AP Chemistry

Chapter 18Electrochemistry

The Voltaic (Galvanic) Cell:How does it work?It produces electrical

energy by carrying out a spontaneous redox reaction.

What is it made of?1. Cathode

a. Where the ion or molecule undergoes reduction.b. The ions that move toward the cathode are called cations.

2. Anodea. Where the ion of molecule undergoes oxidation.b. The ions that move toward the anode are called anions.

Electrons are formed at the anode and move toward the cathode.

3. Salt bridgea. function:

To move anions and cations for each half reaction to the half-cell where it is needed without short-circuiting the cell.

b. What is it? An inverted glass U-tube, plugged with glass wool

at each end. The tube is filled with a solution of a salt that takes no

part in the electrode reactions; potassium nitrate, KNO3, is often used.

H+

MnO4- Fe+2

Connected this way the reaction startsStops immediately because charge builds up.

H+

MnO4- Fe+2

Galvanic Cell

Salt Bridge allows current to flow

Reducing Agent

Oxidizing Agent

e-

e-

e- e-

e-

e-Anode Cathode

Notation:1. A voltaic cell can be abbreviated in the

following manner:Anode reaction ║cathode reaction

2. The salt bridge is indicated by the ║3. The anode and the cathode reactions are

further abbreviated:Reactant │Product

4. The voltaic cellSn (s) + Zn2+

(aq) � Zn (s) + Sn2+(aq)

Is abbreviated:Sn│Sn2+║Zn2+│Zn

5. If no metal is involved in either anode or cathode reaction , or another metal is used as an electrode, the electrode is separated form the reactant│product entry.

If the electrode is used as the anode then the symbol for the metal is written before the reactant for the anode and separated by │.

If the electrode is used as the cathode then the symbol for the metal is written after the product for the cathode and separated by │.

Line Notation Summary

solidAqueousAqueoussolidAnode on the leftCathode on the rightSingle line separates different phases.Double line is porous disk or salt bridge.If all the substances on one side are aqueous, a platinum electrode is indicated.For the last reactionCu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

Drawing a diagram of the cell:(Key points to remember)

There are always two separate half-cells connected by a wire and a salt bridge.

One of the compartments is the anode and the other is the cathode. All species (reactants and products of a half-reaction) except water, are shown in the appropriate compartment.

If a metal precipitates or is used up in a cell half-reaction, (either as product or reactant), it is ordinarily chosen as the electrode for the compartment where that half-reaction occurs.

If no metal is involved in the half-reaction, an electrically conducting non-reactive solid like platinum (Pt) or graphite (C) should be used.

Electron flow from anode to cathode is denoted by an arrow in the external circuit.

The flow of ions through the salt bridge is also indicated by an arrow. Cations move to the cathode, anions move to the anode.

Example #1:

When potassium permanganate and sodium iodide are mixed, a spontaneous

reaction occurs:2 MnO4

-(aq)+10 I-(aq)+16 H+

(aq) � 5 I2 (aq) +2Mn2+(aq) +8H2O

This reaction can serve as a source of energy in a voltaic cell.

a. Write the anode and cathode reactions.b. Draw a diagram of the cell. Indicate which

way the electrons move in the external circuit.c. Write the cell notation.

Solution:Since Mn goes from an oxidation number of 7 to

2, it is being reduced. I goes from an oxidation of -1 to 0, so it is

oxidized. The half-cell reactions are:Anode: 2 I-(aq) � I2 (aq) + 2 e-

Cathode: MnO4

-(aq)+8 H+

(aq) + 5e- � Mn2+(aq) + 4H2O

Example 18L.1

e- �

Pt �

I

anode cathodec. Cell notation: Pt│I-│I2║MnO4

-│Mn2+│Pt

I- I2

MnO4-

H+ Mn2+

Ions

Standard Voltage:

Definition: The voltage that is measured when the current flow is essentially zero, all ions and molecules in solution are at a concentration of 1 M, and all gases are at a pressure of 1 atm.

