Chapter 4 Reactions in Aqueous Solutions

Some typical kinds of chemical reactions:

1. Precipitation reactions:

the formation of a salt of lower solubility causes the precipitation to occur. precipr

2. Acid Base reactions:

the formation of water which is quite stable is a driving force for acid base chemistry.

3. Oxidation Reduction Reactions (reactions where electrons are gained and lost)

the driving force for most reactions including oxidation reduction reactions is the drive to lower the potential energy of the system (that is to convert potential energy to kinetic energy usually in the form of heat) *cca1 glycerine; thermite

Why do these reactions take place?

Electrolyte: the ability of a substance to form ions and conduct electricity

Strong electrolytes:

a substance when dissolved (usually in water) completely (or nearly so), dissociates into ions : MX +H2O = M+ + X- + H2O

ex. NaCl, HCl, H2SO4

Weak electrolytes:

a substance when dissolved (usually in water) partially dissociates into ions : MX +H2O = M+ + X- + MX + H2O

ex. acetic acid (vinegar, HC2H3O2), HF

Non-electrolytes:

a substance when dissolved (usually in water) does not dissociate at all: MX +H2O = MX + H2O

ex. sugar dvd MF:\Media_Assets\Chapter04\\StrongandWeakElectrolytes

Net Ionic Reactions

Electrical neutrality requires the presence of both a cation (+) and anion to be present whenever we deal with any substance. However, some ions, either the cation (+) or anion (-) may merely be spectators in the chemical reaction that occurs but does not appear to play a role in the reaction. In writing net ionic reactions, these ions are removed from the equation if they do not undergo any significant change as a result of the reaction.

Example:

NaOH(aq) + HCl(aq) = NaCl (aq) + H2O

Na+ + OH- + H+ + Cl- = Na+ + Cl- + H2O

nie (net ionic eq) OH- + H+ = H2O

Example:

Bi(OH)3 + HCl(aq) = BiCl3 + H2O

Bi(OH)3 + 3HCl(aq) = BiCl3 + 3H2O

Bi(OH)3(s) + 3H+ + 3Cl- = Bi+3 + 3Cl- + 3H2O

Bi(OH)3(s) + 3H+ = Bi+3 + 3H2O

What do you need to know?

1. Bi(OH)3 is an insoluble precipitate

2. HCl and BiCl3 are strong electrolytes

3. H2O is also mainly in the form of H2O, not H+ + OH-

Solubility Rules:

1. Cations: a compound is probably soluble if it contains the following cations: alkaki metals ( Li+, Na+, K+, Rb+ Cs+).

2. Anions: a compound is probably soluble if it contains the following anions: halogens (except for Ag+, Pb+2, and Hg2

+2 ) nitrate (NO3

-), perchlorate (ClO4-), acetate(C2H3O2

-), sulfate (SO4-2) (except

Ba+2, Hg2+2 and Pb+2 sulfates.

Most other cation-anion combinations form insoluble salts.

Most soluble salts are strong electrolytes

Balance and write net ionic equations in water for each of the following:

NiCl2 + NaOH = Ni(OH)2↓ + NaCl

NiCl2 + 2NaOH = Ni(OH)2↓ + 2NaCl

Ni+2 + 2 OH- = Ni(OH)2↓

AlCl3 + NaOH = NaAl(OH)4 + NaCl

AlCl3 + 4NaOH = NaAl(OH)4 + 3NaCl

Al+3 + 4OH- = Al(OH)4-

KOH + HC2H3O2 (vinegar) = KC2H3O2 + H2O

OH- + HC2H3O2 (vinegar) = C2H3O2- + H2O

AgNO3 + NaCl = AgCl ↓ + NaNO3

Ag+ + Cl- = AgCl

Solubility Rules: 1. Cations: a compound is probably soluble if it contains the

following cation: alkaki metals ( Li+, Na+, K+, Rb+ Cs+).2. Anions: a compound is probably soluble if it contains the

following anions: halogens (except for Ag+, Pb+2, and Hg2+2 )

nitrate (NO3-), perchlorate (ClO4

-), acetate(C2H3O2-), sulfate (SO4

-2) (except Ba+2, Hg2

+2 and Pb+2 sulfates).Most other cation-anion combinations form insoluble salts.Are the following soluble?

K2CrO4ZnCl2Pb(NO3)2Ag2SO4Ca(NO3)2BaSHgSNa2S

Relative reactivity of metals

Na + H2O = NaOH + H2; this reaction occurs with the elements in first two columns and with Al, Mn, Zn, Co, Ni, Sn.

How can we determine which is most and which is least reactive?

Metals

Highly Active

Potassium, K

Lithium, Li

Barium, Ba

Calcium, Ca

Sodium, Na

Magnesium, Mg

Aluminum, Al

Zinc, Zn

Iron, Fe

Nickel, Ni

Tin, Sn

Lead, Pb

Hydrogen, H2

Copper, Cu

Mercury, Hg

Silver, Ag

M + H2O 2MOH +H2

M+ + H2 M + 2 H+

2Al(s) + 3ZnCl2(l) = 2AlCl3(l) + 3Zn(s) ?

