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Chapter 20 CH 182 

 Major Concepts: 

One of the important characteristics of a voltaic cell is its emf, or voltage, which is the difference in the electrical potentials at the two electrodes.  The voltage can be calculated using Standard Reduction Potentials.  

Gibbs free energy can be related to the voltage and spontaneity of the cell  

The voltage associated with cells that run under nonstandard conditions can be calculated using standard voltages and the Nernst equation.   

  20.1 Oxidation States Chemical reactions in which the oxidation state of one or more substances change are called

oxidation-reduction reactions (redox reactions).   Electrochemistry is the branch of chemistry that deals with relationships between electricity and

chemical reactions. Consider the spontaneous reaction that occurs when Zn is added to HCl.

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) • The oxidation numbers of Zn and H+ have changed. • The oxidation number of Zn has increased from 0 to +2.

• The oxidation number of H has decreased from +1 to 0.

• Therefore, Zn is oxidized to Zn2+, while H+ is reduced to H2. • H+ causes Zn to be oxidized. • Thus, H+ is the oxidizing agent, or oxidant. • Zn causes H+ to be reduced. • Thus, Zn is the reducing agent, or reductant.

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Rules for assigning oxidation numbers:  

(1)  All free elements have oxidation states of _________________ 

  

(2) Group 1 elements are always _____________ 

  

(3) Group 2 elements are always _____________ 

  

(4) Halogens are always ____________ 

  

(5) Hydrogen will be  _______________, except when is in a binary compound with a metal, 

  then it will be _________________   

 (6) Oxygen will be _______________, except when it is combined with _________________,  

 then it will be ____________________  

  

(7) The sum of all oxidation numbers in a neutral compound must be ________________ 

  

(8) The sum of all oxidation numbers in an ion must be equal to  

 ___________________________________________________________________________  

  

(9) In chemical reactions, the oxidation numbers may change from reactants to products 

   

  

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Guided Practice H2O Ag KOH NaCrO2

Ca(OH)2

CrCl3

NH3

PbS PbSO4

MnO2

CuCr2O7

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Practice Questions

(1) The nickel-cadmium (nicad) battery, a rechargeable “dry cell” used in battery-operated devices, uses the following redox reaction to generate electricity:

Cd (s) + NiO2 (s) Cd(OH)2 (s) + Ni(OH)2 (s) Identify the substances that are oxidized and reduced, and indicate which are oxidizing agents and which are reducing agents.

(2) Identify the oxidizing and reducing agents in the oxidation-reduction reaction

2 H2O (l) + Al (s) + MnO4- (aq) Al(OH)4

- (aq) + MnO2 (s)

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20.2 Balancing Oxidation-Reduction Equations Recall the law of conservation of mass: The amount of each element present at the beginning of the

reaction must be present at the end. Conservation of charge state that electrons are not lost in a chemical reaction. Steps for Balancing by means of half-reactions for Acidic Solutions:  

1. ___________________________________________________________________

2. ___________________________________________________________________

3. ___________________________________________________________________

4. ___________________________________________________________________

5. ___________________________________________________________________ Practice Questions

(1) Complete and balance this equation by the method of half-reactions:

Cr2O72- (aq) + Cl- (aq) Cr3+ (aq) + Cl2 (g) (acidic solution)

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(2) Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution.

a. Cu (s) + NO3- (aq) Cu2+ (aq) + NO2 (g)

b. Mn2+ (aq) + NaBiO3 (s) Bi3+ (aq) + MnO4- (aq)

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Steps for Balancing by means of half-reactions for Basic Solutions:  

1. ___________________________________________________________________

2. ___________________________________________________________________

3. ___________________________________________________________________

4. ___________________________________________________________________

5. ___________________________________________________________________

6. ___________________________________________________________________ Practice Questions

(1) Complete and balance this equation for a redox reaction that takes place in basic solution

CN- (aq) + MnO4- (aq) CNO- (aq) + MnO2 (s) (basic solution)

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(2) Complete and balance the following equations for oxidation-reduction reactions that occur in basic solution:

a. NO2- (aq) + Al (s) NH3 (aq) + Al(OH)4

- (aq)

b. Cr(OH)3 (s) + ClO- (aq) CrO42- (aq) + Cl2 (g)

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20.3 Voltaic Cells Voltaic, or galvanic cells, are devices in which ___________________________________________ _________________________________________________________________________________ _________________________________________________________________________________

A voltaic cell that uses a salt bridge to complete the electrical circuit.

