for this reaction. 0.56Vd. Calculate ΔGo and K values for this reaction. -3,242,400J, 6.86 x 1056 or e130.87

e. If the concentrations of all the species present are: [Cr2O7

2-]= 0.025M [Cr3+]= 0.45M [Fe3+]= 2.3M [Fe2+]= 1.75MCalculate the ε value. Is the reaction spontaneous under these conditions? 0.16V, yes (don’t forget about H+ concentration)

What is electrolysis

When a current of electricity is forced through the cell to produce a chemical reaction for which the emf is negative.

Electrical energy Chemical Energy

*Opposite of Galvanic Cell which is

Chemical Energy Electrical Energy

Electrolysis half reactionsWrite the half reactions for the electrolysis of

molten NaCl.

Molten NaCl is NaCl liquid (melted). The first reaction is the Na+ becoming Na0 and the second is Cl- becoming Cl20. These types will always take the ions from the solid and form their elements. If they are part of

H O N Cl Br I F they will get a subscript of 2.

Example #11

Write the half reactions for the electrolysis of molten KBr.

Electrolysis in Aqueous Solutions

Sometimes the water reaction will give a larger

cell value than the metal (especially group 1

and some group 2 metals). The larger cell value is the more favorable reaction thus the water reaction will occur instead of the reduction of the metal and oxidation of nonmetal. This means a lower voltage needs to be applied for water reaction.

Let’s Inspect Water Reaction

2H2O O2 + 4H+ + 4e- -1.23V ox

2H2O + 2e- H2 + 2OH- -0.83V red

Water Reaction:

2H2O 2H2 + O2 ε° = -2.06V

Compare to Aqueous: Ex: MgCl2Mg2+ + 2e- Mg -2.37V red

2Cl- Cl2 + 2e- -1.37V ox

Overall: Mg2+ + 2Cl- Mg + Cl2 ε° = -3.74V

Water Reaction costs less and is therefore more favorable.

Example #12

An aqueous solution of sodium chloride has an electrical current running through it. Write the half reactions that will occur at each electrode.

Formulas and relationships

Amps x seconds = Coulombs

1 mole of electrons = 96500C

Electrolysis calculations work kind of like stoichiometry problems from Chem 1.

Example #13

How long must a current of 5A be applied to a solution of Ag+ to produce 10.5g of silver metal?

Example #13

How long must a current of 5A be applied to a solution of Ag+ to produce 10.5g of silver metal?

Ag+ + e- Ag

10.5g Ag/107.87g/mol = 0.097 mol Ag

0.097 mol Ag*(1 mol e-/1 mol Ag) = 0.097 mol e-

0.097 mol e-*(96500C/1 mol e-) = 9393C

1A*1s = 1C

5A*s = 9393C

s = 1879 s

Example #14

A 3A current runs through an Au+3 solution for 1 hour. What mass of gold metal will plate out on the cathode during that time?

Example #14

A 3A current runs through an Au+3 solution for 1 hour. What mass of gold metal will plate out on the cathode during that time?

Au+3 + 3e- Au

1A*1s = 1C

3A*3600s = 10800C

10800C*(1 mol e-/96500C) = 0.112 mol e-

0.112 mol e- * (1 mol Au/3 mol e-) = 0.0373 mol Au

0.0373 mol Au *(196.97g/mol) = 7.35g Au

Example #15

Electrolysis of a molten metal bromide, MBr3, using a 7A current for 30 minutes deposits 2.263g of the metal. What is the metal? (to find the identity of the metal you need to find its atomic weight)

Example #15

Electrolysis of a molten metal bromide, MBr3, using a 7A current for 30 minutes deposits 2.263g of the metal. What is the metal? (to find the identity of the metal you need to find its atomic weight)

M+3 + 3e- M

1A*1s = 1C

7A*1800s = 12600C

12600C*(1 mol e-/96500C) = 0.131 mol e-

0.131 mol e- * (1 mol M/3 mol e-) = 0.0435 mol M

2.263g M/0.0435 mol M = 51.995g/mol Cr

electrochemistry. transfer of electrons oxidation: electrons are lost (product side of equation),...

Documents