Chemistry: Redox Reactions and Electrochemistry

Question 1

Electrochemistry

Answer 1

The study of the relationships between chemical reactions and electrical energy.

Question 2

Electrochemical Reactions

Answer 2

Include spontaneous reactions that produce electrical energy and nonspontaneous reactions that use electrical energy to produce a chemical change. Both types always involve a transfer of electrons with conservation of charge and mass.

Question 3

Oxidation and Reduction

Answer 3

The law of conservation of charge states that an electrical charge can be neither created nor destroyed. Thus, an isolated loss or gain of electrons cannot occur; oxidation (loss of electrons) and reduction (gain of electrons) must occur simultaneously, resulting in an electron transfer called a redox reaction.

Question 4

Oxidizing Agent

Answer 4

Causes another atom in a redox reaction to undergo oxidation, and is itself reduced.

Question 5

Reducing Agent

Answer 5

Causes the other atom to be reduced, and is itself oxidized.

Question 6

Oxidation Numbers

Answer 6

Assigned to atoms to keep track of the redistribution of electrons during a chemical reaction. From the oxidation numbers of the reactants and products, it’s possible to determine how many electrons are gained or lost by each atom. The oxidation number is specifically the number of charges an atom would have in a molecule if electrons were completely transferred in the direction that is indicated by the difference in electronegativity. Along the same lines, an element is said to be oxidized if its oxidation number is increased in a given reaction. An element is said to be reduced if the oxidation number of the element decreases in a given reaction.

Question 7

Assigning Oxidation Numbers: Free Element

Answer 7

The oxidation number of a free element (an element in its elemental state) is zero, irrespective of how complex the molecule is.

Question 8

Assigning Oxidation Numbers: Monoatomic Ion

Answer 8

The oxidation number for a monoatomic ion is equal to the charge of the ion.

For example, the oxidation numbers for Na+, Cu+2, and Fe+3 are +1, +2, and +3.

Question 9

Assigning Oxidation Numbers: Group IA & IIA Elements

Answer 9

The oxidation number of each Group IA element in a compound is +1.

The oxidation number of each Group IIA element in a compound is +2.

Question 10

Assigning Oxidation Numbers: Group VIIA Elements

Answer 10

The oxidation number of each Group VIIA element in a compound is -1, except when combined with an element of higher electronegativity.

For example, in HCl, the oxidation number of Cl is -1; in HOCl however, the oxidation number of Cl is +1.

Question 11

Assigning Oxidation Numbers: Hydrogen

Answer 11

The oxidation number of hydrogen is +1.

However, the oxidation number of hydrogen is -1 in compounds with less electronegative elements than hydrogen (Groups IA and IIA).

Examples include NaH and CaH2. The more common oxidation number for H is +1.

Question 12

Assigning Oxidation Numbers: Oxygen

Answer 12

In most compounds, the oxidation number of oxygen is -2.

This is not the case, however, in molecules such as OF₂. Here, because F is more electronegative than O, the oxidation number of oxygen is +2.

Also, in peroxides such as BaO₂, the oxidation number of O is -1 instead of -2 because of the structure of the peroxide ion, [O–O]-2. (Note that Ba, a group IIA element, cannot be a +4 cation.)

Question 13

Assigning Oxidation Numbers: Neutral Compound

Answer 13

The sum of the oxidation numbers of all atoms present in a neutral compound is zero.

Question 14

Assigning Oxidation Numbers: Polyatomic Ions

Answer 14

The sum of the oxidation numbers of the atoms present in a polyatomic ion is equal to the charge of the ion.

Thus, for SO₄^-2, the sum of the oxidation numbers must be -2.

Question 15

Assigning Oxidation Numbers: Flourine

Answer 15

Fluorine has an oxidation number of -1 in all compounds because it has the highest electronegativity of all elements.

Question 16

Assigning Oxidation Numbers: Metallic & Nonmetallic Elements

Answer 16

Metallic elements have only positive oxidation numbers; however, nonmetallic elements may have a positive or negative oxidation number.

Question 17

Balancing Redox Reactions

Answer 17

By assigning oxidation numbers to the reactants and products, one can determine how many moles of each species are required for conservation of charge and mass, which is necessary to balance the equation.

To balance a redox reaction, both the net charge and the number of atoms must be equal on both sides of the equation.

