Chapter 19



Chapter 19 Dr. Nabil EL-Halabi

Redox Reactions and Electrochemistry

Electrochemistry: is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy.

Electrochemical processes are oxidation-reduction reactions in which:

• The energy released by a spontaneous reaction is converted to electricity or

• Electrical energy is used to cause a nonspontaneous reaction to occur

In redox reactions electrons are transferred from one substance to another.

Consider the following reaction:

Mg(s) + 2HCl(aq) γ MgCl2(aq) + H2(g)

Oxidation No. 0 +1 +2 0

Mg(s) γ Mg2+ + 2e oxidation half-reaction (lose e).

2H+ + 2e γ H2 reduction half-reaction (gain e).

Balancing Redox Equations:

In acidic medium:

Step 1: Write the unbalanced equation for the reaction in ionic form.

Fe2+ +Cr2O72- γ Fe3+ + Cr3+

Step 2: Separate the equation into two half-reactions.

Oxidation Fe2+ γ Fe3+

Reduction Cr2O72- γ Cr3+

Step 3: Balance the atoms other than O and H in each half-reaction.

Cr2O72- γ 2Cr3+

Step 4: For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.

Cr2O72- γ 2Cr3+ + 7H2O

14H+ + Cr2O72- γ 2Cr3+ + 7H2O

Step 5: Add electrons to one side of each half-reaction to balance the charges on the half-reaction.

Fe2+ γ Fe3+ + 1e

6e + 14H+ + Cr2O72- γ 2Cr3+ + 7H2O.

Equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.

6Fe2+ γ 6Fe3+ + 6e

6e + 14H+ + Cr2O72- γ 2Cr3+ + 7H2O.

Step 6: Add the two half-reactions together and balance the final equation by inspection.

The number of electrons on both sides must cancel.

14H+ + Cr2O72- + 6Fe2+ + 6e γ 6Fe3+ + 2Cr3+ + 7H2O 6e

Step 7: Verify that the number of atoms and the charges are balanced in both sides of the equation.

*** For reaction in a basic medium, add OH- to both sides of the equation for every H+ that

appears in the final equation. Where H+ and OH- appeared on the same side of the

equation, we would combine the ions to give H2O.

Example 19.1.

An Overview of Electrochemical Cells

There are two types of electrochemical cells based upon the general thermodynamic nature of the reaction:

1) A galvanic cell (or voltaic cell) uses a spontaneous redox reaction to generate electrical energy. The reacting system does work on the surroundings. All batteries contain voltaic cells.

2) An electrolytic cell uses electrical energy to drive a nonspontaneous reaction (G > 0), the surroundings do work on the reacting system.

All electrochemical cells have several common features:

1) They have two electrodes:

Anode: The electrode at which oxidation half-reaction occurs.

Cathode: The electrode at which reduction half-reaction occurs.

2) The electrodes are dipped into an electrolyte, a solution that contains a mixture of ions and will

conduct electricity.

The difference in electrical potential between the anode and cathode is measured by a voltmeter and reading in volts is called:

• cell voltage

• electromotive force (emf)

• cell potential

Cell diagram

Zn(s) + Cu2+(aq) γ Cu(s) + Zn2+(aq)

[Cu2+] = 1M and [Zn2+] = 1M

Zn / Zn2+(1M) // Cu2+(1M) / Cu

anode cathode

Standard Reduction Potentials (E0)

Is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.

To measure the potential of a single electrode, the hydrogen electrodes shown in the figure, serves as the reference for this purpose. Hydrogen gas is bubbled into HCl solution at 250C.

The platinum electrode has two functions:

1- It provides a surface on which the dissociation of hydrogen molecules can take place:

H2 γ 2H+ + 2e

2- It serves as an electrical conductor to external circuit.

Under standard condition (when the pressure of H2 is 1 atm and the concentration of HCl solution is 1M), the potential for reduction of H+ at 250C is taken as to be exactly zero.

2H+(1 M) + 2e γ H2(1 atm) E0 = 0 V

The standard reduction potential of the hydrogen electrode is defined as zero. The hydrogen electrode is called the standard hydrogen electrode (SHE).

We can use SHE to measure the potentials of other kinds of electrodes. For example,

In this cell the zinc electrode is the anode and SHE is the cathode, we deduce this fact from the decrease in mass of the zinc electrode during the operation of the cell.

The cell diagram is:

Zn(s) / Zn2+(1M) // H+(1 M) / H2(1 atm)/Pt(s)

The emf of the cell is 0.76 V at 250C. We can write the half-cell reactions as:

Cathode (reduction) 2H+(1 M) + 2e γ H2(1 atm)

Anode (oxidation) Zn(s) γ Zn2+(aq) + 2e

-----------------------------------------

Oveall: Zn(s) + 2H+(1 M) γ Zn2+(aq) + H2(1 atm)

E0cell = E0cathode - E0anode E0cell = E0H+/ H2 - E0 Zn2+/ Zn

0.76 = 0 - E0 Zn2+/ Zn E0 Zn2+/ Zn = - 0.76 V

Zn(s) γ Zn2+(aq) + 2e E0 = 0.76 V

Zn2+(aq) + 2e γ Zn(s) E0 = - 0.76 V

Thus the standard reduction potential of zinc is - 0.76 V

***The standard electrode potential of copper can be obtained in a similar fashion, but in this case, the copper electrode is the cathode because its mass increases during the operation of the cell.

