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Chapter 4

Aqueous solutionsTypes of reactions pp

http://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&search=

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4.1 Water, the Common Solvent - Parts of Solutions

Solution - homogeneous mixture. Solute - what gets dissolved. Solvent - what does the dissolving. Soluble - Can be dissolved. Miscible - liquids dissolve in each other.

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Aqueous solutions Dissolved in water. Water is a good solvent

because the molecules are polar.

The oxygen atoms have a partial negative charge.

The hydrogen atoms have a partial positive charge.

The angle is 105º.

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Figure 4.1 The Water Molecule is Polar

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Hydration The process of breaking the ions of

salts apart. Ions have charges and attract the

opposite charges on the water molecules.

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Hydration

H HOH

H OH

HO

H HO

HHO

HH

O

HH

OH

H

O

HH

O

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Figure 4.2Polar Water Molecules Interact with the Positive and Negative Ions of a

Salt

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Solubility How much of a substance will dissolve

in a given amount of water. Usually in g/100 mL Solubility varies greatly, but if they do

dissolve the ions are separated, and they can move around. Water can also dissolve non-ionic

compounds if they have polar bonds.

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Figure 4.3Polar Bond

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4.2 Nature of Aqueous Solutions Strong and Weak Electrolytes

Electricity is moving charges. The ions that are dissolved can move. Solutions of ionic compounds can

conduct electricity. Called “electrolytes”. Solutions are classified three ways.

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Types of solutions Strong electrolytes- completely

dissociate (fall apart into ions). Many ions- Conduct well. Weak electrolytes- Partially fall apart

into ions. Few ions -Conduct electricity slightly. Non-electrolytes- Don’t fall apart. No ions- Don’t conduct.

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Figure 4.5BaCI2 Dissolving

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Types of solutions Acids - form H+ ions when dissolved. Strong acids fall apart completely. Many ions H2SO4 HNO3 HCl HBr HI HClO4 Weak acids- don’t dissociate completely. Bases - form OH- ions when dissolved. Strong bases - many ions. KOH, NaOH, Ca(OH)2

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Figure 4.6HCI (aq) is Completely Ionized

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Figure 4.7An Aqueous Solution of Sodium Hydroxide

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Figure 4.8Acetic Acid in Water

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Figure 4.9The Reaction of NH3 in Water

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4.3 Composition of Solutions Concentration - how much is dissolved.

Molarity = Moles of solute Liters of solution

abbreviated M 1 M = 1 mol solute / 1 liter solution Calculate the molarity of a solution with

34.6 g of NaCl dissolved in 125 mL of solution. Steps . . .

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4.3 Composition of Solutions pp Calculate the molarity of a solution with 34.6 g

of NaCl dissolved in 125 mL of solution. Convert grams to moles. How? Convert mL

to L. How? Divide moles by liters (34.6 g)(1 mol)

(58.5 g)(0.125 L) = 4.73 M

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Molarity How many grams of HCl (M = 36.5)

would be required to make 50.0 mL of a 2.7 M solution? Ans. . .

(0.050 L)(2.7 moles/L)(36.5 g/mole) = 4.9 g HCl

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Molarity What would the concentration of CaCl2 be

if you used 27g of CaCl2 (M = 111) to

make 500. mL of solution? Ans. . . (27 g)(1 mole/110.986)(1/0.500L) = 0.49 M

(about 0.50 molarity) What is the approximate concentration of

each ion? Ca2+ = 0.50 M; Cl1- = 1.00 M

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Molarity Calculate the concentration of a solution

made by dissolving 45.6 g of Fe2(SO4)3

to a total volume of 475 mL. (M of Fe2(SO4)3 = 400. g/mol) Ans. . .

(45.6 g)(1 mole/400. g)/.475 L = 0.240 M What is the concentration of each ion? Fe = 0.48 M; SO4 = 0.72 M

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Figure 4.10 Preparation of a Standard Solution

a.a. Weighed amount of solute is put into Weighed amount of solute is put into volumetric flask and small quantity of volumetric flask and small quantity of H2O addedH2O added

b.b. Solid dissolved by gently swirling Solid dissolved by gently swirling with with stopper in placestopper in place

c.c. More water added until level just More water added until level just reaches mark on neck (d)reaches mark on neck (d)

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Making solutions Describe how to make 100.0 mL of a 1.0 M

K2CrO4 solution.