Symbol for

-the standard voltage of a redox reaction: Eo

-the standard voltage of an oxidation half-reaction: Eo

ox

-the standard voltage of a reduction half-reaction: Eo

red

Quantitative aspects:

1. Eo = Eoox + Eo

red

2 . Eored for the following reaction

2H+(aq,, 1 M) + 2 e- � H2 (g, 1 atm)

Is zero.

Zn+2

SO4-2

1 M HCl

Anode

0.76

1 M ZnSO4

H+

Cl-

H2 in

Cathode

1 M HCl

H+

Cl-

H2 in

Standard Hydrogen

ElectrodeThis is the reference all other oxidations are compared to

Eº = 0

º indicates standard states of 25ºC, 1 atm, 1 M solutions.

Standard potentials:The standard potentials listed in Table 18.1 of the text give you the standard voltages for reduction half-reactions.

� Remember: These are half-reactions in which e-are on the reactant side and there is a reduction in oxidation number.

Standard voltages for oxidation half-reactions are obtained by changing the sign of the reduction half-reaction.

� Do this only if the oxidation half-reaction is the reverse of the reduction half-reaction as written in Table 18.1.

The strength of oxidizing and reducing agents:

1. The larger Eored

is, the stronger the oxidizing agent is.

Remembering the oxidizing agent is the reactant in a reduction half-reaction.

2. The larger Eoox is, the stronger

the reducing agent is.

In order to compare the strength of reducing agents, you must:

*Find the species on the right side of the standard potential table. If the ion or molecule is on the left side, it is not acting as a reducing agent.

*Reverse the reactions, making sure to change Eo

red to Eoos and reverse

the sign.*Compare the magnitude of Eo

oxvalues.

In order to compare the strength of oxidizing agents:

� Find the species on the left-hand side of the standard potential table.

If the ion or molecule is on the right-hand side, it is not acting as an oxidizing agent.

� Compare the magnitude of Eoox values.

� The product formed by a strong oxidizing agent is itself a weak reducing agent.

� The product formed by a strong reducing agent is itself a weak oxidizing agent.

Example #2:

Arrange, in decreasing order of strength, the following oxidizing agents in acidic solution.

I2 (s) ClO3 (aq) F2 (g) Na+(aq) Zn2+

(aq) PbO2 (s)

Solution:Since ranking oxidizing agents, these are the species

that are reduced. They will be found on the left side of the equations listed in Table 18.1. The equations are

I2 (s) + 2 e- � 2 I-(aq) Eored = 0.534 V

ClO3-(aq) + 6 H +(aq) + 5 e- � ½ Cl2 (g) + 3 H2O Eo

red = 1.458 VF2 (g) + 2 e- � 2 F-

(aq) Eored = 2.889 V

Na+(aq) + 2 e- � Na(s) Eo

red = -2.714 VZn2+

(aq) + 2 e- � Zn(s) Eored = -0.762 V

PbO2 (s) + SO42-

(aq) + 4 H+(Aq) + 2e- �PbSO4 (s) + 2H2O

Eored = 1.687 V

From this table the largest Eored is 2.889, which belongs to the

reduction half=reaction of F2. Thus, F2 is the strongest oxidizing agent. Go through the

same reasoning for all other oxidizing agents and get the order

F2 > PbO2 > ClO3- > I2 > Zn2+ > Na+

Example 18L.2

Calculation of Eo from Eored and E

oox

1. Eo = Eored + Eo

ox

2.Given a redox equation, break it up into two half-reactions equations.a. The reduction half-reaction has voltage

Eored. Table 18.1

b. The oxidation half-reaction has voltage Eo

ox. Table 18.1 also except, when the half-reaction is written take the voltage of the reverse reaction,

remembering to reverse the sign!

Example #3: Calculate Eo for the reactionCu2+

(aq) + H2S(g) � S(s) + 2H+(aq) + Cu(s)

Solution:First determine the oxidation numbers on both

sides of the reaction.Cu: +2 to 0 H: +1 both S: -2 to 0

Cu is reduced and H2S is oxidized. Write the half reactions.

