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Assigning Oxidation NumbersAssigning Oxidation NumbersWe will hold off for the time being the formaldefinition of an oxidation number.

The oxidation number (state) of a free oruncombined element is 0.

Fe, Na, O2

The oxidation number of a monatomic ion isequal to its charge.

K+, Al3+, Ca2+, Cu+

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The oxidation number of hydrogen is +1 whenit is covalently bonded.

HCl, H2O2, H2O, H2SO4

The oxidation number of hydrogen is -1 whenit is ionically bonded with a metal.

NaH, CaH2, KH These are examples of hydrides and are named sodium hydride, calcium hydride, and potassium hydride.

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The oxidation number of oxygen is -2 in most of its covalent compounds.

One exception is when oxygen is bonded to fluorine as in OF2.

In OF2, the oxidation number of oxygen is +2.

Fluorine being the most electronegative element is always assigned an oxidation number of -1.

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Another exception is when the oxidation number for oxygen is -1 when it is a peroxide.

O22- can be found covalently bonded as

it is in H2O2 (hydrogen peroxide).

O22- can be found ionically bonded as it

is in K2O2 (potassium peroxide).

Something to remember is that you do not reduce the subscripts in a metallic peroxide (K2O2 ).

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A second exception is when oxygen has an oxidation number of -½.

This occurs when oxygen is the anion, O2

-, and is called a superoxide.

LiO2 is such a compound and is called lithium superoxide.

For binary compounds, the element with thegreater electronegativity is assigned anegative oxidation number equal to itscharge.

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The algebraic sum of the oxidation numbersin a neutral compound must equal zero.

AlCl3 H2O2 NH3

The algebraic sum of the oxidation numbersin a polyatomic ion must equal the charge onthe ion.

H3O+ NH4+ PO4

3-

+3 -1+3 -3

+1 -1+2 -2

-3 +1-3 +3

+1 -2+3 -2

-3 +1-3 +4

+5 -2+5 -8

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Oxidation State Method Oxidation State Method The Oxidation States Method is a means ofbalancing the more simpler redox reactions.

The steps for a variation on this method areas follows:

Assign oxidation numbers to all of the atoms in the equation. Identify the substance reduced (oxidizing agent) and determine the number of electrons gained.

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Identify the substance oxidized (reducing agent) and determine the number of electrons given off. Use coefficients to balance the atoms and electrons in the species reduced. Use coefficients to balance the atoms and electrons in the species oxidized. Use coefficients to balance the remaining atoms by inspection.

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Oxidizing And Reducing AgentsOxidizing And Reducing AgentsThe terminology for oxidation-reduction(redox) can be somewhat confusing.

The oxidizing agent is the species (atom orion) being reduced.

In the following example, elemental chlorine

is said to be reduced. Cl2 → Cl-

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For a species to be reduced, its oxidation is reduced to a smaller number. Another way of saying the same thing is that the oxidation number of chlorine becomes less positive or more

negative. Oxidation-reduction always occur together. Because chlorine is reduced, it is called an oxidizing agent.

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When chlorine is reduced, it is bringing about the oxidizing of another species. Hence, chlorine is called an oxidizing agent. The balanced equation for chlorine is given by:

Cl2 + 2e- → 2Cl-

0 -10 -2-2

-1

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The previous equation shows the conservation of mass and charge.

In the following example, sodium is said to beoxidized.

Na → Na+ + e-

For a species to be oxidized, its oxidation number is increased to a larger number. Another way of saying the same thing is that the oxidation number of sodium becomes more positive or less negative.

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Because sodium is oxidized, it is called a reducing agent. When sodium is oxidized, it is bringing about the reduction of another species. Hence, sodium is called an reducing agent. The balanced equation for sodium is given by:

Na → Na+ + e-0 1 -1

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The previous equation shows the conservation of mass and charge.

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Summary Of Double AgentsSummary Of Double AgentsIn summary,

An oxidizing agent is reduced. A reducing agent is oxidized.

Atoms in Group I have a weak attraction fortheir valence and are easily oxidized.

Atoms in Group VII have a strong attractionfor their valence electrons and are easilyreduced.

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Single Replacement ReactionsSingle Replacement Reactions

A single replacement reaction is one in which

a free element becomes an ion and an ion in

solution becomes a neutral atom.

For example,

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Zn becomes Zn2+ and H+ becomes H2.

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Balancing the net ionic equation could bedone by inspection but we will use theion-electron method.

