This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter Nineteen ELECTROCHEMISTRY
• Balancing Redox Equations
Electrochemical Cells and Standard Electrode Potentials
Spontaneity of Redox Reactions
Effect of Concentration on Cell EMF
Electrolysis and Its Quantitative Aspects BALANCING REDOX EQUATIONS
2. Review the basics of oxidation-reduction reactions.
Balance redox reactions using the ion-electron method. Half-Reactions. Oxidation-reduction reactions were discussed in Chapter 4. It will be useful to review
that material as you start this chapter. Recall that oxidation is a loss of electrons by a chemical species and
reduction is a gain of electrons. Redox reactions are electron transfer reactions involving transfer of electrons
from a reducing agent to an oxidizing agent.
For example, when copper metal is immersed in a solution containing Ag+ ions a reaction occurs in which
electrons are transferred spontaneously from Cu atoms to the Ag+ ions, forming Ag atoms that plate out on the
copper surface. The newly formed Cu2+ ions go into solution.
Cu(s) + 2Ag+(aq) → C u 2+ (aq) + 2Ag(s)
Cu is oxidized to Cu 2+ ion, and Ag+ ion is reduced to Ag.
All redox reactions can be divided into half-reactions. The half-reactions are:
reduction Cu → Cu 2+ + 2e –
2e– + 2Ag+ → 2Ag Copper metal is the reducing agent because it supplies electrons to the silver ion. The silver ion is the oxidizing
agent because it accepts electrons from copper. Two Ag+ ions are required to accept the two electrons from one
Cu atom. Before we introduce electrochemical cells we will discuss how to balance redox equations. Ion-Electron Method. In this method the overall reaction is divided into two half-reactions, one
involving oxidation and the other reduction. Each half-reaction is balanced according to mass and charge, and
then the two half-reactions are added together to give the overall balanced equation. We will balance the equation
below to illustrate the steps involved. This equation shows just the essential changes and is sometimes called a
sketetal equation. To separate the equation into half-reactions, first identify which element is oxidized and which
is reduced. Do this by writing oxidation numbers above each element.
+1 –2 +5 –2
0 +2 –2
H2 S + NO3 → S + NO 3 79
Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 80 / Electrochemistry
Note that S atoms and N atoms undergo oxidation number changes: S (–2 → 0) and N (+5 → +2). Sulfur is
oxidized and nitrogen is reduced. Now we write the half-reactions.
reduction H2 S → S
NO3 → NO Each half-reaction must be balanced by mass. In acidic solution, always add H+ to balance H atoms, and add
H2 O to balance O atoms. Here we add H+ to the oxidation half-reaction, and H2 O to the reduction half-reaction.
oxidation H2 S → S + 2H+ reduction NO3 → NO + 2H2 O – The oxidation half-reaction is now balanced by mass. The reduction half-reaction needs 4H+ on the left-hand
reduction – 4H+ + NO 3 → NO + 2H2 O This half-reaction is now balanced according to mass.
Next the ionic charges must be balanced by adding electrons. In the oxidation half-reaction there is a net
charge of +2 on the right-hand side and zero on the left-hand side. To balance the charge, 2 electrons are added to
oxidation H2 S → S + 2H+ + 2e – Notice that both sides of the equation have the same charge. In this case 0.
The reduction half-reaction can be charge balanced by adding three electrons to the left-hand side.
reduction – 3e– + 4H+ + NO 3 → NO + 2H2 O Note that electrons are added to the reactant side of a reduction half-reaction and to the product side of an
The balanced redox equation is obtained by adding the two half-reactions. But first the number of electrons
shown in the two half-reactions must be the same, since all the electrons lost during oxidation must be gained
during reduction. Multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2 will make the
number of electrons in the two half-reactions equal. That is, 6 electrons are given up during oxidation and 6 are
gained during reduction.
3 × (H2 S → S + 2H+ + 2e – ) gives 3H2 S → 3S + 6H+ + 6e –
– – 2 × (3e– + 4H+ + NO 3 → NO + 2H2 O) gives 6e – + 8H+ + 2NO 3 → 2NO + 4H2 O
The sum of the half-reactions is the overall balanced redox equation.
3H2 S → 3S + 6H+ + 6e –
– + 8H+ + 2NO – → 2NO + 4H O
– 3H2 S + 8H+ + 2NO 3 → 3S + 6H+ + 2NO + 4H2 O
The 6H+ on the RHS will cancel 6 of the 8H+ on the LHS yielding
– 3H2 S + 2H+ + 2NO 3 → 3S + 2NO + 4H2 O Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 81 T he following steps are useful in balancing redox equations by the ion-electron method.
1. Write the skeletal equation containing the oxidizing and reducing agents and the products in ionic form.
2. Separate the equation into two half-reactions.
3. Balance the atoms other than O and H in each half-reaction separately. This is called a mass balance.
4a. For reactions in acidic medium, add H2 O to balance O atoms and H+ to balance H atoms.
4b. For reactions in basic medium, first balance the atoms as you would for an acidic solution. Then, for each
H+ ion, add an OH– ion to both sides of the half-reaction. Whenever H+ and OH– appear on the same side,
combine them to make H 2 O.
5. Add electrons to one side of each half-reaction to equalize the charges. Electrons are added to the reactant side
of a reduction half-reaction, and to the product side of an oxidation half-reaction. The number of electrons
added to one side of a half-reaction should make the total charge of that side equal to the charge on the other
side. This procedure is called a charge balance.
6. Add the two half-reactions together. Before this can be done, the number of electrons shown in both halfreactions must be the same. The number of electrons in the two half-reactions can be equalized by
multiplying one or both half-reactions by appropriate coefficients. Now add the two half-reactions.
7. Check the final equation by inspection. Recall that a properly balanced equation consists of a set of the
smallest possible whole numbers.
_______________________________________________________________________________ EXAMPLE 19.1 Balancing A Redox Reaction
Balance the following redox reaction by the ion-electron (half-reaction) method.
Cu + HNO3 → Cu 2+ + NO 2 (in acidic solution) •Method of Solution
Steps 1 and 2. The skeletal equation is given. Separate the given equation into half-reactions. Note that Cu is
oxidized and nitrogen is reduced. The half-reactions are:
reduction Cu → Cu 2+
HNO3 → NO2 Step 3. The oxidation half-reaction is already balanced by mass.
Step 4a. Balance the reduction half-reaction by adding one H2 O to the right-hand side, and one H+ to the lefthand side.
reduction H+ + HNO 3 → NO2 + H2 O
Step 5. To balance according to charge, add electrons to the product side of the oxidation half-reaction, and the
reactant side of the reduction half-reaction.
reduction Cu → Cu 2+ + 2e –
e– + H+ + HNO 3 → NO2 + H2 O Step 6. Next equalize the number of electrons in the two half-reactions by multiplying the reduction halfreaction by 2.