Metals

Highly Active

Potassium, K

Lithium, Li

Barium, Ba

Calcium, Ca

Sodium, Na

Magnesium, Mg

Aluminum, Al

Zinc, Zn

Iron, Fe

Nickel, Ni

Tin, Sn

Lead, Pb

Hydrogen, H2

Copper, Cu

Mercury, Hg

Silver, Ag

Oxidation and ReductionOxidation: the process by which an element or group of elements loose electronsReduction: the process by which an element or group of elements gain electronsWhat is meant by the term “agent”?

a facilitatorIn order to maintain electrical neutrality, for every electron lost by an element, there must be a gain of an electron by some other reactant. The oxidizing agent is the agent responsible for the loss of electrons. In the process the oxidizing agent get reducedThe agent that looses electrons causes something else to gain electrons and therefore is the agent responsible for reductionOxidizing agent is reducedReducing agent is oxidized

How do I identify an oxidation reduction reaction?

1. Rules in Assigning oxidation states:

All elements are in an oxidation state = 0

Metals usually get oxidized (become cations), non-metals usually get reduced (become anions)

Typical oxidation states

Alkali metals +1 Halogens -1

Alkaline earths +2 Group 6A -2

Group 3A +3 Group 5A -3

H can be –1 or +1

Some typical oxidation reduction reactions

1. Oxidation of “paper”:

C6H12O6 + 6O2 = 6CO2 + 6H2O

2. KMnO4 + C3H8O3 = CO2 + Mn2O3 + K2CO3

3. 2Al + Fe2O3 = Al2O3 + 2 Fe

How do we know that in these reactions, there have been loss andgain of electrons?

2Al + Fe2O3 = Al2O3 + 2 Fe

Aluminum metal is neutralIn Al2O3, Al = +3

Iron metal is neutralIn Fe2O3, Fe = +3

Oxygen remains -2 on both sides

Notice that this reaction could be balanced by mass balance alone.

What is the problem balancing oxidation-reduction reactions by mass balance only?

Let balance this reaction only with regards to mass

Cu + HNO3(aq) → NO2(g) + H2O + Cu(NO3)2 (aq)

Cu + 3HNO3 = Cu(NO3)2 + NO2 + H2O + H+

A reaction that creates or destroys charge needs to be balanced by taking into account electron balance as well as mass balance.

How do you know if mass balancing will not work?

Charge will be created or destroyed by mass balance alone or

you will need to form other products to balance the reaction

Balancing Oxidation and Reduction Reactions

Two steps are involved in balancing oxidation-reduction reactions.

Step 1: First, it is important to balance the loss and gain in electrons

Step 2: Second, it is important to achieve mass balance

How do I identify an oxidation reduction reaction that requires both charge and mass balance?

If charge is created or destroyed when you mass balance an equation, or other products need to be produced, then you have an oxidation reduction equation that requires balancing both charge and mass

Balancing Oxidation and Reduction Reactions

Two steps are involved in balancing oxidation-reduction reactions when the charge on either side of the equation is uneven.

Step 1: First, it is important to balance the loss and gain in electrons

Step 2: Second, it is important to achieve mass balance

What do I do first?

1. Assign oxidation numbers

Assign oxidation states for each of the element in the following:H2SO4

H = +1; O = -2; S = +6H3PO4

H = +1; O = -2; P = +5HClO4

H = +1; O = -2; Cl = +7ZnS

Zn = +2; S = -2HNO3

H = +1; O = -2; N = +5Cr2O7

-2

Cr = +6; O =-2MnO4

-1

Mn = +7; O = -2MnO2

Mn = +4; O = -2C6H12O6

C = 0; H = +1; O = -2H2O2

H = +1; O = -1

What is important in assigning oxidation states is not what oxidation state you give an element but how that state for the element compares on both side of the equation.

An element with the same oxidation state on both sides of the equation is not involved in electron transfer and can be ignored

Cl2 0

Balancing Oxidation and Reduction Reactions

Two steps are involved in balancing oxidation-reduction reactions.

Step 1: First, it is important to balance the loss and gain in electrons

Step 2: Second, it is important to achieve mass balance

What do I do first?

1. Assign oxidation numbers

2. Determine what is oxidized and what is reduced

Which is the reducing agent (what is oxidized)?

Cu + HNO3 = Cu(NO3)2 + NO2

Cu = 0; Cu+2 Cu is oxidized;

Cu → Cu+2 + 2e- (note that the reaction is overall neutral)

Which is the oxidizing agent (gets reduced)?

