Ch.18

Question 1

Electrochemistry

Answer 1

is the study of the interchange of chemical energy and electrical energy
• This involves reduction-oxidation processes, and may involve generation of electrical energy by chemical reactions, or the use of electrical energy to drive chemical reactions

Question 2

Oxidation and reduction

Answer 2

Oxidation is a loss of electrons (increases oxidation number)
• Reduction is a gain of electrons (reduces oxidation number)
• An Oxidising Agent oxidises something else by taking some of its electrons, and is itself reduced in the process
• A Reducing Agent reduces something else by giving away electrons, and is itself oxidised in
the process

Question 3

Oxidation State Rules Review – Textbook rules

Answer 3

1. The oxidation state of an atom in an element (Na(s), O2(g), O3(g), Hg(l) etc.) is zero.
2. The oxidation state of a monoatomic ion is the same as its charge. Na+ is +1, Ba2+ is +2, Cl- is -1, etc.
3. The sum of all oxidation states
   a) Is 0 in a neutral molecule
   b) Is equal to the charge of a polyatomic ion
4. Metals have positive oxidations states
   a) Group 1 metals are always +1
   b) Group 2 metals are always +2
5. Hydrogen is in the +1 oxidation state in most compounds
6. Nonmetals have negative oxidation states in most compounds
   a. Fluorine is always -1
   b. Other halogens are usually -1
   c. Oxygen is usually -2
   d. Other group 16 elements are usually -2
   e. Group 15 elements are usually -3
7. Lower numbered rules take precedence over higher numbered rules if there is a conflict
   a) Conflicts will occasionally happen – note the extensive use of the word “usually”
   b) You will find rule 3 particularly useful

Question 4

Oxidation State Rules Review – Bonding-based rules

Answer 4

The first three rules are the same as the textbook version

4. For covalent molecules and polyatomic ions:
   a) Draw a Lewis structure and consider each atom individually
   b) Ignore any homonuclear covalent bonds
   c) For each heteronuclear covalent bond, the more electronegative element has its oxidation state is decreased by one and the less electronegative element has its oxidation state increased by one
   d) Add any formal charges to determine the final oxidation state

Question 5

Balancing Redox Reactions

Answer 5

We could begin to balance this reaction by looking at oxidation states

We need to account for the possibility that water, or components of water like H+ or OH-, can be reactants or products whenever reactions occur in aqueous solution

To do this we will use the half-reaction method of balancing

Question 6

The Half Reaction Method

Answer 6

We will separate the reaction into two half-reactions: one involving the oxidation and the other one reduction:

Add the half reactions together such that the electrons on each side of the half-reactions exactly cancel

Question 7

Acidic and Basic Media

Answer 7

Some Reactions involve reduction of oxygen-containing polyatomic ions, producing a free metal ion or a polyatomic ion with fewer oxygen atoms

• The freed oxygen atoms remain in the -2 oxidation state in aqueous solution
– Reactions that involve reduction of molecular oxygen also generate O2-

• In acidic media, O2- combines with H+ to form H2O

• In basic media, O2- reacts with H2O to form 2 OH-

• Reaction products may also incorporate O2- from water
– H+ is released in acidic media
– OH- is converted to H2O in basic media

Question 8

Acidic Media

Answer 8

1. Write the equations for the oxidation and reduction half reactions
2. For each half-reaction
   – Balance all the elements except hydrogen and oxygen
   – Balance oxygen using H2O
   – Balance hydrogen using H+
   – Balance the charge using electrons
3. If necessary, multiply one or both balanced half-reactions by integers to equalize the number of electrons transferred in the two half-reactions
4. Add the half reactions, and cancel identical species
5. Check to be sure that the elements and charges balance

Question 9

Basic Media

Answer 9

H+ in not available in basic solution, so you must use H2O as a source of hydrogen to balance half-reactions

– This generates OH-, which will appear on the other side of your reaction Use OH- instead of H2O as your source of oxygen

Remember that polyatomic ions will be ionized, not protonated

Question 10

Basic Media – alternative method

Answer 10

1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present
2. Add OH- ions equal to the number of H+ ions to both sides of the equation
3. Form H2O on the side containing both H+ and OH- ions, and eliminate the number of H2O molecules that appear on both sides of the equation
4. Check that the elements and charges balance

Question 11

Galvanic/Voltaic Cells

Answer 11

Generate electricity through spontaneous chemical reactions

• In order to make a voltaic cell we need two half reactions, the oxidation half cell and reduction half cell

At the anode, oxidation

At the cathode, reduction

Question 12

What is a half cell?