Reduction half-reaction:Cu2-

(aq) + 2 e- � Cu(s)

Eored = 0.339 V

Oxidation half-reaction:H2S(g) � 2H+

(g) + S(s) + 2 e-

Find this equation on the table written:2H+

(g) + S(s) + 2 e-� H2S(g)

Eored = 0.144 V

Reverse the reaction to get Eoox and change the

sign of E.H2S(g) � 2H+

(g) + S(s) + 2 e-

Eoox = -0.144 V

Then: Eocell = Eo

red + Eoox

= 0.339 V + (-0.144 V) = 0.195 V

Example 18L.3, 4

Reminder

For all voltaic (galvanic) cells, Eo is a positive quantity, so choose the oxidation and reduction that will give you a positive value.The quantities Eo

, Eoox, and Eo

red are independent of the cefficients in the equation for the cell reaction. NEVER MULTIPLY by the cefficients – use just as they are written in the table.

Relations between Eo, ∆Go, and K

Indicators of spontaneity1. Eo

a. If Eo > 0, the reaction is spontaneous.b. If Eo < 0, the reaction is nonspontaneous. (The reverse reaction will tend to occur).c. If Eo = 0, the reaction is at equilibrium.

2. ∆Go

a. If ∆Go < 0, the reaction is spontaneous.

b. If ∆Go > 0, the reaction is nonspontaneous. (The reverse reaction will tend to occur).

c. If ∆Go = 0, the reaction is at equilibrium.

3. Ka. If K >1, the reaction occurs spontaneously from left to right at standard conditions.b. If K < 1, the reverse reaction occurs spontaneously.c. If K = 1, the reaction is at equilibrium

Relation between Eo and ∆Go:

Mathematical relationship∆Go = -nFEo

Where n is the number of electrons transferred. F is the Faraday constant 96,485 C/mol

or J/Vmol. (1 C = 1J/V) In kJ…

∆Go = -n(96.5kJ/mol)Eo

Example # 4: Determine ∆Go for the equation4 H+

(aq) + 2 Cl-(aq) + MnO2 (s) � Mn2+(aq) + 2 H2O + Cl2 (g)

Solution:First determine Eo by writing the two half-reactions and

their standard potentials. The oxidation half-reaction is

2 Cl-(aq_ � Cl2 (g) + 2 e-

Eoox = -1.360 V

The reduction half-reaction is2 e- + 4 H+

(aq) + MnO2 (s) � Mn2+(aq) + 2 H2O

E0red = 1.229 V

Calculating Eo, we getEo = -1.360 +1.229 = -0.131 V

Since Eo is negative, the reaction is nonspontaneous and expect ∆Go

to be positive.Note that two electrons are transferred so,

n = 2 mol.To calculate ∆Go, substitute into the equation

∆Go = -96.5nEo

∆Go = -(96.5 kJ/Vmol)(2 mol)(-0.131 V) = 25.3 kJ

Example 18L.5

Relationship between Eo and K:

Mathematical relationship at 25oCEo = RTlnK

nFRT = .0257VF

So ln K= nEo__ 0.0257V

Example # 5:

Write a balanced redox equation for the reaction between sulfide ion and oxygen gas to produce solid sulfur and hydroxide ion. Calculate K for the reaction at 25OC.Solution:The oxidation half-reaction is

S2-(aq) � S(s) + 2 e-

Eoox = 0.445 V

The reduction half-reaction is O2 (g) + 2 H2O + 4 e- � 4 OH-

(aq)Eo

red = 0.401 V

Eo therefore is 0.846 V. Multiply the oxidation half-reaction by 2 to get the same number lost and gained so,

n = 4. Next calculate K.

ln K= (4)(0.846) = 1320.0257

Exponentiate:K = e132 = 2 x 1057

Example 18L.6

Effect of Concentration Upon Voltage:

Qualitative aspects of concentration change:1. A reaction becomes more spontaneous (voltage

increases) ifa. the concentration of the reactant is increased.b. the concentration of the product is decreased .

