Definitions:

 Examples:

CuO _(aq) + H_{2 (g) -->} Cu _(s) + H₂O _(l) (look at the oxygen)

Cu is being reduced

H₂ is being oxidized

Mg + Cl₂ --> MgCl₂

Mg – 2e --> Mg²⁺     (oxidation) loss of electrons and an increase in ox number.

Cl₂ + 2e -->  2Cl^-        (reduction) gain of electrons and an decrease in oxidation number

Calculating oxidation numbers

Write these rules in your books and use them to answer the following examples.

 Examples:

What is the oxidation number of the underlined element in the following compounds?

(a)   Cr₂(SO₄)₃

(b)  Na₂S₂O₃

(c)   K₂MnO₄

(d)  CoCl₃

(e)   MnO₄^-

(f)   CoCl₂ 

Ionic Equations

What happens at the molecular level when ionic compounds dissolve in water?

Recall that water is polar.

Which side of the water molecule will be attracted to the Na+? Why?

Which side of the water molecule will be attracted to the Cl-? Why?

NaCl (aq) can be written as Na+ (aq) and Cl- (aq)

When ionic compounds are dissolved into water, the polar water molecules break apart the solid crystal lattice, resulting in the hydrated ions being evenly distributed through the water. This process is called dissociation, and it is the reason that ionic compounds tend to be strong electrolytes. When two different ionic compounds that have been dissolved in water are mixed, a chemical reaction may occur between certain pairs of the hydrated ions.

Consider the double-replacement reaction that occurs when a solution of sodium chloride is mixed with a solution of silver nitrate. The driving force behind this reaction is the formation of the silver chloride precipitate.

NaCl(aq)+AgNO₃(aq)→NaNO₃(aq)+AgCl(s)

This is called a molecular equation. A molecular equation is an equation in which the formulas of the compounds are written as though all substances exist as molecules. However, there is a better way to show what is happening in this reaction. All of the aqueous compounds can be written as ions because they are actually present in the water as dissociated ions.

Na⁺(aq)+Cl⁻(aq)+Ag⁺(aq)+NO₃⁻(aq)→Na⁺(aq)+NO₃⁻(aq)+AgCl(s)

This equation is called an ionic equation: an equation in which dissolved ionic compounds are shown as free ions.

If you look carefully at the last equation, you will notice that the sodium ion and the nitrate ion appear unchanged on both sides of the equation. When the two solutions are mixed, neither the Na⁺ nor the NO₃⁻ ions participate in the reaction. Although they are still present in the solution, they do not need to be included when describing the chemical reaction that occurs upon mixing.

Na⁺(aq)+Cl⁻(aq)+Ag⁺(aq)+NO₃⁻(aq)→Na⁺(aq)+NO₃⁻(aq)+AgCl(s)

A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction. In the above reaction, the sodium ion and the nitrate ion are both spectator ions. The equation can now be written without the spectator ions.

Ag⁺(aq)+Cl⁻⁽aq)→AgCl(s)

The net ionic equation is the chemical equation that shows only those elements, compounds, and ions that are directly involved in the chemical reaction.

Net ionic equations must be balanced by both mass and charge. An equation that is balanced by mass has equal amounts of each element on both sides. Balancing by charge means that the total charge is the same on both sides of the equation. In the above equation, the overall charge is zero, or neutral, on both sides of the equation. As a general rule, if you balance the molecular equation properly, the net ionic equation will end up being balanced by both mass and charge.

Solubility

Evaluation: Write an ionic and net ionic equation for the following reaction.

manganese(II)chloride(aq) + ammonium carbonate(aq) ® manganese(II)carbonate(s) + ammonium chloride(aq)

Further work

Please complete the following reactions using your solubility rules, and show the total ionic and net ionic forms of the equation:

 K₃PO₄(aq) + Al(NO₃)₃(aq) --> 

BeI₂(aq) + Cu₂SO₄(aq) -->

Ni(NO₃)₃(aq) + KBr(aq) -->

cobalt(III)bromide + potassium sulphide -->

barium nitrate + ammonium phosphate -->

calcium hydroxide + iron(III)chloride -->

rubidium fluoride + copper(II)sulphate -->

Single displacement reactions

Reducing and oxidizing agents

During a redox reaction:

-        A reducing agent loses electrons and gets oxidized

-        An oxidizing agent gains electrons and gets reduced

We can put reagents in order of their oxidizing or reducing ability by carrying out displacement reactions. A displacement reaction is one where one atom replaces another in a chemical reaction.