Answer 12

• A half-cell is a redox couple, two chemical species or collections of chemical species that can be interconverted by the gain or loss of electrons
• In this example, Zn(s) and Zn2+(aq) form a redox couple
– Zn(s) is a reducing agent and Zn2+(aq) is an oxidising agent
• Adding electrons will reduce Zn2+(aq) to Zn(s), removing electrons will oxidise Zn(s) to Zn2+(aq)
• For current to flow, two half cells must be connected by an electrical conductor as well as an ion conductor
– Charge transfer between half cells as a result of electron flow must be balanced by movement of ions between half cells

Question 13

Why do electrochemical cells work?

Answer 13

Consider a cell in which no wire is connecting the two half-cells - no circuit, so no electron flow
• Reduction of ions and/or oxidation of metal can occur at either metal electrode
• Reduction consumes some of the delocalized electrons in the metal, resulting in positive charge, while oxidation releases electrons and increases negative charge

Example

_{Copper has a greater affinity for electrons than Zinc (Cu2+ is a stronger oxidising agent than Zn2+), so the copper electrode ends up slightly positive relative to the zinc electrode}

Connection the zinc and copper electrodes allows electrons to flow from Zn to Cu
• Removing electrons from the zinc promotes oxidation of Zn0 to Zn2+
• Adding electrons to the Cu promotes reduction of Cu2+ to Cu0
• Movement of ions through the salt bridge balances the charge created by movement of electrons through the wire
• Current will flow until the cell reaches equilibrium</sub>

Question 14

Anodes and Cathodes

Answer 14

• The anode is the electrode by which electrons are liberated by an oxidation process and leave the voltaic cell to travel through the electrical circuit
– Anions are attracted to the anode and cations are repelled from the anode when the cell is in operation

• Electrons from the circuit re-enter via the cathode, where they become bound to a new chemical species in a reduction reaction
– Cations are attracted to the cathode and anions are repelled from the cathode when the cell is in operation

Question 15

Cell Potential

Answer 15

The slight excess of electrons at the anode relative to the cathode is the driving force for the electrical current
– This difference in electrical potential between two electrodes is the cell potential (Ecell) or electromotive force (emf)
– Measured in units of volts (V)

• Cell voltage (potential) is a measure of how much electrical work each can be done by each electron travelling through the circuit

• For a spontaneous reaction, there is a positive cell potential
– The electrical potential at the cathode is more positive than the electrical potential at the anode
– The larger the potential difference, the farther the cell is from equilibrium and the more electrical work than can be done by each electron that flows through the circuit
• Cell voltage is zero at equilibrium
– As current flows, reactant concentrations decrease and product concentrations increase
– The reaction quotient increases, approaching the equilibrium constant, and the driving force behind the reaction decreases
– cell voltage drops until equilibrium is reached

Question 16

Standard Cell Potential (E°cell)

Answer 16

Standard cell potential is the potential observed under standard conditions
– Concentrations of 1 mol/L and pressures of 1 bar
– electrical potential is measured in volts (V)

Question 17

Line Notation

Answer 17

• Shorthand description of a voltaic cell
• Electrode | electrolyte || electrolyte | electrode
• Oxidation half-cell (anode) on the left, reduction half-cell (cathode) on the right
• Single | = phase interface
• When multiple electrolytes are in the same phase, a comma is used rather than |
• Double line || = salt bridge or ion-permeable membrane
• Note that the electrodes may or may not participate in the reaction
– When both the reducing and oxidizing agents in a half-cell are in solution, an inert electrode is used

Question 18

Electrical Units

Answer 18

volt (V), the unit of electromotive force or electrical potential
– Also called voltage, after the unit itself, and may be given the symbol V or E

The number of electrons (Q) could be measured in moles, but is more commonly measured in
coulombs (C)
– 1 C = 6.242 x 1018 electrons
Current (I) is the number of electrons that flow through the system per second – Measured in units of Amperes (A), also called amps
– 1 A of current = 1 Coulomb (C) of charge flowing by each second
– 1 A = 1 C/s = 6.242 x 1018 electrons per second