Cu2+(aq) + 2e γ Cu(s)

And the cell diagram is

Pt(s) / H2(1 atm) / H+(1 M) // Cu2+(1M) / Cu(s)

{{

Anode (oxidation) H2(1 atm) γ 2H+(1 M) + 2e

Cathode (reduction) Cu2+(1M) + 2e γ Cu(s)

----------------------------------------------------

Oveall: H2(1 atm) + Cu2+(1M) γ 2H+(1 M) + Cu(s)

E0cell = E0cathode - E0anode

0.34 V = E0Cu2+/ Cu - E0 H+ / H2 0.34 V = E0Cu2+/ Cu - 0

E0Cu2+/ Cu = 0.34 V

Thus the standard reduction potential of copper is 0.34 V.

***For the cell between Cu and Zn, we can write

Anode (oxidation) Zn(s) γ Zn2+(aq) + 2e

Cathode (reduction) Cu2+(1M) + 2e γ Cu(s)

-----------------------------------------------------

Overall: Zn(s) + Cu2+(aq) γ Cu(s) + Zn2+(aq)

E0cell = E0cathode - E0anode E0cell = E0Cu2+/ Cu + E0Zn2+/ Zn

E0cell = 0.34 V - (- 0.76 V) = 1.10 V

Under standard conditions for reactants and products, the redox reaction is spontaneous in the forward direction if the standard emf of the cell is positive. If it is negative, the reaction is spontaneous in the opposite direction.

***In Table 19.1:

By definition, the SHE has an E0 value of 0.00 V.

• E0 is for the reaction as written in the table.

• The more positive E0 the greater the tendency for the substance to be reduced.

F2(1 atm) + 2e γ 2F-(1 M) E0 = 2.87 V

F2 is the strongest oxidizing agent because it has the greatest tendency to be reduced.

Li+(1M) + e γ Li(s) E0 = - 3.05 V

Thus Li+ is the weakest oxidizing agent because it is the most difficult species to reduce.

Conversely, we say that F- is the weakest reducing agent and Li is the strongest reducing agent.

• The diagonal rule: any species on the left of a given half-cell reaction will react spontaneously with a species that appears on the right of any half-cell located below.

• The half-cell reactions are reversible.

• The sign of E0 changes when the reaction is reversed.

• Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 because electrode potentials are intensive properties.

I2(s) + 2e γ 2I-(1 M) E0 = 0.53 V

2I2(s) + 4e γ 4I-(1 M) E0 = 0.53 V

• E0cell > 0 for spontaneous reaction.

Example 19.2, 19.3.

Spontaneity of Redox Reactions:

Electrical energy = volts * coulombs = joules

Coulombs (the total electrical charge) is determined by the number of moles of electrons (n) that pass through the circuit. By definition

Total charge = nF F is faraday constant, is the electrical charge contained in 1 mole of electrons.

1 faraday is equivalent to 96485.3 coulombs, or 96500 coulombs. Thus

1 F = 96500 C/mol because 1J = 1C * 1V

1 F = 96500 J/V. mol

Wmax = Wele = - nFEcell

Wmax = the maximum amount of work that can be done.

Wele = the electrical work.

ΔG = Wmax = - nFEcell

ΔGo = - nFEocell

Since ΔGo = - RTlnK

so - nFEocell = - RTlnK

When T = 298K, Eocell = (0.0257 V/n)lnK

or Eocell = (0.0592 V/n)logK

|Eocell |K |ΔGo |Reaction under standard-state conditions |

|Positive |> 1 |Negative |Spontaneous |

|0 |= 1 |0 |At equilibrium. |

|Negative |< 1 |Positive |Nonspontaneous, Reaction is spontaneous in the reverse direction. |

Example 19.4, 19.5.

The Effect of Concentration on Cell Emf:

The Nernst equation:

Consider a redox reaction of the type:

aA + bB γ cC + dD

ΔG = ΔGo + RTlnQ

ΔG = - nFE and ΔGo = - nFEo

- nFE = - nFEo + RTlnQ E = Eo – RTlnQ/nF.

At 298K E = Eo – (0.0257 V/n)lnQ or E = Eo – (0.0592 V/n)logQ.

***For the reaction Zn(s) + Cu2+(aq) γ Cu(s) + Zn2+(aq)

E = 1.10 V – (0.0257 V/n)ln[Zn2+]/[Cu2+]

Example 19.6, 19.7.

In order to make the reaction spontaneous, we set E equal zero, and we can find the value of K, by which we can adjust the reaction (by changing the concentration of ion solution).

Concentration Cells:

Because electrode potential depends on the ion concentration, it is possible to construct a cell from two half-cells composed of the same material but differing in ion concentration. Such a cell is called a concentration cell.