2. A reaction becomes less spontaneous (voltage decreases) if

a. the concentration of the reactant is decreased .b. the concentration of the product is increased.

(Based on Le Chatelier’s Principle)3. At equilibrium, cell potential is zero. The cell is dead.

There is no driving force in either direction.

Nernst Equation:1. Note that Eo

red, Eoox, and Eo apply only at

standard conditions, that is, 1 M for species in solution, and 1 atm for gases.

2. Mathematical relationship between concentration, Eo, and E, the voltage at a given concentration.

E = Eo – 0.0257 ln Qn

Where E = voltage at given concentrationEo = standard voltagen = number of moles of e- transferred

3. For the reaction:

aA(l) + bB(aq) � cC(aq) + dD(g)

Where Q = [C]c(PD)d

[B]ba. Pure liquids and solids do not appear!

b. The condition for species in solution is expressed in molarity, while the condition for gases is expressed in partial pressure (in atm).

4. The voltage E, at given conditions, is calculated using the equation:

E = Eo – 0.0257 ln [C]c(PD)d

n [B]b

Example #6

Calculate the voltage of the cell at 25oC for the reaction10 Cl-(aq) + 2MnO4

-(aq) + 16 H+

(aq) � 5 Cl2 (g) + 2 Mn2+(aq) +8 H2O

The partial pressures of chlorine gas is 1.00 atm, the pH is 0.50, and the concentrations of all the ionic species except H+ are .10 M.

Solution: Use the Nernst equation to calculate E, since the conditions are no longer the standard conditions of 1 M for ionic species and 1 atm for gases.

First, calculate Eo. Do that by writing the two half-reactions.2Cl-(aq) � Cl2 (g) + 2 e- Eo

ox = -1.360 V5e- + 8 H+

(aq) + MnO4-(aq) � Mn2+

(aq) + 4 H2O Eored = 1.512 V

Note: To calculate Eoox and Eo

red, we do not pay any attention to the coefficients used to balanced independently and look up.

Eo = Eoox + Eo

red

Eo = -1.360 V + 1.512 V = 0.152 V

Then, determine n. Multiply both equations by a number to make the number of electrons gained and lost the same. So…

5[ 2Cl-(aq) � Cl2 (g) + 2 e- ]2[ 5e- + 8 H+

(aq) + MnO4-(aq) � Mn2+

(aq) + 4 H2O ]Which becomes…

10Cl-(aq) � 5Cl2 (g) + 10 e-

10e- + 16 H+(aq) + 2MnO4

-(aq) � 2Mn2+

(aq) + 8 H2O

Since there are 10 electrons gained and lost,

n=10.

Next, determine the conditions for the ionic species in molar concentrations. Determine the pressure in atm for the gases.

All ionic species except H+ have a concentration of 0.10 M. [Cl-] = [MnO4

-] = [Mn2+] = 0.10 MH+ is given in pH. pH = 0.50, using

[H+] = 10-pH = 10-0.50 = 0.32 MChlorine gas has a pressure of 1.0 atm.

Now substitute in.E = 0.152 – 0.0257 ln (PCl2)5[Mn2+]2

10 [Cl-]10[MnO4]2[H+]16

E = 0.152 – 0.0257 ln (1.0)5(0.10)2

(0.10)10(0.10)2(0.32)16

E = 0.046 VExample 18L.7

Review of pH and pOH Equations

pH + pOH = 14 [H+] [OH-] = 10-14

pH = -log[H+] [H+] = 10-pH

pOH = -log[OH-] [OH-] = 10-pOH

Calculating Concentration or

partial pressureExample #7:

Calculate the pH of the cell below with a voltage of 3.14 V and carried out in basic solution.

Ca│Ca2+ (0.10 M) ║ CrO42-(0.10 M)│Cr(OH)3 │Pt

Solution:Write the oxidation ½ reaction.