The most common method for balancing redox reactions is the half-reaction method, or the ion-electron method, in which the equation is separated into two-half reactions–the oxidation and reduction part. Each half-reaction is balanced separately, and they are then added to give a balanced overall reaction.

Question 18

Half-Reaction (Ion-Electron) Method: MnO₄^- + I^- –> I₂ + Mn⁺²

Answer 18

1) Separate the two half-reactions.
I^- –> I₂
MnO₄^- –> Mn⁺²

2) Balance the atoms of each half-reaction. First, balance all atoms except H and O. Next, in an acidic solution, add H₂O to balance the O atoms and then add H⁺ to balance the H atoms. (In a basic solution, use OH^- and H₂O to balance the Os and Hs.)
To balance the iodine atoms, place a coefficient of two before the I^- ion.
2 I^- –> I₂
For the permanganate half-reaction, Mn is already balanced. Next, balance the oxygens by adding 4 H₂O to the right side.
MnO₄^- –> Mn⁺² + 4 H₂O
Finally, add H⁺ to the left side to balance the 4 H₂Os.
MnO₄^- + 8 H⁺ –> Mn⁺² + 4 H2O

3) Balance the charges of each half-reaction. The reduction half-reaction must consume the same number of electrons as are supplied by the oxidation half. For the oxidation reaction, add 2 electrons to the right side of the reaction.
2 I^- –> I₂ + 2 e^-
For the reduction reaction, the charge of +2 must exist on both sides. Add 5 electrons to the left side of reaction.
5 e^- + 8 H⁺ + MnO₄^- –> Mn⁺² + 4 H₂O
Next, both half-reactions must have the same number of electrons so that they will cancel. Multiply the oxidation half by 5 and the reduction half by 2.
5(2 I^- –> I₂ + 2 e^-)
2(5 e^- + 8 H⁺ + MnO₄^- –> Mn⁺² + 4 H₂O)

4) Add the half-reactions.
10 I^- –> 5 I₂ + 10 e^-
16 H⁺ + 2 MnO₄^- + 10 e^- –> 2 Mn⁺² + 8 H₂O
The final equation is:
10 I^- + 16 H⁺ + 2 MnO₄^- + 10 e- –> 5 I₂ + 10 e^- + 2 Mn⁺² + 8 H₂O
To get the overall equation, cancel out the electrons and any H₂Os, H⁺s, or OH^-s that appear on both sides of the equation.
10 I- + 16 H+ + 2 MnO4- –> 5 I2 + 2 Mn+2 + 8 H2O

5) Finally, confirm that mass and charge are balanced. There’s a 4+ net charge on each side of the reaction equation, and atoms are stoichiometrically balanced.

Question 19

Combination Reactions

Answer 19

Type of redox reaction. These types occur with one or more free elements. Example: N2(g) + 3 H2(g) –> 2 NH3 (g) Redox #’s: 0, 0, -3 +1

Question 20

Decomposition Reactions

Answer 20

Type of redox reaction. These lead to the production of one or more free elements. Example: 2 H2O(l) –> H2(g) + O2(g) Redox #’s: +1 -2, 0, 0

Question 21

Displacement Reactions

Answer 21

Type of redox reaction. An atom or an ion of one element is dispalced from a given compound by an atom from a totally different element. Example: 2 Na(s) + 2 H2O(l) –> 2 NaOH + H2(g) Redox #’s: 0, +1, +1 +1, 0

Question 22

Electrochemical Cells

Answer 22

Contained systems in which a redox reaction occurs. Two types are galvanic (voltaic) cells and electrolytic cells. Both contain electrodes at which oxidation and reduction occur. For all electrochemical cells, the electrode at which oxidation occurs is called the anode, and the electrode where reduction occurs is called the cathode.

Question 23

Galvanic (Voltaic) Cells

Answer 23

Electrochemical cell. Spontaneous reactions, with a negative deltaG. Galvanic cell reactions supply energy and are used to do work.

This energy can be harnessed by placing the oxidation and reduction half-reactions in separate containers called half-cells. The half-cells are then connected by an apparatus that allows for the flow of electrons.