• Electrical work is the product of the potential and number of electrons that flow through the circuit
– Work = (number of electrons) x (potential difference) – Like other forms of work, it is measures in joules
– Joules = Volts x Coulombs
• Power is the rate at which work is done
– It is the product of voltage and current – Measured in watts
– Watts = Volts x Amperes

Question 19

Half-Cell Potentials

Answer 19

• Any electrochemical cell requires both an anode and a cathode to operate
– The cell potential is determined by the overall reaction
• Because a voltaic cell can be assembled from any pair of half-cells, it is useful to isolate the
contributions of each half-reaction to the overall potential
– The overall cell potential is then just the sum of the potentials for the oxidation and reduction reactions
• There is no absolute potential reference suitable for measurements of the required precision, so half cell potentials are measured relative to a standard reference, which is arbitrarily assigned a potential of zero (E°reference=0.00 V)

Question 20

Measuring Half-Cell Potentials

Answer 20

The standard reference cell is a standard hydrogen electrode (SHE)
– 1 bar H2 (g) bubbled through 1M H+ solution
– Pt electrode
A cell is constructed using a SHE connected (both electrically and ionically) to a test half-cell

The potential (both the magnitude and direction of electron flow) is then measured

Question 21

Reduction and Oxidation Potentials

Answer 21

The direction of current flow depends on whether the test cell is acting as an anode (the test reducing agent is stronger than H2) or a cathode (the test oxidising agent is stronger than H+)

• If the SHE is the anode and the test cell is the cathode, the oxidising agent in the test cell is being reduced (reduction occurs at the cathode)
– The resulting cell potential is thus the reduction potential of the test redox couple

• If the SHE is the cathode and the test cell is the anode, the reducing agent in the test cell is
being oxidised (oxidation occurs at the anode)
– The resulting cell potential is thus the oxidation potential of the test redox couple

• The oxidation and reduction potentials of a redox couple are related
– One is the negative of the other
• By convention, all of the reactions in tables of half-reactions are written in the reduction
direction, and all of the potentials listed are reduction potentials

Xm+ + n e- → X(m-n)+

• Any redox couple which acts as an anode relative to the SHE thus has a negative reduction potential

• A large negative reduction potential indicates a good reducing agent, while a large positive reduction potential indicates a good oxidising agent
– Remember that the actual reducing agent is the thing on the product side of a reduction half-reaction

Question 22

Using Reduction Potentials

Answer 22

• In a complete cell, the reaction at the cathode is occurring in the reduction direction, but the reaction at the anode is occurring in the oxidation direction
• The overall cell potential is thus the reduction potential for the cathode reaction minus the reduction potential for the anode reaction:
• E°cell = E°cathode(reduction) - E°anode(oxidation)

Overall Reactions and Cell Voltages:

• The E°half-cell refers to the half-reaction written as a reduction reaction
• Remember that the anode reaction is going in the opposite direction

Question 23

Using Tables of Reduction Potentials

Answer 23

• In a standard reduction potential table, all of the half reactions are written as reductions (eg. 2 H+(aq) + 2 e- → H2(g))
– The oxidising agent is on the left, and the reducing agent is on the right
• Half reactions are listed in order of decreasing reduction potentials – The most favorable reduction reactions are at the top
– The best oxidising agents are at the top left
– The least favorable reduction reactions are at the bottom
– The best reducing agents are at the bottom right
• Reactions will be spontaneous (E° > 0.00 V) when the oxidizing agent is located above the
reducing agent in the table

Question 24

Predicting Spontaneous Redox Reactions

Answer 24

• What is the cell voltage?
• Is this reaction spontaneous?
• Identify the half reactions and locate them on the table
• identify the reducing and oxidizing agents
• The cell potential is negative, therefore the reaction is non-spontanous as written
• The cell potential is positive, therefore the reaction is spontanous as written
• Note that the stoichiometry irrelevant to the cell voltage

Question 25

Dissolving Metals in Acid

Answer 25

Any metal with a negative reduction potential can potentially be dissolved in acid according to the reaction
M(s) + n H+(aq) → Mn+(aq) + 1⁄2n H2(g)
– The metal is the anode in this reaction, so the cell potential is positive if E°metal is negative
• Metals with positive reduction potentials will not react spontaneously with H+(aq), but will dissolve in acids that are also strong oxidising agents
• The most common of these is nitric acid, HNO3

