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Chemistry 40S Electrochemistry

(This unit was created by Dr. John Wren)

Name:

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Electrochemistry is any chemistry that involves the exchange of electrons, such as redox chemistry. In redox chemistry one reactant is the reducing agent and the other the oxidizing reagent. Oxidizing agents cause an increase in the oxidation number of the reactant they are interacting with. The target is oxidized, while the compound is reduced. Reducing agents decrease their targets oxidation number. The target is reduced while the chemical is oxidized. Recall: Reduction cannot occur without oxidation, therefore all electrochemistry occurs in pairs. Electrochemical Cells When electrons flow from the reducing agent to the oxidizing agent, an electric current is established. If the flow of electrons is in one direction, they can do work. An electrochemical cell, also known as a voltaic cell is a chemical system in which a spontaneous oxidation-reduction reaction can produce useful electrical work. Nature of Electrochemical Cells Electrochemical cells have 4 main components: 1. Electrode undergoing reduction-labeled the cathode 2. Electrode undergoing oxidation-labeled the anode 3. A salt bridge to allow transport of positive and negative ions 4. Wire to complete the circuit and allow flow of e- Both electrodes must be in solutions of ions that will undergo the expected oxidization or reduction.

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Consider the following set up:

Beaker on the left contains 1M of ZnSO4 and the right contains 1M of CuSO4. The Zn0 electrode is being oxidized to Zn2+ and sending its electrons through the wire to the Cu electrode where Cu2+ ions from the solution are attracted to the plate and reduced to Cu0. Zn is oxidized-therefore it is the anode Cu is reduced-therefore it is the cathode

The solution KNO3 is known as a salt bridge, it transports K+ ions towards the Cathode (Cu) solution to balance the charge from the missing Cu2+ and the NO3

- travels towards the anode solution (Zn) to balance the additional Zn2+. Both solutions must remain neutral.

By running the electrons through the wire, they can be run through a device, like a light bulb, to do “work”, such as making light.

This reaction will continue until equilibrium is reached.

In a standard battery cell, when equilibrium is reached the battery is dead!

To help you remember which is anode and which is cathode remember: “Electrons always flow from A to C”

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Definitions: Electrodes are usually metal strips/wires connected by an electrically conducting wire. Salt Bridge contains a solution of an inert electrolyte. Anode is the electrode where oxidation takes place. Cathode is the electrode where reduction takes place. Convention for expressing the cell:

Anode Half-Cell || Cathode Half-Cell Electrode | Anode Solution || Cathode Solution | Electrode

Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) Pt (s) | H2 (1 atm) | H+ (1 M) || Fe3+ (aq), Fe2+ (aq) | Pt (s)

Standard Potential E0

Half-reactions for the reduction of various compounds and elements are often provided in data tables along with their Standard Electrode Potential, E0. The cell potential shows how much electric potential is available in a chemical reduction. Positive E0 indicate that the reduction is more favoured than the reverse oxidation. Negative E0 indicates that the release of electrons, oxidation, is more favoured 2H+ + 2e- → H2 has E0 = 0 V; this is known as the standard hydrogen electrode. All other electrodes were measured in reference to this value. Standard Potential of a Voltaic Cell The standard potential of any Voltaic, or galvanic, cell is the sum of the standard half-cell potentials for the oxidation and reduction half-cells. E0

cell = E0oxidation + E0

reduction Standard half-cell potentials are always quoted as a reduction process. The sign must be changed for the oxidation process.

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• When selecting two half-cell reactions the more negative value will form the oxidation half-cell.

• Consider the reaction between zinc and silver:

Ag+ (aq) + e- Ag (s) E0 = 0.80 V

Zn2+ (aq) + 2e- Zn (s) E0 = -0.76 V • Therefore, zinc forms the oxidation half-cell:

Zn (s) Zn2+ (aq) + 2e- E0 = -(-0.76 V)

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Complete the following diagram using the following set up: Ag | Ag+ (1 M) || Pb2+ (1 M) | Pb (s) Using a KNO3 (aq) salt bridge. Calculate the Voltage.

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Comparing Electrochemical Cells

There are so many combinations of oxidation and reduction reactions that the number of possible electrochemical cells is enormous.

How could you decide which is best for producing electricity? Electrical concepts:

Let’s compare the flow of electricity to the flow of water through a fire hose.

You want a lot of water to flow to put out fires.

What factors affect the flow of water?

Is there enough water available to put out the fire? The total amount of water.

How fast can the water flow? A high rate of flow is needed to put out the fire.

How high can you reach with the stream of water? There must be enough pressure.