Zn + H2SO4 → ZnSO4 + H2

Zn + 2H+ + SO42- → Zn2+ + SO4

2- + H2

Zn + 2H+ + SO42- → Zn2+ + SO4

2- + H2

Zn → Zn2+ + 2e-

2H+ + 2e- → H2Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

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Alkali metals (Group I) and alkalineearth metals (Group II) are very active metals.

These metals are easily oxidized and are good reducing agents. Such active metals react with water (H2O(l)) or steam (H2O(g)) to replace the hydrogen from the water forming a metallic hydroxide and hydrogen gas. Sometimes water will be written as HOH to show that the first hydrogen is the one that gets replaced.

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Use the ion-electron method to write thebalanced net ionic equation for the reactionbetween sodium and water.

Na + HOH → NaOH + H2

Na + HOH → Na+ + OH- + H2

Na → Na+ + e-

H+ + e- → H2

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2Na → 2Na+ + 2e-

2H+ + 2e- → H2

Very unreactive metals such as Ag, Pt, andAu, are not found in nature chemicallycombined.

2Na(s) + 2H+(aq) → 2Na+(aq) + H2(g)

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When Is There A Reaction?When Is There A Reaction?One example of a single replacement

reactionis when one metal tries replacing another.

For example, what are the products of thefollowing reaction?

Cu + AgNO3 →

To determine the products, one can use thevalues from Standard Reduction PotentialTable.

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From the title, one expects to find the half-reactions to be written as reductions.

From the table, we find that:

Ag+ + e- → Ag E°red = 0.80 V

Cu2+ + 2e- → Cu E°red = 0.34 V

These E° values are measured with respect to the SHE (Standard

Hydrogen Electrode).

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Hydrogen is arbitrarily assigned an E°red = 0 V (where V is volts).

This is covered in much more detail in Electrochemistry.

For now, TP (Think Positive)!

Because E°red(Ag) > E°red(Cu) or E°red(Ag)

is more positive than E°red(Cu), the single

replacement takes place.

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The more formal approach gives the sameresult.

If the reaction was to take place, it would be:

Cu + AgNO3 → Ag + Cu(NO3)2 (unbalanced)

The two half-reactions would be:

Cu → Cu2+ + 2e- E°ox = -0.34 V 2Ag+ + 2e- → 2Ag E°red = 0.80

VCu(s) + 2Ag+(aq) →Ag(s) + Cu2+(aq) E° = 0.46 V

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Important fall out from this problem: Notice that the half-reaction for copper is reversed and the sign changes to negative.

This is necessary because Cu is required

to be oxidized in the reaction, not reduced.

You must multiply the Ag+ half-reaction by 2 so the electrons cancel out in the net ionic equation.

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A very important note is that you do not multiply the E° by 2 because voltage is an intensive property.

For the single replacement to occur, E° must be positive!

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When Is There A No Reaction?When Is There A No Reaction?Not all single replacements occur!

For example, what are the products of thefollowing reaction?

Ag + Cu(NO3)2 →

From the table, we find that:

Ag+ + e- → Ag E°red = 0.80 V

Cu2+ + 2e- → Cu E°red = 0.34 V

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Because E°red(Ag) > E°red(Cu) or E°red(Ag) is more positive than E°red(Cu), the

single replacement does not take place!

This reaction is significantly different.

Ag → Ag+ + e- Eox = -0.80 V

Cu → Cu2+ + 2e- Eox = -0.34 V

TP (Think Positive) still works!

The oxidation potential of copper is more positive than the silver.

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But the problem is that copper needs tobe reduced in this problem!

The more formal approach gives the sameresult.

If the reaction was to take place, it would be:

Ag + Cu(NO3)2 → Cu + AgNO3

The two half-reactions would be:

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2Ag → 2Ag+ + 2e-

Because E° < 0, the reaction does not occur!

E°red = 0.34 VE°ox = -0.80 VE° = -0.46 V

Cu2+ + 2e- → Cu

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When Is There A Reaction?When Is There A Reaction?Another example of a single replacementreaction is when one halogen tries replacinganother.

For example, what are the products of thefollowing reaction?

Cl2 + NaBr →

To determine the products, one can use thevalues from Standard Reduction PotentialTable.

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From the table, we find that:

Cl2 + 2e- → 2Cl- E°red = 1.36 V

Br2 + 2e- → 2Br E°red = 1.07 V

Because E°red(Cl2) > E°red(Br2) or E°red(Cl2) is more positive than

E°red(Br2), the single replacement takes place.