Consider the following cell diagram:

Zn(s) / Zn2+(0.1M) // Zn2+(1M) / Zn(s)

The half-reactions are

Oxidation: Zn(s) γ Zn2+(0.1M) + 2e

Reduction: Zn2+(1M) + 2e γ Zn(s)

------------------------------------

Overall: Zn2+(1M) γ Zn2+(0.1M)

The emf of the cell is

E = Eo – (0.0257 V/n)ln[Zn2+]dil / [ Zn2+]conc

Eo = zero (the same electrode and the same type of ions are involved), so

E = 0 – (0.0257 V/n)ln 0.1 / 1.0 = 0.0296 V

The emf of concentration cells is usually small and decrease continually during the operation of the cell as the concentration in the two compartments approach each other. When the concentrations of the ions in the two compartments are the same, E becomes zero, and no further change occurs.

Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

An electrolytic cell is an apparatus for carrying out electrolysis.

• The external EMF is provided by a source of direct current (sometimes another voltaic cell, or battery)

• This external current acts as an electron pump to push electrons onto one electrode (and drive a reduction reaction) and to withdraw electrons from the other electrode (and drive an oxidation reaction)

Electrolysis of Molten Sodium Chloride:

Anode (oxidation): 2Cl-(l) γ Cl2 + 2e

Cathode (reduction): 2Na+(l) + 2e γ 2Na(l)

---------------------------------

Overall: 2Na+(l) + 2Cl-(l) γ 2Na(l) + Cl2(g)

Electrolysis of Water:

H2SO4 is used because there are a sufficient number of ions to conduct electricity.

The overall reaction is given by:

Anode (oxidation): 2H2O(l) γ O2(g) + 4H+(aq) + 4e

Cathode (reduction): 4H+(aq) + 4e γ 2H2(g)

---------------------------------

Overall: 2H2O(l) γ O2(g) + 2H2(g)

Electrolysis of an Aqueous Sodium Chloride Solutions:

With aqueous solutions of salts it must be considered that the electrolysis might end up driving the reduction or oxidation of H2O(l)

**H2O(l) can be oxidized to produce O2(g) **H2O(l) can be reduced to produce H2(g)

• The oxidation reactions that occur at the anode are:

2Cl-(aq) γ Cl2(g) + 2e E0red = 1.36 V

2H2O(l) γ 4H+(aq) + O2(g) + 4e E0red = 1.23 V

The values do suggest that H2O should be preferentially oxidized at the anode. However, by experiment we find that the gas liberated at the anode is Cl2, not O2. In studying electrolysis process, we sometimes find that the voltage required for a reaction is considerably higher than the electrode potential indicates. The overvoltage is the the difference between the electrode potential and the actual voltage required to cause electrolysis.

• The reduction reactions that occur at the cathode are:

2H2O(l) + 2e γ H2(g) + 2OH-(aq) E0red = -0.83 V

Na+(aq) + e γ Na(s) E0red = -2.71 V

When comparing any two reduction reactions the reaction that is favored is the one with the more positive reduction potential. In this case, the reduction of H2O(l) is the favored reduction reaction

Thus, the half cell-cell reactions in electrolysis of an aqueous solution of NaCl are:

Cathode (reduction) 2H2O(l) + 2e γ H2(g) + 2OH-(aq)

Anode (oxidation) 2Cl-(aq) γ Cl2(g) + 2e

Overall: 2H2O(l) + 2Cl-(aq) γ Cl2(g) + H2(g) + 2OH-(aq)

The electrolysis of an aqueous solution of NaCl produces chlorine gas, hydrogen gas and hydroxide ion. The useful by-product NaOH can be obtained by evaporating the aqueous solution at the end of electrolysis.

Example 19.8.

Quantitative Aspects of Electrolysis:

It takes 2 moles of electrons to reduce 1 mole of Mg2+ ions and 3 moles of electrons to reduce I mole of Al3+ ions:

Mg2+(aq) + 2e γ Mg(s)

Al3+(aq) + 3e γ Al(s)

Charge (C) = current (A) x time (s) or 1 C = 1 A x 1s

Example: For electrolysis of CaCl2, a current of 0.452 A is passed through the cell for 1.50 hr.

Anode (oxidation) 2Cl-(aq) γ Cl2(g) + 2e

Cathode (reduction) Ca2+(aq) + 2e γ Ca(s)

? C = 0.452 A * 1.50 hr * 3600 s * (1 C/1 A s) = 2.44 * 103 C

Because 1 mole of e = 96500 C and 2 mol of e are required to reduce I mole of Ca2+

? g Ca = 2.44 * 103 C * (1 mol e/96500 C) * (1 mol Ca/2 mol e) * (40 g Ca/1 mol Ca) = 0.507 g Ca.

? g Cl2 = 2.44 * 103 C * (1 mol e/96500 C) * (1 mol Cl2/2 mol e) * (70.9 g Cl2/1 mol Cl2) = 0.896 g Cl2

Example 19.9

Selected Problems: 1, 2, 11, 13, 16, 17, 21, 25, 26, 29-34, 45, 46-49, 51, 53, 57-60.

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