Recall the reaction from the previous section:

NaCl (aq) + AgNO₃ (aq) --> NaNO₃ (aq) + AgCl (s)

Is this reaction an example of a redox reaction?

What type of reaction is it?

NB: (Double Displacement reactions are not redox reactions).

Single Displacement Reactions

Single displacement reactions are redox reactions. One species is oxidized while the other is reduced. The loss of electrons in one species is gained by the other species, hence redox. If we react different metals with aqueous solutions of different metal salts we can build a metal reactivity series.

Before you continue ensure that you have a solid grasp of ionic and net ionic equations.

Zn _(s) + Cu ²⁺ _(aq) --> Zn ²⁺ _(aq) + Cu _(s)

Zinc is oxidised and copper is reduced. Zn is a better reducing agent than Cu. A better reducing agent will displace another reagent.

How can you order Zn, Cu and Ag in terms of their reactivity?

Recap:

What is an oxidising agent?

What is a reducing agent?

The more reactive the element, the better a ________________ agent it is.

The more reactive the element, the easier it is to be ____________________.

 Balancing half equations

Writing Half Equations

Consider the following reaction,

CuO (s) + Mg (s) --> Cu (s) + MgO (s)

Let us look at the oxidation states of the copper and magnesium in the compounds.

Copper: Cu²⁺ -->  Cu⁰

Magnesium: Mg⁰ -->  Mg²⁺

Recall this means that copper is reduced and magnesium is oxidized.

You can write two half equations which look at the reaction from the point of view of the Mg2+ and the point of view of Cu.

For Mg to become Mg²⁺, two electrons have to be lost: Mg -2e --> Mg²⁺ or this is typically written as Mg -->  Mg²⁺ +2e (oxidation). It is losing 2 electrons to become the ion. Since it is giving the electrons to the Cu²⁺, it is reducing the copper and is the reducing agent.

For Cu²⁺ to become Cu, two electrons have to be gained: Cu²⁺ +2e -->  Cu (reduction). It is gaining the two electrons lost by the Mg. since it is causing Mg to lose electrons it is the oxidizing agent.

1.     Which of the following are oxidized and reduced? Identify in equation and write short half equations.

(a)   Cu _(s) + ½ O_{2 (g) -->}  CuO (s)

(b)  Fe³⁺ + I^- -->  Fe²⁺ + 1/2I₂

(c)   2CuCl -->  Cu + CuCl₂

(d)  Cl₂ + 2NaOH -->  NaCl + NaOCl + H₂O

Balancing half equations

Half equations are balanced by:

1.     Balancing the number of electrons lost and gained

2.     Putting the two half equations together

Example.

Construct a balanced ionic equation for the reaction of zinc with iodate (V) ions, IO₃^-, in acidic solution using the following:

Zn -->  Zn²⁺ +2e

IO₃ ^- + 6H⁺ +5e -->  ½ I₂ + 3H₂O

Step 1: Balance electrons

5Zn -->  5Zn²⁺ +10e

2IO₃ ^- + 12H⁺ +10e -->   I₂ + 6H₂O

Step 2: Put together

5Zn+ 2IO₃ ^- + 12H⁺  -->   I₂ + 6H₂O + 5Zn²⁺

Extra work:

1.     Balance the following reaction

H₂SO₄ + 8H⁺ + 8e -->  H₂S + 4H₂O

                        2I^- -->  I₂ + 2e

2.     Write half equations and balance the following equations.

Cu⁺(aq)+Fe(s)→Fe³⁺(aq)+Cu(s)

Home work:

Define disproportionate reaction

 Electrochemical cells

We encounter electrochemical cells in all facets of our everyday lives from the disposable AA batteries in our remote controls and the lithium-ion batteries in our iPhones to the nerve cells strewn throughout our bodies. There are two types of electrochemical cells: galvanic (also called Voltaic) and electrolytic. Galvanic cells derives its energy from spontaneous redox reactions, while electrolytic cells involve non-spontaneous reactions and thus require an external electron source like a DC battery or an AC power source. Both galvanic and electrolytic cells will consist of two electrodes (an anode and a cathode), which can be made of the same or different metals, and an electrolyte in which the two electrodes are immersed.

