Topic 9: Oxidation and Reduction

9.1 Introduction to Oxidation and Reduction

9.1.1 Define oxidation and reduction in terms of electron loss and gain.

Oxidation – increasing of oxidation number, signified by a loss of electrons

Example:

Zn(s) → Zn²⁺(aq) + 2e^-

Li (s) → Li⁺(aq) + e^-

Reduction – ‘reduction’or decreasing of oxidation number, signified by a gain of electrons

Example:

Cu²⁺(aq) + 2e^- → Cu(s)

½F₂ + e^- → F^-

9.1.2 Deduce the oxidation number of an element in a compound.

The oxidation numbers must be shown by a sign (+ or -) and a number. The sum of all the oxidation numbers in a compound must add up to 0 (as long as the entire compound is nt charged).

Specie	Oxidation Number	Example
Hydrogen	+1 (except metallic hydrides, where it is -1)	HCl = +1 CaH2 = -1
Oxygen	-2 (except in peroxides, where it is -1)	CO = -2 H2O2 = -1
Element in its Standard State (Mg, O2, F2, S8, Cl2)	0	Mg = 0 S8 = 0
Monatomic Ions (Fe3+, O-2)	Equal to the ionic charge (Fe = +3, O = -2)	Fe2+ = +2

9.1.3 State the names of compounds using oxidation numbers.

Oxidation numbers in names of compounds are expressed by Roman numerals.

Example:

Al₂O₃ – aluminum(III) oxide, ox#(Al)=+3, ox#(O)=-2

FeO – iron(II) oxide, ox#(Fe)=+2, ox#(O)=-2

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

Oxidation – the oxidation number on the left side of a reaction is lower than the oxidation number on the right. The oxidation number increases.

Reduction – the oxidation number on the right side of a reaction is higher than the oxidation number on the right. The oxidation number decreases.

Example:

Cr₂O₇^2- + CH₃OH → Cr³⁺ + CH₂O

Left:

Oxidation # of Cr is +6. Oxidation # of C is -2.

Right:

Oxidation # of Cr³⁺ is +3. Oxidation # of C is 0.

The oxidation number of chromium decreases from left to right, so it is reduced. (reduction)

The oxidation number of carbon increases from left to right, so it is oxidized. (oxidation)

9.2 Redox equations

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

In a redox system, one substance always becomes reduced while the other one oxidized. We write out the half-reactions for the oxidation and reduction.

Example:

Net ionic equation: 3Mg_(s) + 2Fe³⁺ → 2Fe_(s)+ 3Mg²⁺

Oxidation half-reaction: 3Mg_(s) → 3Mg²⁺ + 6e^-

Reduction half-reaction: 2Fe³⁺ + 6e^- → 2Fe_(s)

9.2.2 Deduce redox equations using half-equations.

How to balance redox equations:

1.      Re-write the net ionic equation as half-reaction.

2.      Balance the elements other than H and O.

3.      Balance the oxygen by adding H₂O to appropriate side.

4.      Balance hydrogen and H₂O by adding H⁺ to appropriate side.

5.      Add electrons (e^-) to appropriate side of each half-equation, so that both sides have the same charge.

6.      Combine the two equations. The electrons should cancel out. Cancel out the redundant waters and hydrogens.

9.2.3 Define the terms oxidizing agent and reducing agent.

Oxidizing agent – the species being reduced.

Reducing agent – the species being oxidized.

9.3 Reactivity

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

A more reactive metal will displace a less reactive one from a compound and a more reactive halogen will displace a less reactive one from a compound.

A stronger reducing agent will displace a weaker one from a compound.

Zinc will displace Copper from a solution of Copper Sulphate and therefore it should be placed above Copper in a reactivity series. It can be said that zinc is stronger reducing agent.

Table of Reduction Half-Reactions (Reduced easiest, #1, to reduced with most difficulty, #6)
Position	Reduction Half-Reaction
1	Cu2+(aq) + 2e- → Cu(s)
2	Pb2+(aq) + 2e- → Pb(s)
3	Ni2+(aq) + 2e- → Ni(s)
4	Fe2+(aq) + 2e- → Fe(s)
5	Zn2+(aq) + 2e- → Zn(s)
6	Al2+(aq) + 2e- → Al(s)

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

Reaction will take place if the total cell potential is positive.

Based on the information in the IB data book.
Add the two half equations in question together (one will have to be reversed, invert the sign of the E-zero value also) If the total E-zero value is positive, the reaction is possible.

The element lower in the reduction series will get reduced while the element higher in the series will get oxidized.

9.4 Voltaic cells

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

The reduction and oxidation can be used to construct a voltaic/galvanic/electrochemical cell. In the cell, electrons flow from the oxidized substance to the reduced substance. The elements higher in the Table 14 will donate the electrons to the substances lower in the table. Thus, electricity is created.

The amount of electricity produced by a voltaic cell is calculated as energy per charge. Its unit is Volt.

Example:

Silver is lower in reactivity series than copper is.

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

Anode – oxidation, electrons are released

Cathode – reduction, electrons are accepted

A positive cathode attracts electrons released by an anode.

9.5 Electrolytic cells

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

Electrolysis does the opposite of a voltaic cell. It makes the non-spontaneous reactions happen. The power source is needed.

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

Electrons flow from anode to cathode.

Anode – positive charge, attracts negative ions. Oxidation happens at the anode because substances must lose electrons.

Cathode – negatice charge, attracts positive ions. Reduction happens because the electrons from the power source are being supplied to positive ions.

9.5.3 Describe how current is conducted in an electrolytic cell.

The substance that conducts electricity in an electrolytic cell is known as the electrolyte.  Electrolyte can be dissolved or molten salt. In a solid state, the ions in salt are fixed on a spot, so they cannot conduct the electricity. However, when molten or dissolved the ions can freely move and thus conduct electricity.

The electricity is supplied to the electrolyte through electrodes made of inert materials, so they do not ineract with the solution. Example of such material is graphite or platinum.

9.5.4 Deduce the products of the electrolysis of a molten salt.

In an electrolytic cell, an anode, the positive electrode, will attract negative ions and a cathode, the negative electrode, will attract positive ions. The electrons will go from anode to battery to cathode. In some cases, such as electrolysis of NaCl, gas, such as Cl₂ may be produced. In electrolysis of aqueous solution, water will become reduced and make hydrogen gas.