(1.0 mole/L)(246 g/mole)(0.100 L) = dissolve 24.6 g in 85 ml distilled water; qsad to 100 ml

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Making solutions Describe how to make 250. mL of an 2.0 M

copper(II) sulfate dihydrate solution. What is formula for copper(II) sulfate dihydrate? CuSO4•2H2O (M = 196)

What is the answer? . . . (0.250L)(2.0 mole/L)(196 g/mole) = dissolve

98 g in 200 ml distilled water; qsad to 250 ml

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Dilution Adding more solvent to a known solution. The moles of solute stay the same. moles = M x L M1 V1 = M2 V2 moles = moles Stock solution is a solution of known

concentration used to make more dilute solutions

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Dilution What volume of a What volume of a 1.71.7 MM solution is needed to solution is needed to

make 250. mL of a 0.50 make 250. mL of a 0.50 MM solution? Steps. solution? Steps. VV11MM11 = = VV22MM22 ((xx)()(1.71.7) = () = (250.250.)()(0.500.50) ) So, x = 74 mL, (rounded from 73.5 mL) of the So, x = 74 mL, (rounded from 73.5 mL) of the

1.7 1.7 MM solution combined with enough solution combined with enough distilleddistilled HH220 to make a 0 to make a totaltotal volume of 250. mL volume of 250. mL

Be careful! If you just add it to 250. mL then you Be careful! If you just add it to 250. mL then you get a volume of 323 mL and this throws off your get a volume of 323 mL and this throws off your molarity.molarity.

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Dilution 18.5 mL of 2.3 M HCl is diluted to 250 mL of

water. What’s solution concentration? Ans. (18.5)(2.3) = 250(x) ; x = 0.17 0.17 MM 18.5 mL of 2.3 M HCl is added to 250 mL of

water. Solution concentration? (Tricky) Find moles of HCl

(0.0185 L)(2.3 moles/L) = 0.0426 moles; then M = 0.0426 moles/(0.250 + 0.0185 L) = 0.16 0.16 MM (compare with above)

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Fig. 4.11 p. 145

(a)A Measuring Pipet is graduated & can measure various volumes accurately.

(b)A Volumetric (transfer) Pipet measures one volume accurately.

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More dilution (if time) You have a 4.0 M stock solution. Describe

how to make 1.0L of a .75 M solution. (1)(.75) = 4(x); x = 190 ml diluted with

distilled H20 to 1.0 L

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More dilution (if time) 25 mL 0.67 M of H2SO4 added to 35 mL of

0.40 M CaCl2 . What mass CaSO4 is formed?

Steps. . . 1st get moles of both (moles = V • M) 0.025 L • 0.67 M = 0.017 mol H2SO4;

0.035 L • 0.40 M = 0.014 moles CaCl2 Then, need to find limiting reactant (2 stoichs)

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More dilution (if time) 25 mL 0.67 M of H2SO4 added to 35 mL of 0.40

M CaCl2 . What mass CaSO4 formed?

0.01675 mol H2SO4 and 0.014 moles CaCl2 Then, need to find limiting reactant (2 stoichs) H2SO4 + CaCl2 CaSO4 + 2HCl

CaCl2 is LR and the mole ratio to CaSO4 = 1:1 (0.014 mol CaCl2)(1 mol CaSO4)(136 g CaSO4) = 1.9 g CaSO4

(1 mol CaCl2)(1 mol CaSO4)

IF this were NOT a ppt, then conc = 0.014 moles/(0.025 + 0.035) = 0.23 M

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4.4 Types of Reactions Precipitation reactions When aqueous solutions of ionic

compounds are poured together a solid forms.