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Practice Questions

(1) The oxidation-reduction reaction

Cr2O7- (aq) + 14 H+ (aq) + 6 I- (aq) 2 Cr3+ (aq) + 3 I2 (s) + 7 H2O (l)

is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes.

(2) The two half-reactions in a voltaic cell are

Zn (s) Zn2+ (aq) + 2e-

ClO3- (aq) + 6 H+ (aq) + 6 e- Cl- (aq) + 3 H2O (l)

Indicate which reaction occurs at the anode and which at the cathode. Which electrode is consumed in the cell reaction? Which electrode is positive?

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Lesson #1 Homework 

 (1) Hydrazine (N2H4) and dinitrogen tetroxide  (N2O4) form a self‐igniting mixture that has been used 

as a rocket propellant. The reaction products are N2 and H2O. (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized , and what is being reduced? (c) Which substance serves as the reducing agent, and which is the oxidizing agent? 

        

(2) Complete and balance the following equations, and identify the oxidizing and reducing agents: 

(a) Cr2O72‐(aq)  +  I‐     →    Cr3+(aq)    +   IO3(aq)                                    (Acidic solution) 

(b) MnO4‐(aq)   +   CH3OH(aq)  →   Mn2+(aq)  +HCO2H(aq)                  (acidic solution) 

(c) I2(s)   +   OCl‐(aq)    →    IO3‐(aq)   +  Cl‐                                              (acidic solution) 

(d) As2O3(s)  +  NO3‐      →     H3AsO4(aq)   +   N2O3(aq)                         (acidic solution) 

(e) MnO4‐(aq)   +  Br‐ (aq)    →     MnO2(s)   +   BrO3

‐(aq)                      (basic solution) 

(f) Pb(OH)42‐ (aq)    +   ClO‐      →    PbO2(s)   +   Cl‐(aq)                         (basic solution) 

                    

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(3) A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode compartment consists of an aluminum strip placed in a solution of Al(NO3)3, and the other has a nickel strip placed in a solution of NiSO4. The overall cell reaction is: 

 2Al(s)   +  3Ni2+ (aq)    →      2Al3+(aq)   +   3Ni(s) 

 (a)What is being oxidized, and what is being reduced?  (b) Write the half‐reactions that occur in the two electrode compartments.  (c) Which electrode is the anode, and which is the cathode?  (d) Indicate the signs of the electrodes.  (e) Do electrons flow from the aluminum electrode to the nickel electrode,  or from the  

                     nickel to the aluminum?  (f) In which directions do the cations and anions migrate through the solution? Assume  

                    the Al is not coated with its oxide. 

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20.4 & 20.5 Cell EMF Under Standard Conditions The flow of electrons from anode to cathode is spontaneous. • What is the “driving force”? Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than

the anode.

• Potential difference is the difference in electrical potential. • The potential difference is measured in volts. • One volt (V) is the potential difference required to impart one joule (J) of energy to a charge of

one coulomb (C):

• Electromotive force (emf) is the force required to push electrons through the external circuit. • Cell potential: Ecell is the emf of a cell. • This is known as the cell voltage. • Ecell is > 0 for a spontaneous reaction. For 1M solutions or 1 atm pressure for gases at 25°C (standard conditions), the standard emf

(standard cell potential) is called E°cell. • For example, for the reaction:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

• E°cell = +1.10 V

C

J1V1

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Standard Reduction (Half-Cell) Potentials

Standard reduction potentials, E°cell are measured relative to a standard. The emf of a cell can be calculated from standard reduction potentials:

E°cell = E°red(cathode) – E°red (anode)

The standard reduction potential is an ____________________________ property. • Therefore, changing the stoichiometric coefficient does not affect E°red. Reactions with E°cell > 0 are ________________________________________________

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Practice Questions

(1) For the Zn-Cu2+ voltaic cell we have

Zn (s) + Cu2+ (aq, 1 M) Zn2+ (aq, 1 M) + Cu (s) E°cell = 1.10 V

Given that the standard reduction potential of Zn2+ to Zn (s) is -0.76 V, calculate the E°red for the reduction of Cu2+ to Cu:

Cu2+ (aq, 1 M) + 2 e- Cu (s)

(2) A voltaic cell is based on the half-reactions

In+ (aq) In3+ (aq) + 2e-

Br2 (l) + 2e- 2 Br- (aq)

The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate E°red for the reduction of In3+ to In+.