Ca(S) � Ca2+(aq) + 2 e- Eo

(ox) = 2.869 VWrite the reduction ½ reaction.3e- + 4 H2O + CrO4

2-(aq) � Cr(OH)3 (s) + 5 OH-

(aq) Eo(red) = -0.12 V

Then calculate Eo.Eo = Eo

ox + Eored

Eo = 2.869 V + -0.12 V = 2.75 V

Now balance the ½ reactions electrons.

3[Ca(S) � Ca2+(aq) + 2 e-]

2[3e- + 4 H2O + CrO42-

(aq) � Cr(OH)3 (s) + 5 OH-(aq)]

Then becomes…3Ca(S) � 3Ca2+

(aq) + 6 e-

6e- + 8H2O + 2CrO42-

(aq) � 2Cr(OH)3 (s) + 10 OH-(aq)

Adding the two equations together…3Ca(S) + 8H2O + 2CrO4

2-(aq) � 3Ca2+

(aq) +2Cr(OH)3 (s) + 10 OH-(aq)

Now write the Nernst equation for this reaction.E = Eo – 0.0257 ln [Ca2+]3[OH-]10

6 [CrO42-]2

3.15 = 2.75 – 0.00428 ln (0.10)3[OH-]10

(0.10)2

3.15 = 2.75 – 0.00428 ln [(0.10)[OH-]10]3.15 -2.75 = ln [(0.10)[OH-]10]-0.00428-93.458 = ln [(0.10)[OH-]10]

Now take ex which is the opposite of ln to both sides:2.58 x 10-41 = (0.10)[OH-]10

[OH-]10 = 2.58 x 10-40

[OH-] = 1.10 x 10-4

Now since [H+][OH-] = 10-14

[H+] = 9.096 x 10-11 , and pH = 10.04Example 18L.8

Electrolytic Cells:1. Qualitative Aspects:

a. The reaction in this cell is nonspontaneous.

b. A battery (source of electrical energy) acts as an electron pump, pushing electrons into the cathode and removing them form the anode.

c. An oxidation-reduction carried out in an electrolytic cell is called electrolysis.

Cell Reactions in Aqueous Solutions a. Possible reactions at the cathode:

1. The cation gets reduced to the corresponding metal.2. Water gets reduced to hydrogen gas.

½ reaction:

2H2O + 2e- � H2 (g) + 2OH-(aq) Eo

red = -0.828 V

b. Possible reactions at the anode1. The anion gets oxidized to the corresponding non-metal.2. Water gets oxidized to oxygen gas.

½ reaction: 2H2O � O2 (g) + 4H+

(aq) + 4 e- Eoox = -1.229 V

Quantitative aspects of electrolysis: a. When considering a half-reaction like:

Ni2+(aq) + 2e- � Ni(s)

basic stoichiometry applies (mole to mole ratios).

b. 1 mole e- = 96, 485 coulombs.

Faraday’s constant.

c. 1 coulomb = 1 amp.s

Example # 8:

How many grams of copper metal can be obtained by passing 1.20 amp of electric current for half an hour through an aqueous solution of copper (II) ions?

Solution:Write the ½ reaction.

Cu2+(aq) + 2 e- � Cu (s)

Now given 0.500 hours and 1.20 amp find amp.s:

0.500 hours(60min/hour)(60sec/min)(1.20amp)=2.16 x 103 amp.s2.16 x 103 amp.s = 2.16 x 103 coulombs (C)

Convert to moles(use equation and known conversions):

2.16 x 103 C(1mole e-)(1mole Cu)(63.55 g Cu) = 0.711 g Cu(96485 C)(2 moles e-)(1 mole Cu)

Example 18L.9

Commercial Cells:Electrolysis of aqueous NaCl:

Reaction at the anode:2Cl-(aq) � Cl2 (g) + 2e-

Reaction at the cathode:2 H2O + 2e- � H2 (g) + 2OH-

(aq)

Overall cell reaction :2Cl-(aq) + 2H2O � Cl2 (g) + H2 (g) + 2OH-

(aq)