2e– + 2H+ + 2HNO 3 → 2NO2 + 2H2 O
Now add the two half-reactions to obtain a balanced overall equation.
reduction Cu → Cu 2+ + 2e –
– + 2H+ + 2HNO → 2NO + 2H O
2 ____________________________________________________________ Cu + 2H+ + 2HNO 3 → Cu 2+ + 2NO 2 + 2H2 O Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 82 / Electrochemistry
Step 7. Check that the electrons on both sides cancel, and that the equation is balanced in terms of atoms (mass)
1. Balance the following redox equations.
a. Cr 2 O7 (aq) + Br– (aq) → Cr 3+ (aq) + Br2 (g) (acidic solution)
b. MnO 4 (aq) + I– (aq) → MnO 2 (s) + I 2 (aq) (basic solution) 2. Balance the following redox equations.
a. H 2 C 2 O4 (aq) + MnO4 (aq) → Mn 2+ (aq) + CO2 (g) (acidic solution)
b. Zn(s) + ClO– (aq) → Zn(OH)2 (s) + Cl– (aq) (basic solution) ELECTROCHEMICAL CELLS AND STANDARD
3. Diagram an electrochemical cell, labeling the anode, the cathode, the charges on the electrodes, and the
directions of electron and ion flows.
Calculate the standard emf of an electrochemical cell.
Arrange given redox reagents in order of increasing strength as oxidizing and reducing agents. Electrochemical Cells. E lectrochemistry is the area of chemistry that deals with the interconversion
of electrical energy and chemical energy. Electrochemical processes are redox reactions in which chemical energy
is converted into electricity or in which electricity is used to cause a chemical reaction to occur.
A device, which utilizes a spontaneous redox reaction to supply a constant flow of electrons is called an
electrochemical cell or a voltaic cell. The design of an electrochemical cell is such that the reactants
are prevented from direct contact with each other. The oxidation half-reaction occurs at an electrode called the
anode, and the reduction half-reaction occurs at an electrode called the c athode .
Figure 19.1 shows a copper-silver electrochemical cell. The anode is a bar of copper metal that is partially
immersed in a solution of CuSO4 . The cathode is a small bar of silver that is partially immersed in a solution
of AgNO3 . The reducing agent, Cu metal, does not come into direct contact with the oxidizing agent, Ag+ ion.
Cu atoms loose two electrons and become copper ions that enter the solution. Electrons travel through the outer
circuit to the Ag electrode, where Ag+ ions from the solution are reduced to silver atoms at the surface of the
electrode. These atoms plate out on the electrode. As the reaction proceeds, the copper electrode loses mass and
the silver electrode gains mass.
While electrons travel through the outer circuit from Cu to Ag (from the anode to the cathode), negative
ions move through a porous barrier or a "salt bridge" from the AgNO3 solution into the CuSO4 solution. This
motion resupplies negative charge to the anode compartment and maintains electrical neutrality in the solutions
surrounding the electrodes. An electric current will flow until either the Cu metal or the Ag+ ions are used up. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 83
Anod e V Cu e–
Ag Salt bridge
K+ Ag Cu 2+
Cu SO 4 (aq) Ag Cathode + + Ag NO3 (aq) Figure 19.1 An electrochemical cell consisting of a copper electrode (anode) and a
silver electrode (cathode). The fact that electrons flow from the anode to the cathode means that a difference in electrical potential
energy exists between the two electrodes. Electrons flow from where they are at higher potential to where they
are at lower potential. The difference in electrical potential between the anode and cathode in measured in volts,
and is called the c ell potential or c ell emf which stands for e l ectro m o tive f o rce . The actual difference
in electrical potential depends on the nature of the species involved in the half-reactions. The cell potential is
represented by the symbol Ecell. Standard Electrode Potentials. The emf of an electrochemical cell has been found to be dependent
upon the nature of the half-reactions, the concentrations of the ions, and the temperature. The s tandard cell
emf is the voltage generated under standard state conditions. The standard cell emf has the symbol Ecell . The
superscript ° denotes the standard state conditions. The standard state of all solute ions and molecules is a
concentration of 1 mol/L, and for all gases it is a partial pressure of 1 atm. Unless stated otherwise, the
temperature is assumed to be 25°C. When the concentration of Cu2+ and Ag+ ions in the above cell (Figure
19.1) are both 1.0 M, the cell emf is 0.46 V at 25°C.
Rather than construct a cell and measure its voltage every time a cell emf is needed, a method has been
devised that allows the use o a table of half-reactions and standard electrode potentials. The standard
reduction potential, E red , is the voltage associated with a reduction reaction when all metals, solutes,
gases are in their stantard states.
A number of standard reduction potentials are listed in Table 19.1 of the textbook. Note that the species on
the left side are oxidizing agents and that Ered is related to the tendency of these oxidizing agents to be reduceed.
The values range from +2.87 V for the F2 half-reaction to –3.05 V for the Li+ half-reaction. Note the 0.00
potential assigned to hydrogen ion. The greater the value of Ered , the greater the tendency for the reduction
reaction to occur as written.
reducing agent o
Li+(aq) + e– → Li(s)
Ered = –3.05 V
2H+(aq) + 2e– → H 2 (g)
Ered = 0.00 V
F 2 (g) + 2e– → 2 F – (aq) o Ered = +2.87 V A positive reduction potential means that the oxidizing agent (such as F2 ) has a greater tendency to be
reduced than the hydrogen ion. A negative reduction potential for an oxidizing agent indicates that the hydrogen
ion is more readily reduced than that substance. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 84 / Electrochemistry
The electrode potential due to the loss of electrons at an anode is called an oxidation potential, Eox . If we
reverse the half reactions written above, we have an oxidation half-reaction. The oxidation potential is equal in
magnitude but opposite in sign to the reduction potential.
o 2F – (aq) → F 2 (g) + 2e– o Eox = –E red = –2.87 V Calculation of the Standard Cell EMF. Just as it is convenient to think of an overall cell reaction as
the sum of two half-reactions, the emf of an electrochemical cell is thought of as the sum of two electrode
potentials. One electrode potential is that due to the loss of electrons at the anode Eox , and the other is due to
the gain of electrons at the cathode Ered. The standard emf of the cell, Ecell , is the sum of the standard
oxidation potential and the standard reduction potential.
o o o Ecell = E ox + E red
The standard emf of the silver-copper cell discussed above is calculated as follows. The reaction is
Cu + 2Ag+ → Cu 2+ + 2Ag
The half-reactions are
Cu → Cu 2+ + 2e –
2Ag+ + 2e – → 2Ag oxidation
reduction Therefore, the standard cell emf is given by
o o o Ecell = E ox + E red = E°Cu/Cu2 + + E°Ag+/Ag
Notice that the symbols here represent the direction of the reaction,
o o Therefore, Eox = E°Cu/Cu2 + and Ered = E°Ag+/Ag .