• Nitric acid can readily dissolve metals like copper and silver
• Metals with reduction potentials greater than 0.96 V (like gold and platinum) do not dissolve in nitric acid, but can be dissolved in “aqua regia”, a mixture of nitric and hydrochloric acids
– This works due to formation of a complex ion that reduces [Mn+(aq)] and drives the reaction forward

Question 26

Cell Potential & Gibbs Energy

Answer 26

Spontaneous
-∆G°
+ E°cell
K>1

Nonspontaneous
+∆G°
- E°cell
K<1

Question 27

Relationship between DG° and E°cell

Answer 27

Recall that electrical work is the product of the amount of charge transferred and the electrical potential

W = nFE°cell

W = The maximum amount of work that can be done by a voltaic cell
n = # of moles of e- transferred
F = Faraday’s Constant, is the charge in coulombs of 1 mole of electrons
• F = 96485 C/mol, nF = charge transferred in coulombs

∆G°=-nFE°cell

Question 28

Cell Potential and K

Answer 28

-RTln(K) = -nFE° = = DG°

E° = RTIn(K) / nF Rearrange for E°cell

E° = (0.02569 V/n)ln(K) Combine constants for 298K

E° = (0.05916 VΤn)log(K) Convert to base 10 log

Unit derivation: RT/F = (J/mol•K)(K)/(C/mol) = Joules/Coulomb = Volts

Question 29

Relationship between DrG°, K, and E°cell

Answer 29

DrG° and K
DrG° = -RT In(K)

DrG° and E°cell
DrG° = -nFE°cell

K and E°cell
E°cell = (0.0257 V / n) In(K)

Question 30

Cell Potential and Concentration
And The Nernst Equation

Answer 30

We need to go back to look at ∆G expressions for non-standard reactions

Ecell = E°cell - (0.05916/n)log(Q)

Log(Q)(2.303) = in(Q)

Ecell = E°cell - (0.02569/n)log(Q)

Q closer to Eq than st con —> DG will get smaller and closer to 0
Q farther away from Eq than —> DG will get larger

If before Eq Q < K, than DG become more -, rxn more spon
If past Eq Q > K, DG become more +, rxn less spon

• When Q<1, Ecell>E°cell
• When Q>1, Ecell<E°cell

• When Q = K, (0.05916/n)log(Q) = E°cel and Ecell = 0

Question 31

The Nernst Equation

Answer 31

Ecell = E°cell - (0.02569/n)in(Q)

Ecell = E°cell - (0.05916/n)log(Q)

• When Q<1, Ecell>E°cell
• When Q>1, Ecell<E°cell
• When Q = K, (0.05916 V)log(Q) = E°cell and Ecell = 0 n
• Cell voltage is zero at equilibrium

Question 32

Concentration Cells

Answer 32

• Consider a cell in which both half-cells are identical
• Because the half-reactions are identical, E°cell = 0 (E°cathode – E°anode)
• The reaction is thus at equilibrium under standard conditions
• But what is the half-cells have different concentrations?
• Le Chatelier’s Principle suggests that reaction will occur to increase the lower concentration
and reduce the higher concentration
• For a typical M|Mn+ half cell, oxidation increases [Mn+] and reduction decreases [Mn+] – The less concentrated solution is the anode
– The more concentrated solution is the cathode

• At the anode of a M|Mn+ half cell, Mn+ is a product
• At the cathode, Mn+ is a reactant
• Q = [Mn+anode]/[Mn+cathode]
• Q < 1 when [Mn+anode] < [Mn+cathode]
• Ecell = E°cell - (0.05916 V)log(Q), so Ecell is positive when E° = 0 and Q < 1 n
• The cell potential of a concentration cell depends on the concentration ratio and the number of electrons transferred
• For a single electron reaction, every 10-fold difference in concentration results in a cell potential of 0.05916 V
• allows measurement of extremely low concentrations, as voltage depends linearly on log(Q)
– This can be useful for measuring very large or very small equilibrium constants, like Kf or Ksp

Question 33

Other Cells Under Non-Standard Conditions

Answer 33

• The Nernst equation can be used to calculate any cell voltage under non-standard conditions
• Ecell = E°cell - (0.05916 V/n)log(Q)
• One needs to know the reaction quotient, and the number of electrons transferred in the balanced reaction
– Ensure that the same coefficients are used to calculate both Q and n!
• If the half-reactions involve different numbers of electrons, a change in concentration of the
reaction with fewer electrons results in a larger change in cell voltage
– Less electrons in half-reaction = larger coefficient in balanced reaction and a larger exponent in Q