In electrical cells, there must be a large enough amount of electricity, an adequate flow of electricity, and enough pressure to do work.

Comparing the Flow of Water to Electricity

Quantity measured Units used with water Units used with electricity

Amount of flow Litres Coulombs

Rate of flow Litres/minute Coulombs/second (amperes)

Tendency to flow (pressure) N/m2 Volts

A coulomb (C) is 1.04 × 10-5 moles of electrons.

One coulomb/second is one ampere (A).

Amperes are a measure of current.

Voltage is a measure of the tendency of electrons to flow.

You can increase the electrical power by increasing the voltage, or by increasing the flow.

Electricity from an electrochemical cell can be measured.

A cell with a high voltage can do a large amount of work over time.

Each half-reaction contributes to the total voltage of the cell.

A standard half-cell has hydrogen gas bubbling into a 1M H+ solution, with a platinum electrode.

The other cell usually has a solution of metal ions and an electrode of the same metal.

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If you set up a copper-hydrogen cell, the beaker on the right has copper being reduced: Cu2+ (aq) + 2e- Cu (s)

On the left, hydrogen gas is being oxidized: H2 (g) 2H+ (aq) + 2e-

The voltmeter reads 0.34 V. Brönsted-Lowry acids and base systems show a competition for protons. Electrochemical cells show a competition for electrons.

Copper has a greater tendency to gain electrons than hydrogen ions, so the electrons flow from hydrogen to copper ions, and not the other way.

The spontaneous reaction is: Cu2+ (aq) + H2 (g) 2H+ (aq) + Cu (s)

The Cu2+ ions win in the competition for electrons.

In a zinc-hydrogen cell, the zinc is oxidized and the hydrogen ions from the acid are reduced.

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

2H+ (aq) + 2e- H2 (g)

In this competition, the hydrogen ions win the electrons.

The overall spontaneous reaction is:

2H+ (aq) + Zn (s) H2 (g) + Zn2+ (aq)

The voltage is 0.76 V.

These cells tell us something about the relative ability to gain electrons.

These ions, in order of decreasing tendency to be reduced are: Cu2+ H+ Zn2+

The voltage of a cell is a measure of the tendency of electrons to flow from the substance which is oxidized to the substance which is reduced.

The voltage is composed of two parts: the contribution made by each half-reaction.

Each contribution is called the half-cell potential.

Half-cell potential is represented by E.

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If the solution is 1 M at 25°C and 101.325 kPa, then it is represented by E0. Chemists set the hydrogen half-cell potential as:

H2 (g) 2H+ (aq) + 2e- E0 = 0.00 V This was chosen to make a convenient standard; much like the freezing point of water is 0°C. The reduction of copper ions has this half-cell potential:

Cu2+ (aq) + 2e- Cu (s) E0 = +0.34 V The oxidation of zinc has this half-cell potential:

Zn (s) Zn2+ (aq) + 2e- E0 = +0.76 V However, chemists always compare REDUCTION potentials, so the reduction of zinc ions is:

Zn2+ (aq) + 2e- Zn (s) E0 = -0.76 V This is so that the half-cell potentials are easily compared. Notice that the half-reaction is reversed, and the sign is changed.

If the sign of the half-cell potential is positive, then the atom or ion will more easily accept electrons than hydrogen.

If the sign of the half-cell potential is negative, then the atom or ion will not accept electrons as easily as hydrogen.

Predicting Redox Reactions Chemists can use the half-cell potentials to predict if a redox reaction will occur.

If the cell voltage for the overall reaction is positive, the reaction will proceed as written.

If the cell voltage is negative, the reaction will not occur spontaneously.

P

ositive

roducts

redominate

At equilibrium Silver tarnish is a film of Ag2S, formed when silver is oxidized by H2S in the air. One way to remove tarnish is to place silverware in an electrolyte solution with a piece of metal that is more easily oxidized than silver. Silver sulphide becomes silver and sulphide ions. The more active metal is oxidized and the silver is reduced.

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Example 1: Can aluminum foil immersed in an electrolyte solution clean sliver? Show the balanced equation and the cell potential for the overall reaction.