The reduction potentials of a number of half-reactions are listed in Table 19.1 of the text. The standard
oxidation potential for an anode half-reaction is equal in magnitude, but of opposite sign to that of the reduction
potential for the reverse reaction.
o Ered = E°Ag+/Ag = 0.80 V
o o Eox = E°Cu/Cu2 + = –Ered = – 0.34 V
o o o Ecell = E ox + E red = – 0.34 V + 0.80 V
o Ecell = 0.46 V.
The following list summarizes the information contained in the standard reduction potentials given in Table
19.1 in the text:
2. 3. Back The E° values apply to the half-reactions when read in the forward direstion. The values are reduction
potentials and apply to the reduction half-reactions.
The more positive the value of E°, the greater the tendency for the substance shown on the left to be
reduced. The species listed on the left-hand side of the reactions shown in the table are all capable of acting
as oxidizing agents. Thus F2 is the strongest oxidizing agent and Li+ is the weakest. The species shown on
the right-hand side are reducing agents when the half-reaction occurs in the reverse direction. Therefore, F–
ion is the weakest reducing agent and Li metal is the strongest reducing agent.
All half-cell reactions are reversible. Depending on the reducing strength of the other electrode that is
chosen, any given electrode may act as an anode or as a cathode. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 85
4. 5. When predicting the direction of reaction under standard state conditions, you can use the diagonal rule. This
rule states that any species on the left of a given half-cell reaction will react spontaneously with a species
that is shown on the right of any half-reaction located above it in the table. This rule works because the
sum of any two half-cell potentials, such as those just described, will always give a positive Ecell .
The standard reduction potential is not affected when stoichiometric coefficients of a half-cell reaction are
changed. For example,
o Ag+ + e – → Ag(s)
2Ag+ + 2e – → 2Ag(s) Ered = 0.80 V
Ered = 0.80 V The electrode potentials are the same. The charge on the electrode is related to the number of electrons per
surface area of silver. The ratio of e– to Ag atoms in both cases is the same.
2 Ag atoms 6. Therefore, electrode potentials are intensive properties and do not depend on the size of the electrode or the
manner in which the half-reaction is balanced.
The standard cell emf (Ecell ) changes sign whenever a half-cell reaction is reversed. Cell Diagrams. Rather than always representing a cell by a sketch, we can use a cell diagram. For
instance, the copper-silver cell discussed above and shown in Figure 19.1 is represented by:
Cu(s) | CuSO4 (aq) || AgNO3 (aq) | Ag(s)
In the diagram the anode (where oxidation occurs) is on the left, and the cathode (where reduction occurs) is on
the right. The single vertical lines indicate phase boundaries which in this cell are between the solid electrodes
and the aqueous solutions. The double vertical lines indicate a salt bridge or porous barrier between the two
EXAMPLE 19.2 Electrochemical Cell Reactions
Consider an electrochemical cell constructed from the following half-cells, linked by a KNO3 salt bridge. A Cu
electrode immersed in 1.0 M Cu(NO3 )2 , and a Sn electrode in 1.0 M Sn(NO 3 )2 .
As the reaction proceeds, the Cu electrode gains mass and the Sn electrode loses mass.
a. Which electrode is the anode and which is the cathode?
b. Write a balanced chemical equation for the overall cell reaction.
c. Sketch the half-cells and show the direction of flow of the electrons in the external circuit.
d. Which electrodes do the positive and the negative ions diffuse toward? Include a salt bridge in your sketch.
•Method of Solution
a. Reduction occurs at the cathode and oxidation occurs at the anode. Since the Cu electrode gains mass its
half-reaction must involve the plating out of Cu2+ ions.
2e– + C u 2+ (aq) → Cu(s)
Reduction occurs at the copper electrode; therefore, it is the cathode. Since the Sn electrode loses mass, Sn
metal must be dissolving as it is oxidized, and so Sn is the anode. The Sn 2+ ions go into solution.
Sn(s) → Sn 2+ (aq) + 2e– b. Add the two half-reactions:
Sn → Sn 2+ + 2e – 2 e – + C u 2+ → Cu
cell reaction Back Forward ________________________
Sn + Cu2+ → Sn 2+ + C u Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 86 / Electrochemistry
d. Electrons flow from the anode to the cathode. See Figure 19.2.
The positive ions (cations) diffuse toward the Cu electrode (the cathode) and the negative ions (anions)
diffuse toward the Sn electrode (the anode).
Anod e Sn e– V
K+ Sn 2+ Cu Cathode Cu 2+ Sn (NO 3 ) 2 (aq) Cu (NO 3 ) 2 (aq) Figure 19.2 An electrochemical cell consisting of a tin electrode (anode) and a copper
electrode (cathode). Notice that in Figure 19.1 copper was an anode. _______________________________________________________________________________
EXAMPLE 19.3 Standard Cell EMF
o Calculate Ecell for an electrochemical cell having the following overall cell reaction:
Sn(s) + Pb2+ (aq) → Sn 2+ (aq) + Pb(s)
•Method of Solution
The standardo emf of the cell is the sum of the standard oxidation and reduction potentials of the appropriate halfo
reactions: Ecell = E ox + E red . Separate the reaction into half-reactions, and obtain the oxidation and reduction
potentials from Table 19.1 of the text. Since tin is oxidized, its Eox is equal in magnitude, but of opposite sign
to that of the reduction potential of Sn.
reduction 2e– + P b 2+ (aq) → Pb(s)
E°Pb 2 +/Pb = –0.13V
oxidation Sn(s) → Sn 2+ (aq) + 2e–
E°Sn/Sn2 + = 0.14 V
Sn(s) + Pb2+ (aq) → Sn 2+ (aq) + Pb(s)
Ecell = 0.01 V
The positive value of the standard cell potential indicates that the reaction is spontaneous under standard
conditions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 87
EXAMPLE 19.4 Relative Strength of Oxidizing and Reducing Agents
Consider the following species: Mg, MnO– , C l – , Zn 2+ , and Co 2+ . Using Table 19.1 of the text, separate
these into oxidizing and reducing agents. Then arrange the reducing agents according to increasing strength. Also
arrange the oxidizing agents according to strength.
•Method of Solution
The oxidizing agents appear on the left-hand side of the half reactions in Table 19.1, and the reducing agents
appear on the right-hand side. Therefore, the oxidizing agents are MnO– , C o 2+ , and Zn2+ , and the reducing
agents are Co2+ , Cl – , and Mg.
Recall that oxidizing agents are reduced in redox reactions. The more positive the value of the reduction
potential, the greater the tendency to undergo reduction, and the greater the strength as an oxidizing agent.
