Balancing Redox Reactions: Acid Conditions

Balance the following redox equation:

Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3

- (aq) (acidic)

Oxidation Reaction:Cr2O7

2- (aq) ----> Cr3+ (aq)Cr2O7

2- (aq) ----> 2Cr3+ + 7H2O Balance Cr and O

Cr2O72- (aq) + 14H+ ----> 2Cr3+ + 7H2O Balance H

Cr2O72- (aq) + 14H+ + 6e- ----> 2Cr3+ + 7H2O Balance charge

Reduction Reaction:HNO2 (aq) ---> NO3

- (aq)HNO2 (aq) + H2O ---> NO3

- (aq)HNO2 (aq) + H2O ---> NO3

- (aq) + 3H+

HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ + 2e-

Balancing Redox Reactions: Acid Conditions

Balance the following redox equation:

Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3

- (aq) (acidic)

Step 3: Multiply the half reaction with the lowest number of electrons by the stoichiometric coefficient to make the number of electrons in each half reaction the same.

Cr2O72- (aq) + 14H+ + 6e- ----> 2Cr3+ + 7H2O

HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ + 2e- Multiply by 3

Cr2O72- (aq) + 14H+ + 6e- + 3HNO2 (aq) + 3H2O --->

2Cr3+ + 7H2O + 3NO3- (aq) + 9H+ + 6e-

Balancing Redox Reactions: Acid Conditions

Cr2O72- (aq) + 14H+ + 6e- + 3HNO2 (aq) + 3H2O --->

2Cr3+ + 7H2O + 3NO3- (aq) + 9H+ + 6e-

Step 4: Cancel the compounds found on both side of the arrow

Cr2O72- (aq) + 5H+ + 3HNO2 (aq)--> 2Cr3+ + 4H2O + 3NO3

- (aq)

Step 5: Check work. Done!

Balancing Redox Reactions: Acid Conditions

MnO4- (aq) + Br- (aq) ---> MnO2 (aq) + BrO3

- (aq)

Step 1: Half reactions

Mn: +7 --> +4 Gain 3e-, Reduction

Br: -1 --> +5 Lose 6e-, Oxidation

Step 2: Balance half reactions

Oxidation:

Br- (aq) --> BrO3-

3H2O + Br- (aq) --> BrO3-

3H2O + Br- (aq) --> BrO3- + 6H+

3H2O + Br- (aq) --> BrO3- + 6H+ + 6e-

Reduction:

MnO4- --> MnO2

MnO4- --> MnO2 + 2H2O

MnO4- + 4H+ --> MnO2 + 2H2O

MnO4- + 4H+ + 3e- --> MnO2 + 2H2O

Balancing Redox Reactions: Acid Conditions

MnO4- (aq) + Br- (aq) ---> MnO2 (aq) + BrO3

- (aq)

Step 3: We have 6 electrons transferred in the oxidation reaction and 3 electrons in the reduction reactions. Multiply the reduction reaction by 2 and combine with the oxidation reaction.

2MnO4- + Br- + 3H2O + 8H+ + 6e- --> BrO3

- + 6H+ + 6e- + 2MnO2 + 4H2O

Step 4: Cancel molecules found on both sides of the reaction arrow.

2MnO4- + Br- + 2H+ --> BrO3

- + 2MnO2 + H2O

Step 5: Done!

Balancing Redox Reactions: Basic Conditions

Cr(OH)3 (s) + ClO3

- (aq) ---> CrO42- (aq) + Cl- (aq)

Step 1: Same as acidic conditions. Identify the 2 half reactions

Cr: +3 --> +6 Lose 3 electrons, Oxidation

Cl: +5 --> -1 Gain 6 electrons, Reduction

Step 2: Balance the half reactions

Oxidation reaction:

Cr(OH)3 (s) ---> CrO4

2-

Cr(OH)3 (s) + H2O ---> CrO4

2-

Cr(OH)3 (s) + H2O ---> CrO4

2- + 5H+

Cr(OH)3 (s) + H2O ---> CrO4

2- + 5H+ + 3e-

Balancing Redox Reactions: Basic Conditions

Cr(OH)3 (s) + ClO3

- (aq) ---> CrO42- (aq) + Cl- (aq)

Step 2 (cont’d): Balance the half reactions

Reduction reaction:

ClO3- (aq) ---> Cl-

ClO3- (aq) ---> Cl- + 3H2O

ClO3- (aq) + 6H+ ---> Cl- + 3H2O

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

Step 3: Multiply the oxidation reaction by 2 to get the same # of electrons and combine the equations

2Cr(OH)3 (s) + 2H2O + ClO3

- (aq) + 6H+ + 6e- --> Cl- + 3H2O + 2CrO4

2- + 10H+ + 6e-

Balancing Redox Reactions: Basic Conditions

2Cr(OH)3 (s) + 2H2O + ClO3

- (aq) + 6H+ + 6e- --> Cl- + 3H2O + 2CrO4

2- + 10H+ + 6e-

Step 4: Remove things on both sides of the reaction arrow

2Cr(OH)3 (s) + ClO3

- (aq) --> Cl- + H2O + 2CrO42- + 4H+

Step 5: Now for the tricky part. Count the number of protons and add the same number of hydroxide ions to BOTH sides.

