9. Redox SL

Define oxidation and reduction in terms of electron loss and gain

Reduction- the process of gaining electrons

Oxidation- the process of losing electrons

Deduce the oxidation number of an element in a compound

The oxidation number of…

Flourine      -1

Hydrogen   +1       (except when bonded with a metal à NaH, CaH₂, etc…

Oxygen      -2         (except when bonded in H₂O₂ OR F₂O)

Halogens     -1        (except when bonded to O OR a halogen higher up in the group)

When you’re working with ionic compounds- It’s the charge of the ion

When you’re working with covalent compounds- Use the tricks above to help you figure it out. Keep in mind, POLYATOMICS are covalent compounds (Their total charge just doesn’t equal zero)

State the names of compounds using oxidation numbers

Oxidation numbers are used in compounds involving transition metal ions. They are added as roman numerals in parenthesis

            E.g. FeO is Iron(II) Oxide
            E.g. V­₂O₅ is Vanadium (V) Oxide

Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers

Ex. FIND the oxidation number of H₂ à H­₂O

In H₂, there is no net charge so H has no oxidation number (0)

In H₂O, oxygen has a -2 charge, which means that the 2 H molecules each have a +1 charge in order to balance it out

Therefore, H is REDUCED

A reaction is not redox if the oxidation number does not change for any of the atoms

Disproportionation- a reaction in which the same element is both reduced and oxidized (ex. 2Cu⁺ à Cu­²⁺ + Cu)

            +1      +2        0

Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction

In the equation Zn(s) + I₂(s) à Zn²⁺(aq) + 2I^- (aq) which is reduced?

The easiest way to do this would be to create half equations (deal with one element)

Zn à Zn²⁺                        Add electrons to one side or the other to create an equal charge

Zn à Zn²⁺ + 2e^-               Since the Zinc ion LOST electrons, it was oxidized

I₂ à 2I^-                              In this case, the products have two more electrons, soooo

I₂ + 2e^- à 2I^-                    Since the Iodide ions GAINED electrons, they were reduced

Define the terms oxidizing agent and reducing agent

Oxidizing agent- a substance that GAINS electrons in a redox reaction

            (i.e. it takes the electrons from the other substance, OXIDIZING it)

Reducing agent- a substance that LOSES electrons in a redox reaction

            (i.e. it gives the electrons to the other substance, REDUCING it)

Deduce redox equations using half-equations AND

Identify the oxidizing and reducing agents in redox equations

How does one balance a reaction between Zinc and acidified dichromate ions to produce chromium(III) ions?

Use Half Equations!

Chromium

            The oxidation number of chromium in dichromate is +6

            The oxidation number of chromium in Cr³⁺ is + 3

            THEREFORE, Chromium is being reduced (It is the oxidizing agent)

           

            CrO₇^2- + 3e^- à Cr³⁺

            BUT, realize there are now 7 Oxygen atoms on the left side of the e                                    equation. SO we have to balance it using water molecules (we assume the

            reaction (we assume the reaction is aqueous)

            CrO₇^2- + 3e^- à Cr³⁺ + 7H₂O

            A good start, but we now have 14 hydrogens on the right side of the eq.

            Luckily, the dichromate solution is ACIDIFIED meaning there is plenty

            Of our buddy H⁺ to save us from this disaster

            CrO₇^2- + 14H⁺ à Cr³⁺ + 3e^- + 7H₂O

Zinc

            Since Chromium is being reduced (as we said above), we can take a

            Guess that Zinc must be oxidized (It is the reducing agent)

            Zn à Zn²⁺ + 2e^-

           

            And that’s all you need! (SIMPLE)

Uh oh, we’re not done yet!

            We’re gonna need to balance the number of electrons on each side

            Lets go back to the simple half equations

            The easiest way is to make sure there are 6 electrons on each side

                        2[CrO₇^2- + 3e^- à Cr³⁺] = 2CrO₇^2- + 6e^- à 2Cr³⁺

                        3[Zn à Zn²⁺ + 2e^-] = 3Zn à 3Zn²⁺ + 6e^-

So re-add those coefficients to the original whole equation

            2CrO₇^2- + 14H⁺ + 3Zn + 6e^-à 2Cr³⁺ +6e^-  + 3Zn²⁺ + 7H₂O

Get rid of the electrons (they cancel each other out) and viola!!!

            2CrO₇^2- + 14H⁺ + 3Zn à 2Cr³⁺  + 3Zn²⁺ + 7H₂O

Deduce a reactivity series based on the chemical behavior of a group of oxidizing and reducing agents

Deduce the feasibility of a redox reaction from a given reactivity series

Explain how a redox reaction is used to produce electricity in a voltaic cell. AND

Function of Voltaic cell- convert chemical energy à electrical energy

Draw a voltaic cell-

How does a voltaic cell work? Each electrode establishes equilibrium with the solution, creating a concentration of ions in the solution and excess electrons. The more reactive metal (anode) will be more oxidized than the positive, creating a negative charge. These electrons from the anode travel through the wires to the cathode, where they reduce the ions in solution. This movement of electrons is the electric energy. In order to complete the circuit, the salt bridge is added which will allow negative ions to move in accordance with the increasing/decreasing numbers of positive ions

State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).

OXidation occurs at the ANode (AN OX)

            The anode is NEGATIVE in a voltaic cell

REDuction occurs at the CATthode (RED CAT)

            The cathode is POSITIVE in a voltaic cell

Describe using a diagram, the essential components of an electrolytic cell

Electrolysis- Non-spontaneous reaction driven by application of electrical energy

Electrolyte- Solution containing ions to transport electric current

Draw an Electrolytic Cell

State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode)

OXidation occurs at the ANode (AN OX)

            The anode is POSITIVE in an electrolytic cell

REDuction occurs at the CATthode (RED CAT)

            The cathode is NEGATIVE in an electrolytic cell

Explain how current is conducted in an electrolytic cell

How does an electrolytic cell work?

            Two inert conductors are placed in an ionic solution and a current is applied (usually through a battery.) Electrons leave the battery and move to the negative electrode, where they are used to reduce cations in solution to anions. This negative conductor becomes the Cathode. These negative ions then travel through the solution to the positive ion, where the anions are oxidized back into cations. These electrons move through the wires back to the battery and the cycle is repeated.

Inert conductors are used so that they do not become oxidized/reduced by the reaction.

Examples of inert conductors include- platinum

Deduce the products of the electrolysis of a molten salt

EXAMPLE: CuBr₂

Solid ions form an ionic lattice

Liquid ions are free to move within a solution

At the cathode, the positive ions will be reduced

            Cu²⁺ + 2e^- à Cu

            Cu (s) will bind to the electrode

At the anode, the negative ions will be oxidized

            2Br^- à Br₂

Practice Questions:

http://alpha.chem.umb.edu/chemistry/ch115/Mridula/CHEM%20116/documents/Chapter20PracticeQuestions.pdf