1. Find the reagents on the table. Determine which is more easily oxidized by reversing the reactions and voltage.

Ag (s) Ag+ (aq) + e- E0 = -0.80 V

Al (s) Al3+ (aq) + 3e- E0 = +1.66 V

2. Since the half-cell potential for oxidation of aluminum is greater than that of silver, aluminum can be used to force silver ions to be reduced. If one substance is oxidized, the other must be reduced. Silver can be reduced, so we will use the oxidation of aluminum half-reaction and the reduction of silver half-reaction:

Ag+ (aq) + e- Ag (s) E0 = +0.80 V

Al (s) Al3+ (aq) + 3e- E0 = +1.66 V

3. We can add the two reactions together, once we balance the electrons:

Ag+ (aq) + 3e- 3Ag (s) E0 = +0.80 V

Al (s) Al3+ (aq) + 3e- E0 = +1.66 V

Al (s) + 3Ag+ (aq) Al3+ (aq) + 3Ag (s) E0 = +2.46 V

The half-cell potential is NOT multiplied by 3. Three times the electrons does not mean three times the flow.

Example 2: If silver metal is placed in 1 M HCl, nothing happens. If we placed magnesium in HCl instead, the magnesium oxidizes, producing hydrogen gas. Explain these two phenomena.

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Putting Electrochemical Cells to Work

Both spontaneous and non-spontaneous redox reactions are useful in many ways.

Spontaneous redox reactions are used in batteries.

It is possible to reverse spontaneous redox reactions.

Millions of kilos of materials are made every day this way.

Examples: plating of metals, reduction of metals from ores, separation of valuable elements from compounds.

Electrochemical Cells and Storage Batteries

One of the most practical advantages of a battery is that it can provide electrical energy at a desired time.

Potential energy is not released, and no reactions occur, until the circuit is complete.

Not all electrochemical cells contain water solutions.

Common batteries are dry cells, using solid electrolytes, or electrolytes stored as pastes.

Primary Cells

Primary cells provide electrical energy entirely from chemical reactions.

They are not rechargeable, as the reactions are not reversible.

Some will explode if one attempts to recharge them!

Primary batteries use up the materials in one or both of their electrodes.

Carbon-Zinc Batteries

One common battery uses a manganese oxide paste, water, and ammonium chloride (“Leclanché cells”).

A zinc can acts as the anode and a carbon/manganese dioxide rod acts as the cathode.

Zinc is oxidized and manganese dioxide is reduced to various compounds of Mn3+.

Alkaline-Manganese Batteries

These are the common alkaline cells.

They contain a powdered zinc anode and a carbon/manganese dioxide rod cathode.

The cell uses KOH, an alkaline substance, as the electrolyte.

The key reaction is:

Zn + 2MnO2 ZnO + Mn2O3

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Zinc is oxidized and the manganese in MnO2 is reduced.

This cell is more efficient because the electrical resistance in the cell does not increase with use, and because it yields two to ten times more current.

Secondary Cells

Secondary cells need to be charged before use.

The reactions in secondary cells are reversible, the forward reaction provides electrical energy, and the reverse reaction uses electrical energy.

Rechargeable NiCad and NiMH batteries, as well as car batteries are in this category.

Lead Storage Batteries

These are wet cells.

Car batteries have several compartments, a 12 volt battery has six.

Lead is oxidized to form lead (II) sulphate, at the anode, and lead (IV) oxide is reduced to lead (II) sulphate at the cathode.

Sulphuric acid is the electrolyte used to maintain the charge balance in the cell.

Here are the half-reactions involved:

Pb + SO42– PbSO4 + 2e- E0 = +0.36 V

PbO2 + SO42– + 4H+ + 2e- PbSO4 + 2H2O E0 = +1.68 V

PbO2 + Pb + 4H+ + 2SO42- 2PbSO4 + 2H2O E0 = +2.04 V

Why are there six cells in a car battery? Note that the H2SO4 is consumed during the reaction. As the battery discharges, the sulphuric acid becomes more dilute. Of course, the car’s alternator will recharge the battery, by applying a current, thus reversing the reaction. Simply measuring the density of H2SO4 can test batteries. As the acid becomes more dilute, the freezing point rises, and is more likely to freeze in winter. In winter, you need to keep your battery charged!

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Hydrogen-Oxygen Fuel Cells

A fuel cell “burns” fuel in an unconventional manner.

Gaseous H2 and O2 are circulated in contact with porous carbon electrodes.

These electrodes allow the gases to diffuse towards each other so that the half-reactions occur within the electrodes!

(Anode) 2×[H2 + 2O2- 2H2O + 2e-]

(Cathode) O2 + 4e- 2O2-

(Overall) 2H2 + O2 2H2O

The H2O formed leaves the cell as steam.

Fuel cells have been used in spacecraft to provide fuel, and drinking water!

Unfortunately, fuel cells are expensive, and need to be very large to produce useful amounts of power.