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The more formal approach gives the sameresult.

If the reaction was to take place, it would be:

Cl2 + NaBr → Br2 + NaCl (unbalanced)

The two half-reactions would be:

Cl2 + 2e- → 2Cl- E°red = 1.36 V

2Br- → Br2 + 2e- E°red = -1.07

VCl2(g) + 2Br-(aq) →Br2(l) + 2Cl-(aq) E° = 0.46 V

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When Is There A No Reaction?When Is There A No Reaction?For example, what are the products of thefollowing reaction?

I2 + LiF →

From the table, we find that:

F2 + 2e- → 2F- E°red = 2.87 V

I2 + 2e- → 2I- E°red = 0.53 V

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Because E°red(F2) > E°red(I2) or E°red(F2) is more positive than E°red(I2), the single replacement does not take place!

This reaction is significantly different.

F2 + 2e- → 2F- E°red = 2.87 V

I2 + 2e- → 2I- E°red = 0.53 V

TP (Think Positive) still works!

The oxidation potential of fluorine is more positive than the iodine.

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But the problem is that fluorine needs tobe oxidized in this problem!

The more formal approach gives the sameresult.

If the reaction was to take place, it would be:

I2(s) + 2LiF(aq) → F2(g) + 2LiI(aq)The two half-reactions would be:

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I2 + 2e- → 2I-

Because E° < 0, the reaction does not occur!

E°ox = -2.87 VE°red = 0.53 VE° = -2.34 V

2F- → F2 + 2e-

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Summarizing SRP Rxns For Halogens Summarizing SRP Rxns For Halogens The Standard Reduction Potentials fornonmetals measures the ability of a

nonmetalto accept valence electrons.

The more reactive nonmetals have a greater

tendency to gain an electron or be reduced. Fluorine with its highest electronegativity (4.0) is the most reactive halogen and is the strongest oxidizing agent.

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Iodine has the smallest tendency to gain

electrons because it has the largest radius and smaller electronegativity.

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Balance Rxns Using Oxidation Balance Rxns Using Oxidation NumbersNumbers

When the equations become somewhat moredifficult to balance by inspection, use thefollowing rules.

Assign oxidation numbers to all elements. Identify the element that is reduced and write its half-reaction. Identify the element that is oxidized and write its half-reaction.

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Balance and add the half-reactions so the electrons cancel. Use the resulting coefficients as coefficients in the original equation. Balance the remaining atoms by inspection.

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More Interesting Equations To BalanceMore Interesting Equations To BalanceBalance the following reaction using theoxidation-number method.

HNO3 + H2S → NO + S + H2O

HNO3 + H2S → NO + S + H2O1 -251 -65

1 -2-21

2-2 0 1 -22 -2

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S → S + 2e-

N + 3e- → N

3S → 3S + 6e-

2N + 6e- → 2N

-2 0

5 2

-2 0

5 2

3S + 2N → 3S + 2N-2 5 0 2

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The sum of the reactant oxidation numbers is+4.

The sum of the product oxidation numbers is+4.

Using the coefficients of the net equation inthe original equation gives:

2HNO3 + 3H2S → 2NO + 3S + H2O

2HNO3(aq) + 3H2S(aq) → 2NO(g) + 3S(s) + 4H2O(l)

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Another Interesting ProblemAnother Interesting Problem

Potassium dichromate and hydrochloric acid

react to produce potassium chloride,chromium(III) chloride, water and chlorine.Write the balanced equation for this

reaction.

K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2

1 6 -2 1-1 1 -1 3 -1 1 -2 02 12-14 1-1 1 -1 3 -3 1 -2 0

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Cl is being oxidized. Cr is being reduced.

Cl- → Cl2 Cr2 → Cr3+

2Cl- → Cl2 Cr2 → 2Cr3+

2Cl- → Cl2 + 2e- Cr2 + 6e- → 2Cr3+

6Cl- → 3Cl2 + 6e-

Cr2 + 6e- → 2Cr3+6Cl- + Cr2 → 3Cl2 + 2Cr3+

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Using the coefficients from the net equationgives,

K2Cr2O7 + 6HCl → KCl + 2CrCl3 + H2O + 3Cl2and balancing by inspection gives:

K2Cr2O7(aq) + 14HCl(aq) → 2KCl(aq) + 2CrCl3(aq) + 7H2O(l) +

3Cl2(g)