A solid that forms from mixed solutions is a precipitate

If you’re not a part of the solution, your part of the precipitate

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4.5 Precipitation reactions NaOH(aq) + FeClNaOH(aq) + FeCl33(aq) (aq) NaCl(aq) + Fe(OH)NaCl(aq) + Fe(OH)33(s)(s)

is really . . . is really . . . NaNa++(aq) + OH(aq) + OH--(aq) + Fe(aq) + Fe+3+3(aq) + Cl(aq) + Cl--(aq) (aq) NaNa++(aq) (aq)

+ Cl+ Cl--(aq) + Fe(OH)(aq) + Fe(OH)33(s)(s)

So all that really happens is . . .So all that really happens is . . . NaNa++(aq) + OH(aq) + OH--(aq) + Fe(aq) + Fe+3+3(aq) + Cl(aq) + Cl--(aq) (aq) NaNa++(aq) (aq)

+ Cl+ Cl--(aq) + Fe(OH)(aq) + Fe(OH)33(s)(s)

OHOH--(aq) + Fe(aq) + Fe+3+3(aq) (aq) Fe(OH) Fe(OH)33(s)(s)

Then, balance: Then, balance: 33OHOH--(aq) + Fe(aq) + Fe+3+3(aq) (aq) Fe(OH) Fe(OH)33(s)(s) Double replacement reactionDouble replacement reaction

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Precipitation reaction We can predict the products Can only be certain by experimenting The anion and cation switch partners AgNO3(aq) + KCl(aq) (molecularly) . AgCl + KNO3

Zn(NO3)2(aq) + BaCr2O7(aq) ZnCr2O7 + Ba(NO3)2

CdCl2(aq) + Na2S(aq) CdS + 2NaCl

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Precipitations Reactions Only happen if one of the products is

insoluble Otherwise all the ions stay in solution-

nothing has happened. Need to memorize the rules for solubility

(see, p. 150 or use solubility poem)

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Let’s Sing! Potassium, sodium and ammonium salts,

Whatever they may be,Can always be depended on for solubility.

When asked about the nitratesThe answer is always clear,

They each and all are soluble, Is all we want to hear.

Most every chloride’s solubleAt least we’ve always read

Save silver, mercurous mercuryAnd (slightly) chloride of lead

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Let’s Sing! Every single sulfate

Is soluble, ‘Tis said‘Cept barium and strontium

And calcium and lead.

Hydroxides of metals won’t dissolveThat is, all but threePotassium, sodium and ammoniumDissolve quite readily

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Let’s Sing! And then you must remember

That you must not “forgit”Calcium, barium, strontiumDissolve a little bit.

The carbonates are insoluble,It’s lucky that it’s so,

Or else, our marble buildingsWould melt away like snow.

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Repeat with feeling!!

Potassium, sodium, and ammonium saltsWhatever they may be

Can always be depended onFor solubility

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Solubility Rules All nitrates are soluble

Alkali metals ions and NH4+ ions are

soluble Halides are soluble except Ag+, Pb+2,

and Hg2+2

Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2

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Solubility Rules Most hydroxides are slightly soluble

(insoluble) except NaOH and KOH Sulfides, carbonates, chromates, and

phosphates are insoluble Lower number rules supersede so Na2S

is soluble

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4.6 Describing Reactions in Solution

Molecular Equation - written as whole formulas, not the ions.

K2CrO4(aq) + Ba(NO3)2(aq) NO3(aq) + BaCrO4 (s)

Complete Ionic equation - show dissolved electrolytes as the ions.

2K+ + CrO4-2 + Ba+2 + 2 NO3

- BaCrO4(s) + 2K+ + 2 NO3

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Net Ionic equation - includes only solution components undergoing a change. Spectator ions are those that don’t react.

CrO4-2 + Ba+2 BaCrO4(s)

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Three Type of Equations Write the three types of equations for

the reactions when these solutions are mixed. (Whiteboard)

iron (III) sulfate and potassium sulfide lead (II) nitrate and sulfuric acid.

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4.7 Stoichiometry of Precipitation

pp

Exactly the same, except you may have to figure out what the pieces are.

What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? . . .

Strategy: need to convert both reactants to moles, then determine LR then use stoich.

Have 0.01 moles of both BaCl2 + 2NaOH Ba(OH)2 + 2NaCl

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BaCl2 continued pp

What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide?