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(3) Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction

Cr2O72- (aq) + 14 H+ (aq) + 6 I- (aq) 2 Cr3+ (aq) + 3 I2 (s) + 7 H2O

(4) Using data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:

2 Al (s) + 3 I2 (s) 2 Al3+ (aq) + 6 I- (aq)

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(5) A voltaic cell is based on the following two standard half-reaction:

Cd2+ (aq) + 2e- Cd (s)

Sn2+ (aq) + 2e- Sn (s)

By using the data in the AP references, determine (a) the half-reaction that occur at the cathode and anode, and (b) the standard cell potential.

(6) A voltaic cell is based on Co2+ / Co half-cell and an Fe3+ / Fe2+ half-cell. (a) What reaction occurs at the anode? (b) What is the standard cell potential?

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Strengths of Oxidizing and Reducing Agents Consider a table of standard reduction potentials. We can use this table to determine the relative strengths of reducing (and oxidizing) agents. • The more positive the E°red, the stronger the oxidizing agent (written in the table as a reactant). • The more negative the E°red, the stronger the reducing agent (written as a product in the table). • We can use this to predict if one reactant can spontaneously oxidize another. • For example: • F2 can oxidize H2 or Li. • Ni2+ can oxidize Al(s). • We can use this table to predict if one reactant can spontaneously reduce another. • For example: • Li can reduce F2. Practice Questions

(1) Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3

- (aq), Ag+ (aq), Cr2O72- (aq)

(2) Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I- (aq), Fe (s), Al (s)

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Lesson #2 Homework 

 (1) (a)Why is it impossible to measure the standard reduction potential of a single half‐reaction? 

(b) Describe how the standard reduction potential of a half‐reaction can be determined.  (c) By using data in Appendix E, determine which is the more unfavorable reduction, Cd2+(aq) to        Cd(s) or Ca2+(aq) to Ca(s). 

      

(2) Given the following half‐reactions and associated standard reduction potentials:  

    AuBr4‐(aq)   +  3e‐      →      Au(s)   +   4Br‐(aq)      E°red =  ‐0.858 V     Eu3+(aq)    +  e‐         →     Eu2+(aq)        E°red = ‐0.43 V   IO‐(aq)   +   H2O(l)    +   2e‐     →     I‐(aq)   +   2OH‐(aq)      E °red =  +0.49  V      Sn2+(aq)   +  2e‐            →      Sn(s)         E°red =  ‐0.14 V  

(a)Write the cell reaction for the combination of these half‐cell reactions that leads to the                      largest positive cell emf, and calculate the value.  

(b) Write the cell reaction for the combination of half‐cell reactions that leads to the smallest                      positive cell emf, and calculate that value.       

(3) A voltaic cell  consists of a strip of lead metal in a solution of Pb(NO3)2 in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with Cl2 gas bubbled around the electrode. The two beakers are connected by a salt bridge.  

 (a) Which electrode serves as the anode, and which is the cathode?  (b) Does the Pb electrode gain or lose mass as the cell reaction proceeds?  (c) Write the equation for the overall cell reaction.  (d) What is the emf generated by the cell under standard conditions? 

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mol-V485,96

mol485,961

JCF

EMF and ΔG We can show that:

ΔG = nFE

• where ΔG is the change in free energy, n is the number of moles of electrons transferred, F is

Faraday's constant, and E is the emf of the cell. We define a faraday (F) as:

• Since n and F are positive, if ΔG < 0 then E > 0 and the reaction will be spontaneous. • When the reactants and products are in their standard states:

ΔGo = –nFEo

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Practice Questions

(1) (a) Use the standard reduction potentials to calculate the standard free-energy change. ΔG°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction

4 Ag (s) + O2 (g) + 4 H+ (aq) 4 Ag+ (aq) + 2 H2O (l)

(b) Suppose the reaction in part (a) was written

2 Ag (s) + ½ O2 (g) + 2 H+ (aq) 2 Ag+ (aq) + H2O (l)

What are the values of E°, ΔG, and K when the reaction is written this way?

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(2) For the reaction

3 Ni2+ (aq) + 2 Cr(OH)3 (s) + 10 OH- (aq) 3 Ni (s) + 2 CrO42- (aq) + 8 H2O (l)

(a) What is the value of n? (b) Use the data in Appendix E to calculate ΔG°. (c) Calculate K at T=298 K.

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20.6 Cell EMF Under Nonstandard Conditions A voltaic cell is functional until E = 0 at which point equilibrium has been reached. • The cell is then “dead.” The point at which E = 0 is determined by the concentrations of the species involved in the redox

reaction. The Nernst Equation We can calculate the cell potential under nonstandard conditions. Recall that:

ΔG = ΔG° + RT lnQ

We can substitute in our expression for the free energy change:

nFE = nFE° + RT lnQ

Rearranging, we get the Nernst equation:

or

• Note that there is a change from natural logarithm to log base 10.