Zn2+ 1.51 V (in acid solution)
–0.76 V The order of increasing oxidizing strength is:
Zn2+ < C o 2+ < MnO–
For a substance on the right-hand side of the half-reactions in Table 19.1 to act as a reducing agent it must react
in the reverse direction, whereby it is oxidized. Write the oxidation potentials for each:
Co 2+ 2.37 V
–1.82 V The greater the value of the oxidation potential, the greater the tendency to be oxidized. Substances that are
easily oxidized are good reducing agents. The order of increasing reducing strength is:
Co 2+ < C l – < Mg
The Co2+ ion can be both a reducing agent and an oxidizing agent, because Co2+ participates in two different
Co 2+ + 2e – → Co
Co3+ + e – → Co 2+
3. Back o Calculate the standard emf (Ecell ) of cells that have the following overall reactions:
a. 2Co 2+ (aq) + Zn2+ (aq) → 2Co 3+ (aq) + Zn(s)
b. Cl 2 (g) + 2Br– (aq) → 2Cl– (aq) + Br2 (l) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 88 / Electrochemistry
4. Consider a voltaic cell constructed from the following half cells, linked by a KCl salt bridge.
• a Cu electrode in 1.0 M Cu(NO3 )2 solution
• a Pb electrode in 1.0 M Pb(NO3 )2 solution.
Answer the following questions:
a. Which electrode is the cathode?
b. Write a balanced chemical equation for the reaction occurring when the cell is operating.
c. Which electrode gains mass?
d. Which direction do electrons flow in the external circuit?
e. Which electrode will K+ of the KCl salt bridge move toward? 5. Given the following standard reduction potentials in acid solution:
Al3+ + 3e –
Sn 4+ + 2e –
S n 2+
Fe3+ + e –
a. Write the formula of the strongest reducing agent.
b. Write the formula of the strongest oxidizing agent. SPONTANEITY OF REDOX REACTIONS
2. Predict whether a redox reaction will be spontaneous or nonspontaneous.
Calculate ∆ G° and K for a redox reaction, given Ecell . Criterion for Spontaneity. For a process carried out at constant temperature and pressure, the Gibbs free
energy change is equal to the maximum amount of work (wmax) that can be done by the process.
∆ G = wmax
Any reaction that occurs spontaneously can be utilized to perform useful work. Combustion of fuels and the
oxidation of food components both provide energy to do work. The difference in free energy between the
reactants and the products is the energy available to do work.
Electrical work is equal to the product of the cell emf (Ecell) and the total charge in coulombs carried
through the circuit. The total charge in coulombs is nF, where n is the number of moles of electrons transferred
from the reducing agent to the oxidizing agent according to the balanced equation, and F is the Faraday constant.
One Faraday equals the charge carried by one mole of electrons which is 96,500 C/mol of electrons.
The electrical work is given by
wele = –nFE cell
The maximum work from an electrochemical process is
wmax = wele = –nFE cell
The negative sign is in accord with the sign convention that when work is done on the surroundings, the system
loses energy. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 89
The Gibbs free energy change for a redox reaction is
∆ G = –nFE
For a spontaneous redox reaction, ∆ G will be negative, and E must be positive. For reactions in which reactants
and products are in their standard states:
∆ G° = –nFE° The Equilibrium Constant. In Chapter 18 you learned that ∆ G° is related to the equilibrium constant K
for the reaction.
∆ G° = – RT ln K
Therefore, the equilibrium constant of a redox reaction is related to the standard cell emf (E°) by the equation.
–nFE° = – RT ln K
and E° = RT
nF Substituting for R and F at 25°C gives the term RT/F which equals 0.0257 V.
E° = 0.0257 V
n Therefore reactions with large equilibrium constants generate higher standard cell emf's. When E° > 1, K > 1.
ln K = nE°
0.0257 V If any one of the three quantities ∆ G°, K, or E° is known, both of the others can be calculated. Table 19.1
summarizes the criteria for spontaneous redox reactions.
Table 19.1 Criteria for Spontaneous Redox Reactions
Reaction under Standard
Positive < 1
NegativeNonspontaneous. Reaction is
spontaneous in the reverse
EXAMPLE 19.5 Predicting Spontaneous Redox Reactions
Predict whether a spontaneous reaction will occur when the following reactants and products are in their standard
b. Back 2Fe3+ (aq) + 2I– (aq) → 2Fe 2+ (aq) + I2 (s)
Cu(s) + 2H +(aq) → Cu 2+ (aq) + H2 (g) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 90 / Electrochemistry
•Method of Solution
For a redox reaction to be spontaneous, it must have a positive cell emf. For the reaction in part (a) we will
apply the diagonal rule. In part (b) we will calculate the emf of the cell.
a. According to the diagonal rule any species on the left (oxidizing agent) of a given half-rxn will react
spontaneously with a species that appears on the right (reducing agent) of any half-cell reaction located
above it in Table 19.1. In the above reaction we ask, can Fe3+ oxidize I– ion? I – ion appears above and to
the right of Fe3+ in Table 19.1 in the textbook. Thus Fe3+ will oxidize I– . b. To calculate the cell emf separate the reaction into half-reactions:
2H + + 2e – → H2 (g)
Cu(s) → Cu 2+ (aq) + 2e– reduction
o o o Ecell = E ox + E red
Ecell = –0.34 V + 0.0 = –0.34 V
The negative standard cell emf indicates that this reaction is not spontaneous at standard conditions. Cu will
not dissolve in 1 M HCl. Also Cu appears to the right, but below H+. Therefore H+ cannot oxidize Cu.
Since the cell emf changes sign when the reaction is reversed, the reaction Cu2+ + H2 → Cu + 2H+ will have a
positive cell emf, and the reduction of Cu2+ by H 2 will be spontaneous under standard conditions.
EXAMPLE 19.6 Free Energy Change and the Standard Cell EMF
Calculate ∆ G° and the equilibrium constant at 25°C for the reaction
2Br– (aq) + I2 (s) → Br2 (l) + 2I – (aq)
•Method of Solution
First find the standard cell emf as we have described previously. Then the standard Gibbs free energy change is
∆ G° = –nFE°
o The equilibrium constant at 25°C is related to Ecell by
ln K =
The half-reactions are:
o reduction 2e– + I2 (s) → 2I– (aq)
Ered = 0.53 V
2Br– (aq) → Br2 (l) + 2e –
Eox = –1.07 V
2Br– (aq) + I2 (s) → Br2 (l) + 2I – (aq) Ecell = –0.54 V
Substituting into ∆ G° = –nFE°,
∆ G° = (–2 mol)(96,500 J/V·mol)(–0.54 V) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 91
where n = 2 mol of electrons transferred in the balanced equation. The units of the Faraday constant must be
expressed in J/(V·mol) in order to cancel the units of volts from Ecell . Remember that 1 J = 1 C·V.
Continuing with our calculation:
o ∆ Grxn = 1.04 × 10 5 J = 104 kJ
The positive value of ∆ G° indicates the reaction is not spontaneous under standard conditions. Next we calculate
the equilibrium constant.
ln K = nE°
= – 18.3 Taking the antilog of both sides we get
K = 10–18.3
= 5 × 10 –19
6. Predict whether Br– ion can reduce I2 under standard state conditions.
2Br– (aq) + I2 (s) → Br2 (l) + 2I – (aq) 7. Predict whether the following reactions will occur spontaneously in aqueous solution at 25°C under standard
a. Ag(s) + Fe3+ (aq) → Ag+(aq) + Fe2+ (aq)
b. Cu2+ (aq) + H2 (g) → Cu(s) + 2H+(aq) 8. Calculate the standard Gibbs free energy changes for the two reactions in problem 7. 9. Calculate the equilibrium constant for the following redox reaction at 25°C.