Primary Cells (non-rechargeable)Dry Cell (Leclanche cell)

� The anode is zinc wall.� The cathode is graphite rod.� The space between electrodes is filled with a paste that

contains MnO2 , ZnCl2, and NH4Cl.� The half reaction at the anode is Zn(s) � Zn2+

(aq) + 2e-.� The half reaction at the cathode is

2MnO2 (s) + 2NH4+

(aq) + 2e- � Mn2O3 (s) + 2NH3 (aq) + H2O� The overall reaction is

Zn(s)+ 2MnO(s)+ 2NH4+

(aq) � Zn2+(aq)+ Mn2O3 (s) + 2NH3 (aq) + H2O

� In an alkaline cell KOH rather that NH4Cl is used. The overall reaction for an alkaline cell is

Zn(s) + 2MnO2 (s) � ZnO(s) + Mn2O3 (s) .

Mercury Cell� It is used in hearing aides, watches, cameras, etc.� The anode is zinc-mercury amaolgam. The

reacting species is zinc.� The cathode is a plate of mercury (II) oxide, HgO.� The electrolyte is paste containing HgO and

sodium or potassium hydroxide.� The half-reaction at the anode is

Zn(s) + 2 OH-(aq) � Zn(OH)2 (s) + 2e-

� The half reaction at the cathode is HgO (s) + H2O + 2e- � Hg(l) + 2OH-

(aq)

� The overall reaction is Zn(s) + HgO(s) + H2O � Zn(OH)2 + Hg(l).

Storage (rechargeable) cellsLead storage battery

The anode is made up of a group of lead plates, the grills of which are filled with spongy gray lead.

The cathode is made up of another group of plates of similar design filled with lead (IV) oxide, PbO2.

The anode and the cathode are immersed in a water solution of sulfuric acid, H2SO4.

When a lead storage battery supplies current, it acts as a voltaic cell.

The half-reaction at the anode isPb(s) + SO3

2-(aq) � PbSO4 (s) + 2e-.

The half reaction at the cathode is PbO2(s) + 4H+

(aq) + SO42-

(aq) + 2e- � PbSO4 (s) + 2H2O . The overall reaction is

Pb(s) + PbO2(s) + 4H+(aq) + SO4

2-(aq) � 2PbSO4(s)

+2H2O.

When a lead storage battery is being recharged, it acts as a electrolytic cell.

The half-reaction at the anode is PbSO4 (s) + 2H2O � PbO2(s) + 4H+

(aq) + SO42-

(aq) + 2e-. The half reaction at the cathode is

PbSO4 (s) + 2e- � Pb(s) + SO32-

(aq)

The overall reaction is 2PbSO4(s) +2H2O � Pb(s) + PbO2(s) + 4H+

(aq) + SO42-

(aq

Nicad battery� The anode is cadmium metal.� The cathode contains nickel (IV) oxide, NiO2.� The electrolyte is a concentrated solution of

potassium hydroxide.� When the battery supplies energy,

The half-reaction at the anode is Cd(s) + 2OH-

(aq) � Cd(OH)2 (s) + 2e-..� The half reaction at the cathode is

NiO2(s) + 2H2O + 2e- � Ni(OH)2(s) + 2OH-(aq).

� The overall reaction is Cd(s) + NiO2 (s) + 2H2O � Cd(OH) 2(s) + Ni(OH)2(s).

Corrosion

Rusting - spontaneous oxidation.Most structural metals have reduction potentials that are less positive than O2 .

Fe → Fe+2 +2e- Eº= 0.44 VO2 + 2H2O + 4e- → 4OH-Eº= 0.40 VFe+2 + O2 + H2O → Fe2 O3 + H+

WaterRust

Iron Dissolves- Fe → Fe+2

e-

Salt speeds up process by increasing conductivity

Preventing Corrosion

Coating to keep out air and water.Galvanizing - Putting on a zinc coatHas a lower reduction potential, so it is more easily oxidized.Alloying with metals that form oxide coats.Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

End of Chapter 18

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