Question 24

Daniel Cell

Answer 24

A Galvanic cell. In it, a zinc bar is placed in an aqueous ZnSO4 solution, and a copper bar is placed in aqueous CuSO4 solution. The anode of this cell is the zinc bar where Zn(s) is oxidized to Zn+2(aq). The cathode is the copper bar, and it’s the site of the reduction of Cu+2(aq) to Cu(s). The half-cell reactions are written as follows. Zn(s) –> Zn+2(aq) + 2 e- –> (anode) Cu+2(aq) + 2e- –> Cu(s) –> (cathode) If the two half-cells weren’t separated, the Cu+2 ions would react directly with the zinc bar, and no useful electrical work would be obtained. To complete the circuit, the two solutions must be connected. Without connection,t he electrons from the zinc oxidation half-reaction wouldn’t be able to get to the copper ions, thus a wire (or other conductor) is necessary. If only a wire were provided for this electron flow, however, the reaction would soon cease anyway because an excess negative charge would build up in the solution surrounding the cathode and an excess positive charge would build up in the solution surrounding the anode. This charge gradient is dissipated by the presence of a salt bridge, which permits the exchange of cations and anions. The salt bridge contains an inert electrolyte, usually KCl or NH4NO3, whose ions will not react with the electrodes or with the ions in solution. At the same time, the anions from the salt bridge (e.g., Cl-) diffuse from the salt bridge of the Daniel cell into the ZnSO4 solution to balance out the charge of the newly created Zn+2 ions, and the cations of the salt bridge (e.g., K+) flow into the CuSO4 solution to balance out the charge of the SO4-2 ions left in solution where the Cu+2 ions deposit as copper metal. During the course of the reaction, electrons flow from the zinc bar (anode) through the wire and the voltmeter, toward the copper bar (cathode). The anions (Cl-) flow externally (via the salt bridge) into the ZnSO4, and the cations (K+) flow intot he CuSO4. This flow depletes the salt bridge and, along with the finite quantity of Cu+2 in the solution, accounts for the relatively short lifetime of the cell.

Question 25

Cell Diagram

Answer 25

A shorthand notation representing the reactions in an electrochemical cell. A cell diagram for the Daniel cell is as follows

Zn(s) | Zn⁺² (xM SO₄^-2) || Cu⁺² (yM SO₄^-2) | Cu(s)

The following rules are used in constructing a cell diagram:

• The reactants and products are always listed from left to right in the form: anode | anode solution || cathode solution | cathode
• A single vertical line indicates a phase boundary.
• A double vertical line indicates the presence of a salt bridge or some other type of barrier.

Question 26

Electrolytic Cells

Answer 26

Electrochemical cell. Nonspontaneous reactions. Positive deltaG. In electrolysis, electrical energy is required to induce reaction. The oxidation and reduction half reactions are usually placed in one container.

Michael Faraday was first to define certain quantitative principles governing behavior of electrolytic cells. He theorized that the amount of chemical change induced in an electrolytic cell is directly proportional to the number of moles of electrons that are exchanged during a redox reaction. The number of moles exchanged can be determined from the balanced half-reaction.

In general, for a reaction that involves the transfer of n electrons per atom:

Mⁿ⁺ + n e^- –> M(s)

One mole of M(s) will be produced if n moles of electrons are supplied.

The number of moles needed to produce a certain amount of M(s) can now be related to a measurable electrical property. One electron carries a charge of 1.6 x 10^-19 coulombs (C). The charge carried by one mole of electrons can be calculated by multiplying this number by Avogadro’s number:

(1.6 x 10^-19)(6.022 x 10²³) = 96,487 C/mol e^-

This number is called Faraday’s constant, and one Faraday (F) is equivalent to the amount of charge contained in one mole of electrions (1 F = 96,487 coulombs, or J/V).

Question 27

Electrode Charge Designations

Answer 27

The anode of an electrolytic cell is considered positive, since it’s attached to the positive pole of the battery and so attracts anions from the solution. The anode of a voltaic cell is considered negative because the spontaneous oxidation reaction that takes place the voltaic cell’s anode is the original source of that cell’s negative charge (i.e., source of electrons). In spire of this difference in designating charge, oxidation takes place at the anode in both types of cells, and electrons always flow through the wire from the anode to the cathode.

In a voltaic cell, change is spontaneously created as electrons are released by the oxidizing species at the anode; since this is the source of electrons, the anode of a voltaic cell is considered the negative electrode. In an electrolytic cell, the electrons are forced through the cathode, where they encounter the species that’s to be reduced. Here it’s the cathode that’s providing electrons, and thus the cathode of an electrolytic cell is considered the negative electrode. Alternatively, one can think of the cathode as the electrode attached to the negative pole of the battery used for the electrolyte.