2Fe2+ (aq) + Ni2+ (aq) → 2Fe 3+ (aq) + Ni(s) EFFECT OF CONCENTRATION ON CELL EMF
2. Calculate the emf of an electrochemical cell in which the reactants and products are present at nonstandard
From a knowledge of E and E°, calculate the concentrations of a given ion in a half- cell. The Nernst Equation. We mentioned earlier that the emf of a cell depends on the nature of the reactants
and products, and on their concentrations. Since the cell emf is a measure of the spontaneity of the cell reaction,
we might reasonably expect the voltage to fall as reactants are consumed and products accumulate.
The equation that relates the cell emf to the concentrations of reactants and products is named after Walter
Nernst. At 298 K, for a redox reaction of the type: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 92 / Electrochemistry aA + bB → cC + dD
The Nernst equation is
E = E° – 0.0257 V
n where E° is the standard cell emf, E is the nonstandard cell emf, 0.0257 V is a constant at 298 K, and n is the
number of moles of electrons transferred according to the balanced equation. The reaction quotient Q contains the
concentrations of reactants and products.
[A]a[B]b The Nernst equation predicts that E will decrease as reactant concentrations decrease and as product
concentrations increase. As Q increases, ln Q increases. Therefore, an increasingly larger number is subtracted
from E°. Therefore, E is not a constant , but decreases as a reaction proceeds. E° is a constant for a reaction and is
characteristic of the reaction.
At equilibrium, this equation reduces to one we have seen before. When the reaction is at equilibrium no net
reaction occurs and no net transfer of electrons occurs, and E = 0. Also, at equilibrium Q = K, and the Nernst
n 0 = E° – Therefore, the standard cell o
emf is related to the equilibrium constant.
When both Ecell and Ecell are known, the Nernst equation can be used to calculate an unknown
concentration. This will be illustrated in Example 19.8.
EXAMPLE 19.7 EMF for a Nonstandard Cell
Calculate the voltage of a cell at 25°C in which the following reaction occurs at the concentrations given:
Zn(s) + 2H +(aq, 1 × 10 –4 M) → Zn 2+ (aq, 1.5 M) + H2 (g, 1 atm)
•Method of Solution
The cell emf can be calculated by use of the Nernst equation.
E = E° –
Ecell = E ox + E red = 0.76 V + 0.00 V = 0.76 V
n = 2 (the number of moles of electrons transferred according to the balanced equation)
P H2 = 1, because hydrogen is in its standard state, 1 atm.
[Zn2+ ] = 1.5 M and [H+] = 1 × 10 –4 M
Substitution into the Nernst equation above yields:
E = 0.76 V –
(1 × 1 0 –4 )2
= 0.76 V – 0.24 V
E = 0.52 V Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 93
The low concentration of H+ (1 × 10 –4 M) compared to its standard state value (1 M) means that the driving
force for reaction will be less than in the standard state, and as we see E < E°.
EXAMPLE 19.8 Determining Ion Concentrations
An electrochemical cell is constructed from a silver half-cell and a copper half-cell. The copper half-cell contains
0.10 M Cu(NO3 )2 and the concentration of silver ions in the other half-cell is unknown. If the Ag electrode is
the cathode, and the cell emf is measured and found to be 0.10 V at 25°C, what is the Ag+ ion concentration?
•Method of Solution
o Since both E cell and Ecell are known, the Nernst equation can be used to calculate an unknown concentration.
To write the reaction quotient Q we need the cell reaction. If Ag is the cathode, the reduction half-reaction is
cathode Ag+ + e – → Ag
Then Cu must be oxidized:
anode Cu → Cu 2+ + 2e – The net reaction is
Cu + 2Ag+ → Cu 2+ + 2Ag
where n = 2. The Nernst equation is
E = E° – 0.0257 V
[Ag+]2 Substitution gives:
0.10 V = 0.46 V – 0.0257 V
[Ag+]2 This equation must be solved for the unknown concentration [Ag+].
2(0.10 V – 0.46 V)
12.2 = ln
Take the antilog of both sides:
1012.2 = (0.10)
[Ag+]2 1.5 × 10 12 =
[Ag+] = (0.10)
1.5 × 1 0 12 = 2 .6 × 10 –7 M
_______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 94 / Electrochemistry
10. Determine the cell potential (Ecell) for the following reaction where the concentrations are as shown.
2Ag+(1 M) + H2 (1 atm) → 2Ag(s) + 2H +(pH = 7.12)
11. The cell emf for the following reaction is 0.98 V.
2Ag+(aq, 1 M) + H 2 (g, 1 atm) → 2Ag(s) + 2H +(aq, ? M)
Calculate the concentration of hydrogen ions.
o 12. Calculate E°, Ecell , and ∆ G for the following cell reaction.
3Zn(s) + 2Cr 3+ (0.0010 M) → 3Zn 2+ (0.010 M) + 2Cr(s) ELECTROLYSIS AND ITS QUANTITATIVE ASPECTS
3. Diagram an electrolytic cell, showing the reactions that occur at the anode and cathode.
Choose the most likely oxidation and reduction processes to be involved in the electrolysis of a given
Calculate the amount of a substance produced by the flow a given electrical current for a given time, or
calculate the time required for a given current to produce a given amount of product. Electrolysis of Molten Salts. E lectrolysis is the process in which electrical energy is used to cause
nonspontaneous redox reactions to occur. An e lectrolytic cell is the apparatus for carrying out electrolysis.
A schematic diagram of an electrolytic cell is shown in Figure 19.3. A battery or other DC power supply serves
as an "electron pump" that supplies electrons to the cathode, where chemical species are reduced. Electrons
resulting from the oxidation of chemical species are withdrawn from the anode and return to the battery. In the
electrolytic cell the cathode is negative and the anode is positive. This is the opposite of an electrochemical cell.
Molten calcium chloride can be decomposed into the elements calcium and chlorine in an electrolytic cell.
Let's examine the redox processes and the movement of ions and electrons in the apparatus. First, no current can
flow through the cell unless it is filled with a liquid because ions in a solid crystal do not move from place to
place. CaCl 2 melts at 780°C. The electrode in the electrolysis cell that is attached to the battery's negative
terminal is the cathode. Electrons flow from the battery onto the cathode. Ca2+ ions are attracted to the cathode
and are reduced to Ca metal. The anode of the electrolysis cell is attached to the positive terminal of the battery.
Chloride ions drift toward the anode where they are oxidized. The electrons travel from the anode to the battery.