2Cr(OH)3 (s) + ClO3

- (aq) + 4 OH- --> Cl- + H2O + 2CrO42- + 4H+ + 4 OH-

OR

2Cr(OH)3 (s) + ClO3

- (aq) + 4 OH- --> Cl- + 5H2O + 2CrO42-

12.3: The Structure of Galvanic Cells

A Galvanic Cell is a device in which electric current is produced by a spontaneous chemical reaction

A battery is a series of galvanic cells connected in series so that the voltage of the battery is the sum of the voltages of the individual cells

•Voltage is the ability to push electric current through a circuit

12.3: The Structure of Galvanic Cells

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

Electrons from the zinc are transferred to the copper

If we put a wire between the two conductive electrodes (one of which will be Zn), we could capture the flow of electronsThe solution between the two electrodes is an ELECTROLYTE (perhaps copper nitrate?)

12.3: The Structure of Galvanic Cells

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

The ANODE is the electrode where oxidation takes place

The CATHODE is the electrode where reduction takes place

Electrons are PUSHED from the ANODE to the CATHODE

Galvanic cells have the cathode marked with a ‘+’ sign and the anode with a ‘-’ sign•Electrons come from the anode, hence the ‘-’ sign •Electrons are going to the cathode, so they are drawn to the ‘+’ sign

12.3: The Daniel Cell

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

By separating the electrodes from each other and the electrolyte solution, the current has to flow through the wire

The anode is pure zinc surrounded by zinc sulfate (and the clay pot)

The Cathode is pure copper

The electrolyte solution is copper (II) sulfate

Galvanic Cells

We can use other materials at the electrode

Take your car battery for example.•We can dip a platinum wire into a solution of sulfuric acid to complete the reduction half reaction

2H+ + 2e- --> H2

This arrangement of an electrode is called a half cell. The Zn electrode is a Daniel cell is another example of a half cell

12.4: Cell Potential and Gibbs Free Energy

We describe the Cell Potential, E, of a cell as the ability to force electrons through a circuit

•Cells with high cell potentials generate high voltages and vice versa for cells with low potentials

•On you consume all of the ions or metals in the electrodes, the battery is dead

•The redox reaction is at equilibrium

12.4: Measuring Cell Potential

The SI Unit of cell potential is the Volt

A volt is defined as the potential necessary for one Coulomb (a unit of charge times time) falling through a 1V potential to generate 1J of energy

1V•C = 1J

This means that cell potential is an example of non-expansion work

•We need to relate G to E

Cell Potential and Gibbs Free EnergyAt constant T & P, we know that the free energy is equal to the maximum non-expansion work that can be done by a system

G = we

The work done when a number of moles, n, of electrons travels through a potential difference, E, is their charge times the cell potential

we = - neNAE n = # of molese = 1.602x10-19 CNA= 6.022 x1023 e-/mole

But, we can simplify eNA to Faraday’s constant, F, which is equal to 9.6485x104 C/mole e-

we = -nFE

Combine the 2 equations and you get G = -nFE

Cell Potential and Gibbs Free EnergyG = -nFE

•If the cell potential is positive, G is negative and the cell will form products in the redox reaction spontaneously•If the cell potential is negative, G is positive and the cell will form the reactants of the redox reaction spontaneously•We will cell the ideal or maximum potential of a cell the Electromotive Force, emf of the cell

•Real life cells always produce a cell potential that is smaller than the emf

ExampleThe emf of the Daniel cell for certain concentrations of copper and zinc ions is 1.04V. What is the G under these conditions?

ExampleThe reaction taking place in the solver cell used in cameras and watches is:

Ag2O (s) + Zn (s) --> 2Ag (s) + ZnO (s)

The emf when the battery is new is 1.6V. What is the G of the cell?