In either case, a simple mnemonic is that the CAThode attracts the CATions. In the Daniel Cell for example, the electrons created at the anode as the zinc oxidizes travel through the wire to the copper half-cell, where they attract copper(II) cations to the cathode.

Question 28

Electrphoresis

Answer 28

A technique used to separate amino acids based on their isoelectronic points, or pI’s. The positively charged amino acids (i.e., those are protonated at the pH of the solution) will migrate toward the cathode; negatively charged amino acids (i.e., those that are deprotonated at the solution pH) migrate instead toward the anode.

Question 29

Reduction Potentials

Answer 29

Sometimes when electrolysis is carried out in an aqueous solution, water rather than the solute is oxidized or reduced. For example, if an aqueous solution of NaCl is electrolyzed, water may be reduced at the cathode to produce H2(g) and OH- ions, instead of Na+ being reduced to Na(s), as occurs in the absence of water. The species in a reaction that will be oxidized or reduced can be determined from the reduction potential of each species, defined as the tendency of a species to acquire electrons and be reduced. Each species has its own intrinsic reduction potential; the more positive the potential, the greater the species’ tendency to be reduced.

A reduction potential is measured in volts (V) and is defined relative to the standard hydrogen electrode (SHE), which is arbitrarily given a potential of 0.00 volts. Standard reduction potential (E*) is measured under standard conditions: 25 degrees Celsius, a 1 M concentration for each ion participating in the reaction, a partial pressure of 1 atm for each gas that’s part of the reaction, and metals in their pure state. The relative reactivities of different half-cells can be compared to predict the direction of electron flow. A higher E* means a greater tendency for reduction to occur, while a lower E* means a greater tendency for oxidation to occur.

Question 30

Standard Electromotive Force

Answer 30

Standard reduction potentials are also used to calculate the standard electromotive force (EMF or E*cell) of a reaction, the difference in potential between two half-cells. The EMF of a reaction is determined by adding the standard reduction potential of the reduced species and the standard oxidation potential of the oxidized species. When adding standard potentials, don’t multiply by the number of moles oxidized or reduced.

EMF = E*red + E*ox

The standard EMF of a galvanic cell is positive, while electrolytic cell negative.

Question 31

EMF and Gibbs Free ENergy

Answer 31

The thermodynamic criterion for determining the spontaneity of a reaction is deltaG, Gibbs free energy, the max amount of useful work produced by a chemical reaction. In an electrochemical cell, the work done is dependent on the number of coulombs and the energy available. Thus, deltaG and EMF are related as follows:

deltaG = -nFE_cell

where n is number of moles of electrons exchanged, F is Faraday’s constant, and E_cell is EMF of cell.

Note: Keep in mind that if Faraday’s constant is expressed in coulombs (J/V), then deltaG must be in J, not kJ.

If the reaction is under standard conditions, then the deltaF is standard Gibbs free energy and Ecell is standard cell potential. deltaG = -nFE*_cell

Question 32

Effect of Concentration on EMF

Answer 32

Concentration does have an effect on the EMF of a cell: EMF varies with the changing concentrations of the species involved. It can also be determined by use of the Nernst equation:

E_cell = E*_cell - (RT / nF)(ln Q)

Q is the reaction quotient for a given reaction. For example, in:

a A + b B –> c C + d D

Q = ([C}^c x [D]^d) / ([A]^a x [B]^b)

The EMF of a cell can be measured by a voltmeter. A potentiometer is a kind of voltmeter that draws no current and gives a more accurate reading of the difference in potential between two electrodes.

Question 33

EMF and the Equilibrium Constant (Keq)

Answer 33

For reactions in solution deltaG* can be determined in another manner, as follows:

deltaG* = -RTlnK_eq

where R is gas constant 8.314 J/(K x mol), T is temperature in K, and Keq is equilibrium constant for reaction.

If you combine this equation with the standard Gibbs free energy one, then:

deltaG* = -nFE*_cell = -RTlnK_eq

or simply

nFE*_cell = RTlnK_eq

If the values for n, T, and Keq are known, then the E*cell for the redox reaction can be readily calculated.