For every electron leaving the battery, one must return. The reactions at the electrodes are:
Ca2+ (l) + 2e – → Ca(s)
– (l) → C l (g) + 2e–
Overall Back Forward Ca2+ (l) + 2Cl– (l) → Ca(s) + Cl2 (g) Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 95 e – e Anod e
(+) – Cathode
(–) Battery –
2e Cl 2 Ca –
2e 2+ Ca
Cl Molten CaCl 2 consists of Ca 2+ and Cl – ions Figure 19.3 Schematic diagram of an electrolytic cell. During electrolysis the cations
(M+) migrate to the cathode where they are reduced. Anions (X– ) migrate to the anode,
where they are oxidized. Voltage Requirement. The minimum voltage required to bring about electrolysis can be estimated from
the standard reduction potentials. For the above reaction:
o o o Ecell = E ox + E red
Ecell = –1.36 V + (–2.87 V) = – 4.23 V
The negative emf of the cell reflects that this is a nonspontaneous reaction.
This reaction can be forced to occur by connecting the electrodes to an external source of electrical energy
such as a battery, as in Figure 19.3. The electrolysis will occur when the external voltage is greater than the
negative voltage of the nonspontaneous reaction. The external source of electrical energy must generate greater
than 4.23 V. Competing Electrode Reactions in Aqueous Solution. A complicating factor in the electrolysis of
species in aqueous solution is that water molecules may be oxidized or reduced in preference to the solute
Reduction of water:
Oxidation of water: 2H2 O(l) + 2e – → H 2 (g) + 2OH–
H2 O(l) → 2 O 2 (g) + 2H + + 2e – Eox = –0.83 V
Eox = –1.23 V In the electrolysis of KI solution, for example, H2 (g) is formed at the cathode, and I2 at the anode. H2 (g) is
coming from the reduction of water. Water has a higher (more positive) standard reduction potential than K+,
which indicates that it will be reduced more readily than potassium ion.
K+(aq) + e– → K(s)
2H2 O(l) + 2e– → H2 (g) + 2OH– (aq) o Ered = –2.93 V
Ered = –0.83 V I2 is formed at the anode during electrolysis of aqueous KI. This means the anode reaction is
2I– (aq) → I2 + 2e –
1 H2 O(l) → 2 O 2 (g) + 2H +(aq) + 2e– Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 96 / Electrochemistry
This preference is consistent with the respective oxidation potentials.
2I– → I2 (s) + 2e –
H2 O → 2 O 2 (g) + 2H + + 2e – o Eox = –0.53 V
Eox = –1.23 V The higher oxidation potential of I– indicates that it should be more readily oxidized than H2 O. However, keep
in mind that standard oxidation potentials do not always make the correct prediction of which species will be
oxidized in an electrolysis reaction. The existence of an overvoltage is a complicating factor. Overvoltage is the
additional voltage over and above the standard potential required to cause electrolysis. Overvoltage affects water
oxidation the most. According to the oxidation potentials, I– ion and Br– ion, will be oxidized in the presence of
H2 O, and Cl – ion would not be. Due to overvoltage effects that reduce the tendency of H2 O to be oxidized, Cl–
ion will also be oxidized in aqueous solution. Quantitative Aspects of Electrolysis. Our knowledge of the quantitative relationships in electrolysis is
due mostly to the work of Michael Faraday. He observed that the quantity of a substance undergoing chemical
change during electrolysis is proportional to the quantity of electrical charge that passes through the cell. The
quantity of electrical charge is expressed in coulombs. The electrical charge of 1 mole of electrons is called 1
Faraday which is equal to 96,500 coulombs. 1 mole e– = 1 F = 96,500 C.
The number of electrons shown in the half-reaction is the link between the number of moles of substance
reacted and the number of faradays of electricity required. We obtain the following conversion factors for the two
half-reactions shown below:
K+ + e – → K Ca2+ + 2e – → Ca 1 mol K
1 m ol K
1 mol e – 1 m ol Ca
1 m ol Ca
2 mol e – The number of coulombs passing through an electrolytic cell in a given period of time is related to the electrical
current in amperes (A) where 1 ampere is equal to 1 coulomb per second.
1 A = 1 C/s
In an electrolysis experiment, the measured quantities are current and the time that the current flows. The
number of coulombs passing through the cell is
Total charge in coulombs (C) = current( A) × time (t)
Charge in coulombs = C
s When an electrical current of known amperes flows for a known time the amount of chemical change can be
calculated according to the following road map.
current × time → coulombs → faradays → moles of → grams of
The number of faradays is the key term. It links the experimental variables to the theoretical yield. Reversing
the road map allows one to calculate the time required to produce a specific amount of chemical change with a
given electrical current.
EXAMPLE 19.9 Electrolysis of Aqueous Solutions
Predict the products of the electrolysis of aqueous MgCl2 solution.
•Method of Solution Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 97
In aqueous solution the metal ion is not the only species the can be reduced. H 2 O can be reduced as well. The
two possible reduction reactions are
Mg2+ + 2e – → Mg
2H2 O + 2e – → H2 + 2OH – o Ered = –2.37 V
Ered = –0.83 V Of the two, water has a higher standard reduction potential, and therefore has a greater tendency to be reduced
than Mg 2+ . The two possible oxidation reactions and their oxidation potentials are:
o 2Cl – → C l 2 + 2e – Eox = –1.36 V 1 H2 O → 2 O 2 + 2H+ + 2e – o Eox = –1.23 V The oxidation potentials indicate that H2 O is more readily oxidized than chloride ion. However, in this particular
case, as discussed in Section 19.7 of the text, Cl– is actually oxidized. The large overvoltage for O2 formation
prevents its production when Cl– ion is there to compete. The overall reaction is
2H2 O + 2Cl– → H2 + 2OH – + C l 2
Keep in mind that when two standard oxidation potentials have similar values as in this case, they are not a
reliable predictive tool.
EXAMPLE 19.10 Quantitative Aspects of Electrolysis
How many Faradays are transferred in an electrolytic cell when a current of 12 amps flows for 16 hours?
•Method of Solution
First find the number of coulombs passing through the cell in 16 hours, and convert to faradays. Our road map
current × time → coulombs → Faradays
Recall 12 A = 12 C/s
The number of coulombs in 16 hours is given by multiplying the current times the time (in seconds).
3 600 s
× 16 h ×
= 691,000 C
One Faraday is the charge of one mole of electrons.
1 Faraday = 96,500 C = 9.65 × 10 4 C
= 7.2 Faradays
9.65 × 1 0 4 C
EXAMPLE 19.11 Quantitative Aspects of Electrolysis
691,000 C × How many grams of copper metal would be deposited from a solution of CuSO4 by the passage of 3.0 A of
electrical current through an electrolytic cell for 2.0 hours?
•Method of Solution
Cu 2+ is reduced according to the half-equation
Cu 2+ + 2e – → Cu Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 98 / Electrochemistry
This half-equation tells us that 2 moles of electrons are required to produce 1 mol of Cu. In terms of faradays
2 mol e –
1 mol Cu
1 mol Cu
The number of coulombs passing through a cell is given by the current times the time.