Have 0.010 moles of bothBaCl2 + 2NaOH Ba(OH)2 +

2NaCl NaOH is limiting & makes 0.005 mol Ba(OH)2

0.005 mol BaCl2 • 208 g/mol = 0.85 g formed

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Another Stoich of Precipitation problem What volume of 0.204 M HCl is needed to

precipitate the silver from 50.ml of 0.0500 M silver nitrate solution?

HCl + AgNO3 AgCl + HNO3

50.ml of 0.0500 M AgNO3 = 0.0025 moles Mole ratio = 1:1; need 0.0025 mol HCl also Since M = mol/vol; vol = mol/M Vol = 0.0025 ÷ 0.204 = 0.0122 L = 12.2 mL

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Types of Reactions4.8 Acid-Base

For our purposes an acid is a proton donor.

A base is a proton acceptor (usually OH) What is the net ionic equation for the

reaction of HCl(aq) and KOH(aq)? Acid + Base salt + water H+ + OH- H2O

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4.8 Acid - Base Reactions Often called a neutralization reaction because Often called a neutralization reaction because

the acid neutralizes the base.the acid neutralizes the base. Often Often titratetitrate to determine concentrations. to determine concentrations. Solution of known concentration (titrant) . . .Solution of known concentration (titrant) . . . is added to the unknown (analyte) . . .is added to the unknown (analyte) . . . until the equivalence point is reached where until the equivalence point is reached where

enough titrant has been added to neutralize it. enough titrant has been added to neutralize it.

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Titration pp

Where indicator changes color is Where indicator changes color is endpointendpoint.. Not always at the Not always at the equivalenceequivalence point. point.

A 50.00 mL sample of aqueous Ca(OH)A 50.00 mL sample of aqueous Ca(OH)22

requires 34.66 mL of 0.0980 requires 34.66 mL of 0.0980 MM Nitric acid for Nitric acid for

neutralization. What is [Ca(OH)neutralization. What is [Ca(OH)22 ]? ]? # of H# of H++ x x MMA A x Vx VA A = = # of OH# of OH-- x x MMB B x Vx VBB

(look at formulas, not at the balanced reaction)(look at formulas, not at the balanced reaction)

(1)(0.0980)(34.66) = (2)(50)(x) = 0.0340 (1)(0.0980)(34.66) = (2)(50)(x) = 0.0340 MM but look at Zumdahl method page 150!but look at Zumdahl method page 150!

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Acid-Base Reaction pp

75 mL of 0.25 75 mL of 0.25 MM HCl is mixed with 225 mL of 0.055 HCl is mixed with 225 mL of 0.055 MM Ba(OH) Ba(OH)22. What is the concentration of the . What is the concentration of the excessexcess HH++ or OH or OH-- ? We can do this in mmoles since the ? We can do this in mmoles since the units will cancel. Use mL and mmoles is same as L units will cancel. Use mL and mmoles is same as L and moles. Steps. . .and moles. Steps. . .

Mmole HCl = 75 mL • 0.25 Mmole HCl = 75 mL • 0.25 MM = 18.75 mmole; = 18.75 mmole; Since 1:1 mole ratio of HSince 1:1 mole ratio of H++ : HCl; H : HCl; H1+1+ = = 18.75 mmole18.75 mmole Mmole Ba(OH)Mmole Ba(OH)22 = 225 mL • 0.055 M = 12.37 mmole = 225 mL • 0.055 M = 12.37 mmole 2:1 mole ratio OH2:1 mole ratio OH-- : Ba(OH) : Ba(OH)22 ; OH ; OH-- = = 24.75 mmole24.75 mmole H : OH mole ratio = 1:1; so 6.00 mmoles xs OH in H : OH mole ratio = 1:1; so 6.00 mmoles xs OH in

(75 + 225) ml = 0.0200 (75 + 225) ml = 0.0200 MM OH OH1-1-

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Types of Reaction4.9 Oxidation-Reduction called Redox

Ionic compounds are formed through the transfer of electrons.

An Oxidation-reduction reaction involves the transfer of electrons.