Galvanic Cells

Galvanic cells traditionally are used as sources of DC electrical power. A simple galvanic cell may contain only one electrolyte separated by a semi-porous membrane, while a more complex version involves two separate half-cells connected by a salt bridge. The salt bridge contains an inert electrolyte like potassium sulfate whose ions will diffuse into the separate half-cells to balance the building charges at the electrodes. According to the mnemonic “Red Cat An Ox”, oxidation occurs at the anode and reduction occurs at the cathode. Since the reaction at the anode is the source of electrons for the current, the anode is the negative terminal for the galvanic cell.

Electrolytic Cells

Let’s take a look at the final type of cell, the electrolytic cell. An electrolytic cell is an electrochemical cell in which the energy from an external power source is used to drive a normally non-spontaneous reaction, i.e. apply a reverse voltage to a voltaic cell. We encounter electrolytic cells during the charging phase of any type of rechargeable battery from the lead-acid battery in automobiles to the lithium-ion battery in smartphones.

In comparison to the galvanic cell, the electrodes of an electrolytic cell can be placed in a single compartment containing the molten or aqueous electrolyte. In addition, since the external battery source is what drives the electrons through the circuit, the electrodes will match the positive and negative terminal of the battery. While the anode remains the site of oxidation, it becomes the positive terminal, and the cathode becomes the negative terminal.

Another use for an electrolytic cell is for the decomposition of compounds, i.e. water, sodium chloride, into simpler compounds. Industrial processes take advantage of this in the production of chlorine or sodium hydroxide. Since electrolytic cells can be conducted in molten or aqueous electrolyte, depending on the cation and anion, the products from using a molten electrolyte may be different from the products from using an aqueous electrolyte.

Electrolysis of Brine

Brine is a solution of sodium chloride (NaCl) and water (H₂O). The process of electrolysis involves using an electric current to bring about a chemical change and make new chemicals. The electrolysis of brine is a large-scale process used to manufacture chlorine from salt. Two other useful chemicals are obtained during the process, sodium hydroxide (NaOH) and hydrogen (H₂).

It is important that the chlorine and sodium hydroxide produced in the process are separated they react when they come into contact with each other.

Uses in industry

Product of electrolysis of brine

• Chlorine - disinfectant and purifier, manufacture of hydrochloric acid and making plastics

• Sodium Hydroxide - processing food products, removing pollutants from water and manufacture of paper

• Hydrogen - manufacture of hydrochloric acid and potential as a pollution-free fuel

Environmental impacts of the electrolysis of brine on a large scale must be considered. The process uses a lot of electricity that is mainly produced by the burning of fossil fuels. During the actual process of electrolysis, metal must be in contact with the solution of brine. A metal commonly used is mercury which is toxic. Some mercury escapes into the solution and into the environment.

Electrode potentials

When a metal is placed in a solution of its ions a voltage is established between the metal ions and the metal ions in solution. Similarly if a piece of metal is placed in a beaker of water, there will be a tendency for metal atoms to shed electrons and enter into solution as metal ions. Eventually a metal ion will pick back up its electrons and stick to the surface of the metal again. This continues until the rate of ions leaving the surface is equal to the rate of them returning to the surface of the metal.

A piece of magnesium/zinc would form more ions in solutions that a piece of copper which is less reactive.

Therefore,

Mg²⁺ (aq) +2e ↔ Mg (s) the tendency is greater for Mg to form the ion so this equation would shift more towards the left

Cu²⁺ (aq) + 2e ↔ Cu (s) the tendency is not as great (as Mg) for Cu to form the ion

In magnesium there is a lot of difference between the negativeness of the metal and the positiveness of the solution around it. This potential difference (difference between one metal-metal ion system and another system) can be measured as voltage.