3.0 A × 2.0 h × 1 C /s
= 2.16 × 10 4 C
1h From this number of coulombs we can find the number of Faradays of charge passing through the cell. The road
map will be:
current × time → coulombs →[faradays] → mol Cu → g Cu
The faraday is the link from the current measurements to the molar amount of Cu formed. The number of
2.16 × 10 4 C × 1F
= 0.224 F
9.65 × 1 0 4 C The number of moles of Cu is
0.224 F × 1 m ol Cu
= 0.112 mol Cu
2F The number of grams of Cu is
6 3.5 g
= 7.11 g Cu
1 mol Cu
Of course, this calculation could be carried out using the factor label method.
0.112 mol Cu × 1 C /s
3 600 s
1 m ol Cu
6 3.5 g
× 2 .0 h ×
= 7.11 g Cu
1 mole Cu
9.65 × 1 0
3.0 A × EXERCISES
13. a. Write the half-reaction that occurs at the anode during the electrolysis of aqueous lithium bromide.
b. Write the half-reaction that occurs at the cathode during the electrolysis of aqueous lithium bromide?
14. How many coulombs are required to cause reduction of 0.20 mol of Cr3+ to Cr?
15. How many faradays are transferred in an electrolytic cell when a current of 5.0 amps flows for 10.0 hours?
16. How many faradays are required to electroplate 29.4 g of nickel from a solution containing Ni2+ ion?
17. How many minutes are required to electroplate 25.0 g of metallic Cu using a constant current of 20.0 A?
18. A current of 0.80 A was applied to an electrolytic cell containing molten CdCl2 for 2.5 hours. Decide on
the appropriate half-reaction, and calculate the mass of cadmium metal deposited.
_________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 3 99
5. a. Consider the solution in the anode half-cell of Figure 19.1. How is the concentration of Cu2+ ions
related to the concentration of SO4 ions before any reaction occurs? How are the concentrations of these
two ions related after reaction occurs? b. How does the solution stay electrically neutral?
What electrodes would you use to build an electrochemical cell that will produce 2.0 – 2.5 V?
Electrolysis experiments provide one of the most accurate ways to determine Avogadro's number.
Electrolysis of a NiSO 4 solution for 252 s by a current of 1.02 A yielded a deposit of 0.1531 g Ni. Use
this information to calculate a value for Avogadro's number. Do not use the Faraday constant, but you will
need the electron charge: 1.60 × 10 –19 C .
Why are different products obtained when molten AlBr3 and aqueous AlBr3 are electrolyzed using inert
What is the difference between Ecell and Ecell ? PRACTICE TEST
1. Draw the cell diagrams for each of the redox reactions given below. You may use platinum as an inert
a. 2Al(s) + 3H2 SO4 (aq) → Al 2 (SO4 )3 (aq) + 3H2 (g)
b. Fe 2 (SO4 )3 + 3Pb(s) → 3PbSO4 (s) + 2Fe(s)
c. CuSO4 (aq) + H2 (g) → Cu(s) + H2 SO4 (aq)
d. 2NaBr(aq) + I2 (g) → Br2 (l) + 2NaI(aq)
e. 2FeCl2 (aq) + SnCl2 (aq) → 2FeCl3 (aq) + Sn(s)
f. 2FeCl2 (aq) + SnCl4 (aq) → 2FeCl3 (aq) + SnCl2 (aq) 2. Consider an electrochemical cell constructed from the following half-cells that are linked by a porous
membrane. (1) an Au electrode dipped into 1.0 M Au(NO3 )3 and (2) an Fe electrode dipped into 1.0 M
FeSO4 . Answer the following questions:
a. Which electrode is the cathode?
b. Write a balanced equation for the reaction occurring while the cell is discharging.
c. What emf should the cell generate?
d. In what direction will electrons flow in the outer circuit?
e. Toward which electrode will positive ions migrate? 3. Arrange the following species in order of increasing strength as oxidizing agents:
Ce4+ , O 2 , H 2 O2 , SO 4 .
4. Arrange the following species in order of increasing strength as reducing agents:
Zn, Ni, H2 , F 2 .
5. a. Will O2 (g) oxidize I– ion in acid solution under standard conditions?
b. Will O2 (g) oxidize Br– ?
c. Will O2 (g) oxidize Cl2 ?
6. Calculate ∆ G° and the equilibrium constant (K) for the following reaction.
Fe3+ (aq) + Ag(s) → Fe 2+ (aq) + Ag+(aq)
7. At what ratio of [Fe 2+ ]/[Fe3+ ] would the reaction given in problem 6 become spontaneous if [Ag+] = 1.0 M?
8. Calculate the cell emf for the following reaction:
2Ag+(0.10 M) + H2 (1 atm) → 2Ag(s) + 2H +(pH = 8) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 00 / Electrochemistry
9. What is the minimum voltage required to bring about electrolysis of a solution of Cu2+ and Br– at standard
10. What change in the emf of a hydrogen-copper cell will occur when NaOH(aq) is added to the solution in the
11. Consider a uranium-bromine cell in which U is oxidized and Br2 is reduced. The half reactions are
U3+ (aq) + 3e– → U(s)
E°U3+ /U = ?
– → 2Br– (aq)
Br2 (l) + 2e
E°Br2 /Br – = 1.07 V
If the standard cell emf is 2.91 V, what is the standard reduction potential for uranium?
12. When the concentration of Zn2+ ion is 0.15 M, the measured voltage of the Zn-Cu cell is 0.40 V. What is
the Cu 2+ ion concentration?
13. A hydrogen electrode is immersed in an acetic acid solution. This electrode is connected to another
consisting of an iron nail dipping into 0.10 M FeCl2 . If E cell is found to be 0.24 V, what is the pH of the
acetic acid solution?
14. How many faradays are transferred in an electrolytic cell when a current of 2.0 amps flows for 6 hours?
15. How many faradays are required to electroplate 6.0 g of chromium from a solution containing Cr3+ ?
16. a. How many grams of nickel can be electroplated by passing a constant current of 5.2 A through a solution
of NiSO 4 for 60.0 min?
b. How many grams of cobalt can be electroplated by passing a constant current of 5.2 A through a
solution of CoCl3 for 60.0 min?
17. How long will it take to produce 54 kg of Al metal by the reduction of Al3+ in an electrolytic cell using a
current of 500 amps?
18. Balance the following redox reactions by the ion-electron method.
a. H 2 O2 + I– → I2 (acidic solution)
b. Cr 2 O7 + H 3 AsO3 → Cr 3+ + H3 AsO4 (acidic solution)
c. Cl2 → ClO 4 + C l – (basic solution)
d. Ag(s) + NO3 (aq) → NO2 (g) + Ag +(aq) (acidic solution)
19. Given the reduction potential of the SHE, and the half-equation
2H2 O + 2e – → H2 (g) + 2OH– (aq) E° = –0.83 V calculate the value of the ionization constant for water.