The Nernst equation can be simplified by collecting all the constants together and using a temperature

of 298 K:

QnF

RTEE ln

QnF

RTEE log

303.2

Qn

EE log0592.0

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Example: If we have the reaction:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

• If [Cu2+] = 5.0 M and [Zn2+] = 0.050M:

Practice Questions

(1) Calculate the emf at 298 K generated by the cell described when [Cr2O72-] = 2.0 M,

[H+] = 1.0 M, [I-] = 1.0 M, and [Cr3+] = 1.0 x 10-5 M.

Cr2O72- (aq) + 14 H+ (aq) + 6 I- (aq) 2 Cr3+ (aq) + 3 I2 (s) + 7 H2O (l)

VVEcell 16.10.5

050.0log

2

0592.010.1

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(2) Calculate the emf generated by the cell described in the practice exercise accompanying reaction when [Al3+] = 4.0 x 10-3 M and [I-] = 0.010 M.

2 Al (s) + 3 I2 (s) 2 Al3+ (aq) + 6 I- (aq)

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(3) What is the pH of the solution in the cathode compartment of the cell pictured in Figure 20.11 when PH2 = 1.0 atm, [Zn2+] in the anode compartment 0.10 M, and cell emf is 0.542 V.

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Lesson #3 Homework 

 (1) From each of the following pairs of substances, use data in Appendix E to choose the one that is 

stronger oxidizing agent:  

(a) Cl2(g)  or Br2(l) 

(b) Ni2+(aq) or Cd2+(aq) 

(c) BrO3‐(aq) or IO3

‐(aq) 

(d) H2O2(aq) or O3(g) 

   

(2) For each of the following reactions, write a balanced equation, calculate the standard emf, 

calculate the G ° at 298K, and calculate the equilibrium constant K at 298 K.  

(a) Aqueous iodide ion is oxidized to I2(s) by Hg22+(aq).  (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion.  (c) In basic solution Cr(OH)3(s) is oxidized to CrO4

2‐(aq) by ClO‐(aq).       

(3) A voltaic cell utilizes the following reaction: 

 Al(s)   +   3Ag+(aq)    →    Al3+(aq)    +   3Ag(s) 

 What is the effect on the cell emf of each of the following changes?  (a) Water is added to the anode compartment, diluting the solution.  (b) The size of aluminum electrode is increased.  (c) A solution of AgNO3 is added to the cathode compartment, increasing the quantity of Ag+ but  

                    not changing its concentration.  (d) HCl is added to the AgNO3 solution, precipitating some of the Ag+ as AgCl. 

         

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(4) A voltaic cell utilizes the following reaction and operates at 298 K: 

3Ce4+(aq)  +  Cr(s)   →   3Ce3+(aq)    +   Cr3+(aq) (a)What is the emf of this cell under standard conditions?  (b) What is the emf of this cell when [Ce4+] = 2.0 M, [Ce3+] = 0.010 M, and [Cr3+] =  0.010 M?  (c) What is the emf of the cell when [Ce4+] = 0.35 M, [Ce3+] = 0.85 M, and [Cr3+] = 1.2M? 

         

(5) A voltaic cell is constructed with two silver‐silver chloride electrodes, each of which is based on the following half‐reaction: 

 AgCl(s)  +  e‐    →      Ag(s)   +  Cl‐(aq) 

 The two cell compartments have [Cl‐] = 2.55 M, respectively.  

(a) Which electrode is the cathode of the cell?  (b) What is the standard emf of the cell?  (c) What is the cell emf for the concentrations given?  (d) For each electrode predict whether [Cl‐] will increase, decrease, or stay the same as        the cell operates. 

      

(6) A voltaic cell is constructed that is based on the following reaction:  

 Sn2+(aq)  +  Pb(s)    →      Sn(s)  +  Pb2+(aq) 

 (a)If the concentration of Sn2+ in the cathode compartment is 1.00 M and the cell  

                   generates and emf of +0.22 V, what is the concentration of Pb2+ in the anode                     compartment?  

(b) If the anode compartment contains [SO42‐] = 1.00 M in equilibrium with PbSO4(s),  

                    what is the Ksp of PbSO4? 

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Concentration Cells A concentration cell is one whose emf is generated solely because of a concentration difference. Example: Consider a cell with two compartments, each with a Ni(s) electrode but with different

concentrations of Ni2+(aq). • One cell has [Ni2+] = 1.0 M and the other has [Ni2+] = 0.001 M. • The standard cell potential is zero. • But this cell is operating under nonstandard conditions! • The driving force is the difference in Ni2+ concentrations. • Anode (dilute Ni2+): Ni(s) Ni2+(aq) + 2e

• Cathode (concentrated Ni2+): Ni2+(aq) + 2e Ni(s) Using the Nernst equation we can calculate a cell potential of +0.0888 V for this concentration cell.