Demo: Set up 2 beakers, one containing a solution of Cu²⁺ ions and a piece of Cu metal, and another containing Zn²⁺ ions and a piece of Zn metal. Use alligator clips to attach the 2 metal strips to a voltmeter. No voltage is observed.

Standard Cell Potential.

In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.

An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell.

Each beaker containing a metal in a solution of its ions is called a half cell. We can calculate the voltage of an electrochemical cell made up of two half cells. This voltage is the potential difference between the two half cells. This is the standard cell potential. Std conditions must be used.

Demo: Soak a piece of paper towel in a salt solution (e.g., NaCl(aq)) and place one end in the Cu²⁺ solution and one end in the Zn²⁺ solution. A voltage is observed.

The two half cells are connected by a salt bridge. This allows the movement of ions between the two half cells, completing the circuit so that the ionic balance is maintained. The salt bridge can be a strip of filter paper or string soaked in potassium nitrate or potassium chloride which serves as an electrolyte.

Using the electrochemical cell produced, the direction of flow of electrons can be demonstrated. The anode can be identified as the negative electrode where oxidation occurs. The cathode can be identified as the positive electrode where reduction occurs.

Zn (s) -2e ↔ Zn²⁺ (aq) or Zn (s) ↔ Zn²⁺ +2e   oxidation (more reactive so greater tendency to form the ion)

Cu²⁺ (aq) + 2e ↔ Cu (s) reduction

Cell diagrams

Electrochemical cell notation or a cell diagram is a shorthand method of summarizing the anode and cathode reactions in an electrochemical cell. We will use the convention where the anode reaction (ox) appears on the left and the cathode reaction (red) appears on the right. In the notation the half-cells are separated by double vertical lines which represent the salt bridge. Single lines separate anode and cathode from their ion forms in solution. For example: anode/anodic ion solution//cathodic ion solution/cathode for the previous electrochemical cell example the cell notation is:

Zn(s)/Zn²⁺(aq)//Cu²⁺(aq)/Cu(s)

Standard Hydrogen Electrode

If we want to compare the ability of different metals to release electrons, we need to utilize a standard electrode for comparison. The half-cell chosen for this is the hydrogen electrode. Since hydrogen is a gas, it is bubbled into the solution. Hydrogen doesn’t conduct so we make electrical contact via a piece of unreactive metal Pt, coated with finely divided Pt to increase the surface area. The electrode is under std conditions [H+ (aq)] = 1.00M; pressure = 100kPa (1atm) and 298k.

The potential of hydrogen is defined as zero so when connected to another half-cell, the measured voltage called the electromotive force (emf) is the electrode potential of the cell.

Measuring std electrode potential

To measure the std electrode potential for

Zn²⁺ (aq) +2e ↔ Zn (s)

We place a rod of pure zinc into a 1.00M solution of Zn²⁺ (aq). This is the Zn(s)/Zn²⁺(aq) half-cell. We then connect it to a std hydrogen electrode.

When the Zn²⁺ + 2e- → Zn is connected the voltage is -0.76 volts. This is the voltage of the Zn half-cell.

When the Cu²⁺ + 2e-→ Cu is connected to hydrogen half-cell the voltage is + 0.34 volts.

If the voltage of many half-cells are measured they can be listed from most positive voltage to most negative to produce a Table of Standard Half-Cell Potentials. The sign on the cell voltage depends on whether the half-cell donates or receives electrons with respect to the std hydrogen electrode. Half cells with more negative values (+0.4 is more negative than +1.20) are better at releasing or losing electrons since they are more easily oxidized. This makes them better reducing agents than hydrogen.

When connected to the hydrogen cell, the voltage recorded on the voltmeter (the voltage relative to the Hydrogen Half-Cell) is the voltage assigned to each half-cell.

For example, when the Cu²⁺ + 2e-→ Cu is connected to H₂ hydrogen half-cell the voltage is + 0.34 volts. This is the voltage of the Cu half-cell. When the Zn²⁺ + 2e- → Zn is connected the voltage is -0.76 volts. This is the voltage of the Zn half-cell.

 Types of half cells

1.     Metal- metal ion (as mentioned before)

e.g.