We need a way of keeping track.

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Oxidation States - Memorize!A way of keeping track of the electrons.Not necessarily true of what is in nature,

but it works (i.e., a model only).Need the rules for assigning (memorize).

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Oxidation States - Memorize!Elements & Ions

The oxidation state of elements in their standard states is zero.

Na(s) = 0; Cl2(g) = 0, but O2(s) ≠ 0!Oxidation states for monoatomic ions are

the same as their charge.Na1+ = 1; Cl1- = 1-; N3- = 3-

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Oxidation states - memorize!

•Oxygen & HydrogenOxygen is assigned an oxidation state of

-2 in its covalent compounds (H2O)except as peroxide (H2O2) where = -1 or with fluorine (OF2) where = +2

Hydrogen in compounds with: nonmetals is assigned +1 (H2O)with metals it is -1 (CaH2)

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Oxidation states - memorize! fluorine is always -1 in compounds (HF).The sum of the oxidation states must be:

zero in compounds (Na2SO4) or equal the charge of the ion (SO4

2-)

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Tr103 Assigning Oxidation NumbersWhat is the oxidation # of Mn in KMnO4?

1•1 + 1•Mn + 4•-2 = 0; Mn = +7

What is the oxidation # of Cr in BaCr2O7?

1•2 + 2•Cr + 7•-2 = 0; Cr = +6

What is the oxidation # of Cl in Sr(ClO4)2?

1•2 + 2*Cl + 8•-2 = 0; Cl = +7! Even though as an ion is would be Cl1-

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Assign the oxidation states to each element

CO2

NO3-1

H2SO4

Fe2O3

Fe3O4

C = +4; O = -2 N = +5; O = -2 H = +1; S = +6; O = -2 Fe = +3; O = -2 Fe = +2.67; O = -2 Remember: This isn’t real.

It’s only a model to help us.

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Oxidation-Reduction Electrons transfer, so oxidation states change. 2Na + Cl2 2NaCl

CH4 + 2O2 CO2 + 2H2O Oxidation is the loss of electrons (an increase

in the oxidation state). Reduction is the gain of electrons (a decrease

in the oxidation state). OIL RIG LEO GER Coefficients or subscripts do NOT change the

oxidation values.

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Oxidation-ReductionOxidation means an increase in oxidation

state - lose electrons.Na Na1+ + 1 e-

Reduction means a decrease in oxidation state - gain electrons.

2 e- + Cl2 2Cl1-

The substance that is oxidized is called the reducing agent.

The substance that is reduced is called the oxidizing agent.

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A Summary of an Oxidation-Reduction Process

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AgentsOxidizing agent - has the potential to

cause another substance to get oxidized.Other substance loses electrons, but . . . In causing that other substance to get

reduced the oxidizing agent gets reduced.The oxidizing agent gains electrons (gets

a more negative oxidation state).MnO4

1- will oxidize Fe2+ to Fe3+ so here it is an oxidizing agent.

But the Mn in MnO41- gets reduced from

+7 to +2 (MnO41- + 5e- Mn2+)

+7 +2

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AgentsReducing agent - has the potential to

cause another substance to get reduced.Other substance gains electrons, but . . . In causing that other substance to get

reduced the reducing agent gets oxidized.The reducing agent loses electrons (gets

a more positive oxidation state).Na will reduce Zn2+ to Zn so here it is a

reducing agent. But the Na gets oxidized from 0 to +1

(Na Na1+ + 1e-)

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Identify the Substance oxidized Substance

reduced Oxidizing agentReducing agent in the

following reactions:Fe(s) + O2(g) Fe2O3(s) oxidation #s?… 0 0 +3 -2 O, R, OA, RA? O R

RA OA

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Identify the O, R, OA, RA in the following reactions:Fe2O3(s)+ 3 CO(g) 2 Fe(s) + 3 CO2(g)Fe3+ reduced to Feo so Fe2O3 is the OA

C2+ oxidized to C4+ so CO is the RA. SO3

2- + H+ + MnO41- SO4

2- + H2O + Mn+2

Sulfur goes from +4 to +6 (from SO32- to

SO42-) so it is oxidized and its compound,

SO32- is the reducing agent which reduces Mn

in MnO41- from +7 to +2.