Question 34

Effect of Reaction Progress on Ecell

Answer 34

• What happens when a voltaic cell is operated to do electrical work.
• For a cell consisting of M|Mn+ half cells, the Mn+ concentration declines at the cathode,
while the Mn+ concentration increases at the anode
• This increases Q: Q = [Mn+anode]n [Mn+cathode]n
• Which decreases Ecell: Ecell = E°cell - (0.05916 V/n)log(Q)
• Eventually, Q = K and Ecell = 0, indication that the cell has come to equilibrium and can do no more work
• Behavior during discharge is affected by the states of the reactants and products
– think about how Q is calculated

Question 35

Electrolysis

Answer 35

is the process of using electrical energy to drive a chemical reaction in the non-spontaneous direction

• An electrolytic cell must be connected to a source of electrical energy in order to operate
• Electrolytic cells can be used to separate elements from their compounds
• Electrolysis can be used to recharge spent voltaic cells by reversing the reaction and regenerating the reactants from the products

Voltaic cell allows us to use free energy from a rxn to produce electricity
Electrolysis cell is opposite, use electricity to increase free energy

Force reduction and oxidation to occur. Cathode connected to negative on voltage source and anode to positive. Is nonspon direction.

Question 36

Electrolysis of Aqueous Solutions

Answer 36

• As we have seen, an electrolytic cell can be made by using an electrical power source with a voltage > Ecell to reverse the reaction in a voltaic cell
• However, electrolytic cells are not always constructed from a pair of half-cells
• Any combination of electrodes and mixtures of aqueous ions may be electrolysed
• We will need to be able to predict what reaction will take place when an electrical current is passed through the cell

• If we have a mixture of oxidising agents in solution, the order in which they are reduced when subjected to electrolysis is based on their reduction potentials

• The easiest cation to reduce (the one with the highest reduction potential) will be reduced first

• If we have a mixture of reducing agents in solution, the order in which they are oxidised when subjected to electrolysis is based on their reduction potentials

• One also needs to consider the possibility that the electrode itself may be oxidised

• Note that reduction potentials are given for the anode reactions, even though the reactions are written as oxidations
• Note that these reactions are not under standard conditions in neutral solution
• These potentials must be considered in addition to those of any solutes to determine what will be reduced first

Question 37

Electroplating

Answer 37

• Metal cations can be reduced to the 0 oxidation state by electrolysis
• The reduction reaction produces a coating of the metal on the cathode, which can be useful for decorative purposes and/or to provide corrosion resistance
• A solution containing a salt of the metal to be plated is used with the object to be plated as the cathode
• The anode is made of the plating metal, and dissolves to maintain the metal concentration
• E°cell is zero, (cell is at equilibrium) but some voltage is needed to make the reaction proceed in the desired direction

Question 38

Limitations of electroplating in aqueous solution:

Answer 38

• Cannot be used to plate electropositive metals
• If E° is less than -0.41 V (in neutral solution), hydrogen
will be produced instead of reducing the metal ions
• Using high pH to lower the reduction potential of hydrogen is usually not possible due to low solubility of most metal hydroxides
• Very electronegative metals (like Au) can be plated easily, but electrodes made of these metals will not dissolve, so the metal ions will need to be replenished by adding a salt instead of dissolving the electrode

Question 39

Electrolysis of Molten Salts

Answer 39

• Metals can be isolated and purified through the electrolysis of their molten (melted) salts
• In the electrolysis of pure molten salts, the anion is oxidized and the cation reduced
• There is no water present, so metals with strongly negative reduction potentials can be reduced
to free metals
• This is very useful for manufacture of electropositive metals
• Metals in use before the invention of electrolysis were either found “native” (gold, silver) or were produced by reduction of their compounds with organic materials like coal or charcoal (tin, lead, copper, iron)
• More electropositive metals like aluminum cannot be produced this way, as carbon is a not sufficiently strong reducing agent
– Before introduction of electrolysis for aluminum production, aluminum cost more than gold!

Question 40

Stoichiometry of Electrolysis

Answer 40

• Since we are providing an external current to our electrolytic cell we need to consider how much electrical energy is required to produce a given amount of metal
• We can determine the number of moles of electrons transferred by considering how long we provide the current for
• 1A = 1C/s
• F = 96485 C/mol
• One can quickly determine that large scale electrolysis requires a lot of electricity