Standard Reaction Free EnergyG = -nFE°

E° is the standard EMF of the cell•All gases are 1bar•All solutes are 1M•All liquids and solids are pure

•The free energy, G°, will change based up the molar amounts of materials used, but E° doesn’t

•The ability to push/pull electrons in independent of how many materials there are

•Make sense? (There’s the ‘n’ term in the G° equation)

12.5: Cell Notation1. The cell voltage is affected by

the liquids/electrolytes mixing2. In order to harness the

maximum potential of a cell, we separate the electrodes/solutions and use a salt bridge to complete the circuit

3. A salt bridge is a gel of concentrated aqueous ions that allows the circuit to complete• Allows the ions to maintain

a balance of charge between the cells without actually touching

• The electrolyte in the salt bridge doesn’t participate in the redox reaction

12.5: Cell Notation

1. We show the presence of a salt bridge in the notation of a reaction by a double line: ||

For example: In the Daniel cell previously described, we describe it as:

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

This doesn’t show how the electrodes are separated though. The salt bridge is between the 2 half cells or electrodes, so we write it as:

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

Typically, the anode (-) is on the left and the cathode (+) is on the right

Oxidation occurs on the left and reduction occurs on the right

Example

Write the cell reaction occurring for the cell:

Pt (s) | H2 (g) | HCl (aq) | HgCl2 (s) | Hg (s)

Example

Write the cell reaction occurring for the cell:

Pt (s) | H2 (g) | H+ (aq) || Co3+ (aq) | Co2+ (aq)

The sign of the emf is the same as the sign of the right hand electrode in the cell diagram

What does this mean?

• If the emf is negative, then the reverse reaction is spontaneous

We can build a galvanic cell out of nearly any combination of electrodes

Each electrode has a Standard Potential, E°, which is a measure of the electron pulling power of the electrode• Because of this, standard potentials are always written as

reduction reactions

12.6: Standard Potentials

12.6: Standard Potentials

As we go down the table, you’ll notice that the E° values become negative. What does this mean?

If E° is negative, what is G°?

The reverse reaction is spontaneous

12.5: Building a Cell with Standard Potentials

The standard potential of a cell is equal to the difference of the standard potentials of the 2 electrodes.

E° (cell) = E°(right electrode) - E°(left electrode)

OrE° = E°R - E°L

OrE° = E°reduction - E°oxidation

If E° is greater than zero, then the cell works as written

12.5: Building a Cell with Standard Potentials

The more positive the potential, the greater the electron pulling power of the reduction half reaction

• This means the redox couple is a strongly oxidizing pair

The more negative the potential, the greater the electron donating power of the oxidation half reaction

• This means the redox couple is a strongly reducing pair

Example

The standard potential of the Ag+/Ag electrode is 0.80V and the standard EMF of the cell

Pt (s) | I2 (s) | I- (aq) || Ag+ (aq) | Ag (s)

at the same temperature. What is the standard potential of the I2/I- electrode?

Non-traditional Standard Potentials

In general, the most negative E° values are on the left and the most positive are on the right side of the periodic table.

Does this make sense given our understanding of periodic trends?

• Especially, Zeff and electronegativity

The Electrochemical Series

The more negative the standard potential, the greater its reducing strength

Only redox couples with negative standard potential values can reduce hydrogen

The Electrochemical Series

The species on the left of each equation are potential oxidizing agents

The species on the right are potential reducing agents

The relationship trend is called the Electrochemical Series

F2 is a strong oxidizing agent

Li+ is an extremely poor oxidizing agent

Lithium metal is a strong reducing agent

Predicting the feasibility of a possible redox reaction

Standard electrode potentials (redox potentials) are one way of measuring how easily a substance loses electrons. In particular, they give a measure of relative positions of equilibrium in reactions such as:

Zn2+ + 2e- <--> Zn (s) E°= -0.76VCu2+ + 2e- <--> Cu (s) E°= +0.34V

The more negative the E° value, the further the position of equilibrium lies to the left. Remember that this is always relative to the hydrogen equilibrium - and not in absolute terms.

The negative sign of the zinc E° value shows that it releases electrons more readily than hydrogen does. The positive sign of the copper E° value shows that it releases electrons less readily than hydrogen.

Whenever you link two of these equilibria together (either via a bit of wire, or by allowing one of the substances to give electrons directly to another one in a test tube) electrons flow from one equilibrium to the other. That upsets the equilibria, and Le Chatelier's Principle applies. The positions of equilibrium move - and keep on moving if the electrons continue to be transferred.

Using the Electrochemical Series

Predicting the feasibility of a possible redox reaction

The two equilibria essentially turn into two one-way reactions:

•The equilibrium with the more negative (or less positive) E° value will move to the left.

•The equilibrium with the more positive (or less negative) E° value will move to the right.

Using the Electrochemical Series

Will magnesium react with dilute sulfuric acid?

Mg2+ + 2e- <--> Mg (s)E°= -2.37V2H+ + 2e- <--> H2 (s) E°= 0.0V

Using the Electrochemical Series

Will copper react with dilute sulfuric acid?

Cu2+ + 2e- <--> Cu (s) E°= +2.37V2H+ + 2e- <--> H2 (s) E°= 0.0V

Using the Electrochemical Series