H2 O H+(aq) + OH– (aq) 20. A cell was constructed using the standard hydrogen electrode as one half-cell, and a lead electrode in a 0.10
M K 2 CrO4 solution in contact with undissolved PbCrO4 as the other half-cell. The potential of the cell
was measured to be 0.50 V with the Pb electrode as the anode. From these data determine the Ksp of
21. A 40 watt light bulb is powered by a lead storage battery that has 20.0 g of Pb available for the anode? For
how many hours will the bulb provide illumination? Assume the voltage is constant at 1.5 V. Recall: 1
watt = 1 J/s. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Electrochemistry / 4 01
2– 1. a. 14H + + Cr2 O7 (aq) + 6Br– (aq) → 2Cr 3+ (aq) + 3Br2 (g) + 7H 2 O
b. 2MnO 4 (aq) + 6I– (aq) + 4H2 O → 2MnO 2 (s) + 3I 2 (aq) + 8OH– (aq)
– 2. a. 5H 2 C 2 O4 (aq) + 2MnO4 (aq) + 6H+(aq) → 2Mn 2+ (aq) + 10CO2 (g) + 8H 2 O(l)
b. Zn(s) + ClO– (aq) + H2 O(l) → Zn(OH)2 (s) + Cl– (aq)
o o 3. a. E cell = –2.58 V
b. Ecell = 0.29 V
2+ (aq) + Pb(s) → Cu(s) + Pb2+ (aq)
4. a. Cu b. Cu
d. From Pb to Cu. e. Cu
5. a. Al(s) b. Fe3+
7. a. No b. Yes
8. a. ∆ G° = 2.90 kJ b. ∆ G° = – 65.6 kJ
9. Kc = 3.0 × 10–35
10 E = 1.22 V
11. 9.0 × 10–4 M
12. Ecell = 0.02 V Ecell = 0.01 V ∆ G = –5.79 kJ
13. a. 2Br– → Br2 + 2e –
b. 2H 2 O + 2e – → H2 (g) + 2OH– (aq)
14. 57,900 C
15. 1.86 F
16. 0.808 F
17. 63.3 min
18. 4.19 g Cd c. Cu Conceptual Questions
2. 3. 4. Back 2– a. After: [Cu2+ ] > [SO 4 ] b . Cl – ions migrate into the solution from the salt bridge. Two chloride ions
for each excess Cu2+ ion.
The idea is to choose two electrodes whose reduction potentials differ by 2.0 – 2.5 V. The text points out
that an alternate way to calculate the standard emf of a cell is to calculate the difference between the
reduction potential of the cathode and the reduction potential of the anode.
Ecell = E cathode – Eanode
There are many possibilities. Pick an electrode such as Zn and look in the table for an electrode about
2.0 – 2.5 V above the reduction potential of Zn. Au and Ce4+ ao suitable. A Zn/Au cell has
Ecell = 1.50 V – (–0.76 V) = 2.26 V, and a Zn/Ce4+ cell has E cell = 1.61 V – (–0.76 V) = 2.37 V. You
don't have to pick zinc. There are many possible cells with a standard cell emf in this range.
Avogadro's number is a ratio. Its units are "atoms per mole." We are given the mass of Ni so we can use it
to find the number of moles. You should get 2.609 × 10 –3 mol Ni. Now to count the atoms. Actually we
will count the electrons and then divide by 2 to get the number of Ni atoms. As usual the total charge in
coulombs flowing through the cell is equal to the current (A) times the time (t). Dividing the total charge
by the charge on one electron gives the total number of electrons passing through the cell. The number of
Ni atoms produced by the reduction of Ni2+ ions is just half the number of electrons. Divide the number of
Ni atoms by its corresponding number of moles in order to get Avogadro's number.
Molten AlBr3 consists of Al3+ ions and Br– ions. When it undergoes electrolysis the Al3+ ion is reduced to
Al metal, and the Br– ion is oxidized to Br2 . An aqueous solution of AlBr 3 consists of H2 O molecules, as Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 02 / Electrochemistry 5. well as Al 3+ ions and Br– ions. As you can tell from oxidation and reduction potentials, Al3+ ion will not
be reduced in aqueous solution. Rather the H2 O molecules are and the products are H2 and OH– ion. In
aqueous solution Br– ion will be oxidized just as it was in the molten state.
Ecell refers to the cell potential when the reactants at standard state concentrations are converted into
products at their standard state concentrations. When the reactants at concentrations other than standard state
concentrations are converted into products at concentrations other their standard state concentrations the cell
potential is just Ecell. E cell is a constant while Ecell is not. Practice Test
1. a. Al(s) | Al3+ (aq) || H+(aq) | H2 (g) | Pt(s)
b. Pb(s) | Pb 2+ (aq) || Fe3+ (aq) | Fe(s)
c. Pt(s) | H2 (g) | H+(aq) || Cu 2+ (aq) | Cu(s)
d. Pt(s) | Br– (aq) | Br2 (l) || I2 (s) | I– (aq) | Pt(s)
e. Pt(s) | Fe 2+ (aq), Fe3+ (aq) || Sn 2+ (aq) | Sn(s)
f. Pt(s) | Fe 2+ (aq), Fe3+ (aq) || Sn 4+ (aq), Sn2+ (aq) | Pt(s)
2. a. Au b. 2Au3+ (aq) + 3Fe(s) → 2Au(s) + 3Fe2+ (aq)
c. 1.94 V d. from Fe to Au e. Au
3. SO2 – < O 2 < Ce4+ < H2 O2
4. F 2 < H2 < Ni < Zn
5. a. Yes b. Yes c. No
6. ∆ Grxn = 2.9 kJ; K = 0.31
7. [Fe2+ ]/[Fe3+ ] = 0.31
8. E = 1.21 V
9. 0.73 V
10. The cell emf will increase due to a decrease in [H+]. The cell reaction is Cu2+ + H2 → Cu + 2H+
11. –1.84 V
12. [Cu2+ ] = 3.1 × 10 –25 M
13. pH = 3.88
14. 0.45 F
15. 0.35 F
16. a. 5.69 g Ni b. 3.81 g Co
17. 322 h
18. a. H 2 O2 + 2I– + 2H+ → I2 + 2H2 O
b. Cr 2 O7 + 3H3 AsO3 + 8H+ → 2Cr 3+ + 3H3 AsO4 + 4H2 O
c. 4Cl2 + 8OH – → ClO 4 + 7 Cl – + 4H2 O
d. Ag(s) + 2H +(aq) + NO3 (aq) → NO2 (g) + Ag +(aq) + H2 O
19. Kw = 9.0 × 10 –15
20. [Pb2+ ] = 3.0 × 10 –13 M; Ksp = 3.0 × 10 –14
21. 0.19 h
_______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
View Full Document