Quantitative Aspects of Electrochemical Cells

In an electrochemical cell with a copper anode in a copper solution and a silver cathode in a silver solution, we get this:

Cu Cu2+ + 2e- E0 = -0.34 V

2Ag+ + 2e- 2Ag E0 = +0.80 V

2Ag+ + 2Cu Cu2+ + 2Ag E0 = +0.46 V

As the cell discharges, the silver electrode increases in mass and the copper electrode decreases in mass.

Is the change in mass the same at both electrodes? o NO!

In one experiment, the silver electrode gained 2.15 g and the copper electrode lost 0.635 g.

The relationship in a chemical equation is not expressed in grams, it is in moles!

Convert the grams to moles:

2.15 g 108 g/mole = 0.0199 moles of Ag reacted

0.635 g 63.5 g/mole = 0.0100 moles of Cu reacted

The ratio of moles of silver to moles of copper reacted is:

0.0199 0.0100 2:1 ratio

For every 2 moles of silver, one of copper reacts.

Same as the redox equation!

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Example 3: By what mass in grams would a chromium cathode increase when it is coupled to a magnesium half-cell in which the magnesium anode loses 1.53 g?

1. Write the half-reactions, balance them and add them: Mg Mg2+ + 2e- Cr3+ + 3e- Cr

Balanced: 3Mg 3Mg2+ + 6e-

2Cr3+ + 6e- 2Cr

2Cr3+ + 3Mg 3Mg2+ + 2Cr

So, three moles of magnesium would yield 2 moles of chromium.

2. Calculate chromium’s mass of increase:

Electrolysis Electrolysis is when the anode in a cell is replaced with an external, typically non-chemical, electron source. These electrons are effectively pumped into the cell forcing the reduction of the target chemical. This is one way that we can reduce compounds that are weak oxidizing agents, like Cr3+ to Cr metal. This is also how electroplating works, the target object is electrified, making it an anode, and it is directly placed in a solution of the cathode metal, which deposits directly on the object.

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Electrolysis is the process by which a chemical reduction occurs by passing an electrical current through a solution of ions. Faradays Law of Electrolysis

Michael Faraday (1791-1867) was a British chemist and physicist who contributed significantly to the fields of electromagnetism and electrochemistry.

He also invented the earliest form of the Bunsen burner.

Some historians of science refer to him as the greatest experimentalist in the history of science.

It was largely due to his efforts that electricity became viable for use in technology.

Faraday's Law of Induction states that a magnetic field changing in time creates a proportional electromotive force.

But we are interested in his work with electrochemical cells.

The SI unit of capacitance, the farad, is named after him, as is the Faraday constant, the charge on a mole of electrons, about 96 485 coulombs.

Before, we were using the number of grams lost by an electrode to predict the number of grams gained by the other one.

We can also look at the number of electrons.

For example: Cu2+ (aq) + 2e- Cu (s)

We would predict that: 2 mol e- = 1 mol Cu

If we knew how many electrons were supplied, we could tell how many moles of copper would be deposited; such as 4 mol e- gives 2 mol Cu.

This leads to our modern version of Faraday’s law of electrolysis: The number of moles of product formed in an electrolysis cell by an electric current is chemically equivalent to the number of moles of electrons supplied.

From the number of moles of product formed, we can calculate the mass of products or, if they are gases, their volumes.

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Example 4: Calculate the volume of hydrogen produced at STP by the electrolysis of water when 0.050 moles of electrons are provided.

Write the reduction half-reaction to find the relationship between the number of moles of electrons and of product.

2H2O + 2e- H2 + 2OH-

To go further, Faraday’s law suggests that we can determine the number of moles of product by the quantity of electricity applied.

We can use the Faraday constant, F = 96.5 kC/mol, the charge of a mole of electrons, to convert the coulombs supplied to moles of electrons.

Once we know the coulombs supplied, we can calculate:

To find the charge that is supplied, we need to know that an electrical current of 1 ampere supplies 1 coulomb of charge per second, that is 1 A = 1C/s, so:

Coulombs supplied = current in amperes × time in seconds

For instance, the total charge passing through a TV that uses 2.0 A for 1 h is:

Coulombs supplied = 2.0 A × 3600 s = 7200 C or 7.2 kC

So how many electrons is that?

Note that even large currents over long times supply very few moles of electrons.

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Example 5: Aluminum is produced by electrolysis of aluminum oxide dissolved in molten cryolite (Na3AlF6). Calculate the mass of aluminum that can be produced in one day in an electrolysis cell operating continuously at 100 000 A.

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Example 6: Calculate the mass of magnesium produced by the electrolysis of magnesium chloride in a cell operating all day at 50 000 A.

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