Zn(s)/Zn²⁺(aq)

2.     Half cells containing non metals and metal ions.

e.g. ½ Cl_2(g) +e ↔ Cl^- (aq)

The gas is bubbled through and the setup is similar to the hydrogen electrode.

NB.

3.     Half cells containing ions in different oxidation states.

Both ions of the same element in different ox states must be present and have conc 1M. a Pt electrode is used to make the electrical connection with the solution. E.g. Fe3+/Fe2+ half cell.

 Calculate standard cell potentials from standard electrode potentials of two half cells.

2.     Use standard electrode potentials to determine the direction of electron flow.

3.     Use standard electrode potentials to determine the feasibility of a reaction

E^◦ values and Redox Reactions

E^◦ refers to a reduction reaction, so it is written as a reduction and electrons appear on the left.

The more positive the value (or less negative) the easier it is to gain electrons and they are reduced making them better oxidizing agents.

The more negative the value of E^◦ (or less positive), the easier it is to lose electrons making them better reducing agents.

Direction of electron flow in cells.

When we connect two half cells:

·       The half-cell with the more positive value of E^◦ is the positive pole/cathode.

·       The half-cell with the less positive value of E^◦ is the negative pole/anode.

The direction of the electron flow is from the negative pole to the positive pole.

How can we predict if a reaction will take place or not?

Step one

Invert the more negative equation. Add the two half equations. This is the feasible equation.

Step two

Use the feasible equation to calculate E_cell

Calculating E^◦ of the cell.

E_cell = E_cathode - E_{anode   or}   E^◦_cell = E^◦_red - E^◦_ox

If this value is positive then the reaction is feasible.

NB: there are several ways to do this calculation but since we are using the data from the table in your books this is the easiest way to use values from that table. 

Examples

Example 1

For example, what is the cell voltage for the following reaction? Is the reaction spontaneous?

Cu(s) + Fe²⁺ (aq) → Cu²⁺ (aq) + Fe(s)

Cu²⁺ + 2e- → Cu    E⁰ = + 0.34 V

 Fe²⁺ + 2e- → Fe    E⁰ = - 0.45 V

E^◦_cell = E^◦_red - E^◦_ox

            = -0.45-0.34 = -0.79 (not spontaneous)

The reverse of this reaction is spontaneous. (+0.79V)

Spontaneity could be demonstrated using the original Zn metal in a Cu²⁺ ion solution, which is spontaneous and produces a precipitate of Cu metal. Then try Cu metal in a Zn²⁺ ion solution and this reaction gives no reaction.

Zn + Cu²⁺ → Zn²⁺ + Cu    E⁰ = + 1.10 V Spontaneous

 Zn²⁺ + Cu → Zn + Cu²⁺    E⁰ = - 1.10 V Non-spontaneous

What is the feasible reaction for the following?

Zn²⁺ + 2e ↔ Zn     E^◦= -0.76V  

Cu²⁺ + 2e ↔ Cu     E^◦= +0.34V  

Step one

Zn + Cu²⁺ à Cu + Zn²⁺

Step two

E^◦_cell = E^◦_red - E^◦_ox

0.34 – (-0.76) = +1.10V (feasible)

Example 2

Will chlorine oxidise Fe²⁺ ions to Fe³⁺ ions?

Step one

Fe³⁺ + e ↔ Fe²⁺    E^◦= +0.77V  

1/2Cl₂ + e ↔ Cl^-    E^◦ = +1.36V

The feasible reaction can be found by reversing the more negative value i.e

Fe²⁺ + ½ Cl₂ ↔Cl^- + Fe³⁺ is more feasible… this time the chlorine is reduced

Step two

E^◦_cell = E^◦_red - E^◦_ox

E^◦ _cell = 1.36-0.77 = +0.59V (positive therefore feasible)

Example 3

Will aqueous iodine oxidise silver to Ag⁺? Use the table to determine the electrode potentials and which reaction is feasible.

Step one

I₂+ 2e ↔ 2I^-      +0.54V

Ag⁺ + e ↔ Ag     +0.80V

The feasible reaction is:

2I^- + Ag⁺ ↔ I₂ + Ag

Therefore the question asked is not feasible.