In HNO3, N = +5; NO2 , N = +4

e- + HNO3 → NO2 + OH- (the reaction is overall neutral)

N is reduced; note that the charge on oxygen is still -2,

hydrogen is still +1

Balancing Oxidation and Reduction Reactions

Two steps are involved in balancing oxidation-reduction reactions.

Step 1: First, it is important to balance the loss and gain in electrons

Step 2: Second, it is important to achieve mass balance

What do I do first?

1. Assign oxidation numbers

2. Determine what is oxidized and what is reduced

3. Mass balance the oxidation half reaction; mass balance the reduction half reaction

4. Combine the two half reactions; for reactions taking place in H2O, it is permissible to break up water to form OH- and H+ as necessary or to form water from OH- and H+.

Cu + HNO3 = Cu(NO3)2 + NO2

Cu → Cu+2 + 2e-

e- + HNO3 → NO2 + OH-

Cu → Cu+2 + 2e-

2e- + 2HNO3 → 2NO2 + 2OH-

Cu + 2HNO3 → 2NO2 + Cu(OH)2

In acid medium:

Cu(OH)2 + 2 HNO3 = Cu(NO3)2 + 2 H2O

So:

Cu +4 HNO3 = Cu(NO3)2 + 2 H2O + 2 NO2

Balanced equation

Cu(s) + 4HNO3 (aq)→ 2NO2(g) + 2H2O (l) + Cu(NO3)2(aq)Net ionic equation

Cu (s); NO2 (gas)

HNO3 exists in H2O as H+ + NO3-

Cu(NO3)2 is a strong electrolyte and water soluble

Cu(s) + 4H+(aq) + 2NO3-(aq) → 2NO2(g) + 2H2O(l) + Cu+2(aq)

Balance the following equations:

Fe(CN)6-3 (aq) + N2H4 (aq) = Fe(CN)6

-4 (aq) + N2(g)

Oxidation

H = +1; N in N2H4 is -2; in N2: N = 0

Reduction

Fe in Fe(CN)6-3 = +3; Fe(CN)6

-4 = +2

N2H4 (aq) = N2(g) + 4 H+ + 4 e-

4e- + 4Fe(CN)6-3 = 4Fe(CN)6

-4

N2H4 (aq) + 4Fe(CN)6-3(aq) = 4Fe(CN)6

-4(aq) + N2(g) + 4 H+(aq)

Let’s look at an oxidation reduction reaction that can be balanced by mass balance

CH4 + 2O2 = CO2 + 2H2O

Let’s assign H as H+1 then C is C-4 both in CH4 and in C+4 in CO2

CH4 = CO2 + 8 e-1 + 4H+

2O20 + 8 e-1 = 4O-2

CH4 + 2O2 = CO2 + 4H+ + 2O-2 = CO2 + 2 H2O

If we assign H as H-1, then C is C+4 in both CH4 and CO2

C+4H4 = C+4 + 4H+ +8 e-1

4H-1 goes to 4H+1 + 8 e-1

2O2 + 8 e- = 2O-2

CH4 + 2O2 = CO2 + 2H2O

PbO2 (s) + Mn+2 (aq) = Pb+2 (aq) + MnO4- (aq) in acid solution

PbO2, Pb = +4; Pb+2 reduction What’s the half reaction?

2e- + PbO2 (s) = Pb+2 (aq) +2O-2

Mn+2; MnO4-, Mn = +7; oxidation What’s the half reaction?

4O-2 + Mn+2 (aq) = MnO4- + 5 e-

10e- + 5PbO2 (s) = 5Pb+2 (aq) + 10 O-2

8O-2 + 2Mn+2 (aq) = 2MnO4- + 10e-

5PbO2 (s) + 2Mn+2 (aq) = 2MnO4- + 5Pb+2 + 2O-2

4H+(aq) + 5PbO2 (s) + 2Mn+2 (aq) = 2MnO4-(aq) + 5Pb+2(aq) + 2H2O

KMnO4 + C3H8O3 = CO2 + Mn2O3 + K2CO3 + H2O

MnO4-, Mn = +7 Mn2O3, Mn = +3

8e- + 2KMnO4 → Mn2O3 + 5O-2 + 2K+

C3H8O3, C = -2/3; CO2, C = +4

1C-2/3 = 1C+4 + 4.666 e- ; 3 C-2/3 = 3C+4 + 14e-

3O-2 + C3H8O3 → 3CO2 + 8H+ + 14e-

multiply 14*4 = 56; 7*8 = 56

56e- + 14KMnO4 → 7Mn2O3 + 35O-2 + 14K+

12O-2 + 4C3H8O3 → 12CO2 + 32H+ + 56e-

14 KMnO4+ 4C3H8O3→12CO2+14K+ +32H+ +7Mn2O3+ 23O-2 +

14KMnO4+4C3H8O3→12CO2+7K2O +16H2O +7Mn2O3

14KMnO4+4C3H8O3→5CO2+7K2CO3 +16H2O +7Mn2O3