Concentration cell based on the Ni2+ - Ni cell reaction

In (a) the concentration of Ni2+ (aq) in the two compartments are unequal, and the cell generates an electrical current. The cell operates until the concentration of Ni2+ (aq) in the two

compartments become equal, (b) at which point the cell has reached equilibrium and is “dead”.

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Practice Questions

(1) A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and an unknown concentration of H+ (aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, PH2 = 1.00 atm). At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate [H+] for the solution at electrode 1. What is its pH?

(2) A concentration cell is constructed with two Zn (s) – Zn2+ (aq) half-cells. The first half-cell has [Zn2+] = 1.35 M, and the second half-cell has [Zn2+] = 3.75 x 10-4 M.

a. Which half-cell is the anode of the cell? b. What is the emf of the cell?

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20.9 Electrolysis Electrolysis reactions are nonspontaneous reactions that require an external current in order to force

the reaction to proceed. • They take place in electrolytic cells.

In voltaic and electrolytic cells, reduction occurs at the cathode and oxidation occurs at the anode. • However, in electrolytic cells, electrons are forced to flow from the anode to the cathode. • In electrolytic cells the anode is positive and the cathode is negative. • In voltaic cells the anode is negative and the cathode is positive. Example: The decomposition of molten NaCl. • Cathode: 2Na+(l) + 2e 2Na(l) • Anode: 2Cl(l) Cl2(g) + 2e. • Industrially, electrolysis is used to produce metals like Al. Electrolysis of high-melting ionic substances requires very high temperatures. • Do we get the same products if we electrolyze an aqueous solution of the salt? • Water complicates the issue! • Example: Consider the electrolysis of NaF(aq):

Na+ (aq) + e Na(s) E°red = 2.71 V

2H2O(l) + 2e H2(g) + 2OH(aq) E°red = 0.83 V • Thus, water is more easily reduced than the sodium ion.

2F(aq) F2(g) + 2e E°red = +2.87 V

2H2O(l) O2(g) + 4H+(aq) + 4e E°red = +1.23 V • Thus, it is easier to oxidize water than the fluoride ion.

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Electrolysis does not always involve inert electrodes.

• Active electrodes are electrodes that take part in electrolysis. • An example is electroplating.

Consider an active Ni electrode and another metallic electrode (steel) placed in an aqueous solution of

NiSO4: • Anode (nickel strip): Ni(s) Ni2+(aq) + 2e • Cathode (steel strip): Ni2+(aq) + 2e Ni(s) Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion.

Electrolysis of Molten sodium chloride Cl- ions are oxidized to Cl2 (g) at the anode, and Na+ ions are reduced to Na (l) at the cathode.

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Quantitative Aspects of Electrolysis We want to know how much material we obtain with electrolysis. Consider the reduction of Cu2+ to Cu.

Cu2+(aq) + 2e Cu(s). 2 mol of electrons will plate 1 mol of Cu. The charge of one mol of electrons is 96,500 C (1 F). • A coulomb is the amount of charge passing a point in one second when the current is one ampere. The amount of Cu can be calculated from the current (I) and time (t) required to plate. q

I = ------------ t

Practice Questions

(1) Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.

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(2) (a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2 e Mg. Calculate the required mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 x 103 s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is 100.0 A?

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Lesson #4 Homework 

 (1) (a)  Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.5 x 104 A  

       flowing for a period of 24 h.  Assume the electrolytic cell is 85% efficient. 

(b)  What is the energy requirement for this electrolysis per mol Li formed in the applied          emf is +7.5 V 

            (2) Some years ago a unique proposal was made to raise the Titanic.  The plan involved placing 

pontoons within the ship using a surface‐controlled submarine‐type vessel.  The pontoons would contain cathodes and would be filled with hydrogen has formed by the electrolysis od water.  It has been estimated that it would require about 7 x 108 mol of H2 to provide the buoyancy to lift the ship. 

a. How many coulombs of electrical charge would be required? 

b. What is the minimum voltage required to generate H2 and O2 if the pressure on the gases at the depth of the wreckage (2 mi) is 300 atm?  

c. What is the minimum electrical energy required to raise the ship by electrolysis? 

d. What is the minimum cost of electrical energy required to generate the necessary H2 if the electricity costs 85 cents per kilowatt‐hour to generate at the site?