Mn is reduced & MnO41- is the Oxidizing

agent (it oxidizes S from +4 to +6).

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Half-ReactionsAll redox reactions can be thought of as

happening in two halves.One produces electrons - Oxidation half.Other requires electrons - Reduction half.We write the half-reactions to help us

balance the equation.Then we add the half-reactions to cancel

out the electrons.Every redox reaction has an oxidation

half-reaction & a reduction half-reaction.

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Half-ReactionsWrite the half reactions for the following.

Na + Cl2 Na+ + Cl- Steps. . . Na Na1+ + 1e- 2e- + Cl2 2Cl1-

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Half-Reactions (more steps)Write the half reactions for:

SO3-2 + H+ + MnO4

- SO4-2 + H2O + Mn+2 (steps …)

1st, identify the elements that are changing their oxidation numbers, they are . . .

S from +4 to +6 so it is losing 2 e- (it is being oxidized so it is the reducing agent)

Mn from +7 to +2 so it is gaining 5 e- (being reduced so it is the oxidizing agent).

SO3-2 + H+ + MnO4

- SO4-2 + H2O + Mn+2 +4

+7 +6 +2

So, the half reactions are the stuff with sulfur and manganese.

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Half-Reactions (more steps) SO3

-2 + H+ + MnO4- SO4

-2 + H2O + Mn+2 +4 +7 +6 +7 S

from +4 to +6 (losing) and . . . Mn from +7 to +2 (gaining)

The half-reactions are . . . SO3

-2 SO4-2 + 2 e-

+4 +6

5 e- + MnO4- Mn+2

+7 +2

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Not every reaction is redox Only reactions that change oxidation state (gain

or lose electrons) are redox. Is SO2 + H2O H2SO3 a redox? +4 -2 +1 -2 +1+4-2 No! Is H2 + CuO Cu + H2O a redox? 0 +2 -2 0 +1-2 Yes! (Discuss Rainbow Matrix )

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4.10 Balancing Redox Equations In aqueous solutions the key is the

number of electrons produced must be the same as those required.

For reactions in acidic solution: 8-step procedure.

Write separate half-reactions (assign oxidation numbers, temporarily delete substances that don’t change)

For each half reaction balance all reactants except H and O

Balance O using H2O

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Acidic SolutionBalance H using H+

Balance charge using e- Multiply equations to make electrons

equalAdd equations and cancel identical

species (cancel coefficients as needed)Check that charges and elements are

balanced.

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Example pp

A breathalyzer uses potassium dichromate to test for ethanol because the orange potassium dichromate changes to the green chromium 3+ ion in the presence of alcohol.

Write and balance the Breathalyzer reaction given the following:

The reactants are K2Cr2O7, HCl and C2H5OH.

The products are CrCl3, CO2, KCl & H2O.

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Example pp

Write and balance the Breathalyzer equation given the following: Reactants: K2Cr2O7, HCl and C2H5OH. Products: CrCl3, CO2, KCl & H2O.

Write the formula equation: K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O

Write the ionic equation (solubility poem): 2K1+ Cr2O7

2- + H1+ + Cl1- + C2H5OH Cr3+ 3Cl1- + CO2 + K1+ + Cl1- + H2O

Assign oxidation numbers 2K1+ Cr2O7

2- + H1+ + Cl1- + C2H5OH Cr3+ 3Cl1- + CO2 + K1+ + Cl1- + H2O +1 +6 -2 +1 -1 -2+1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2

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Example pp

Write the formula equation: K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O Write the ionic equation (solubility poem): 2K1+ Cr2O7

2- + H1+ + Cl1- +

C2H5OH Cr3+ 3Cl1- + CO2 + K1+ + Cl1- + H2O Assign oxidation numbers 2K1+ Cr2O7

2- + H1+ + Cl1- +

C2H5OH Cr3+ + 3Cl1- + CO2 + K1+ + Cl1- + H2O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2

Delete substances where no element changes its oxidation number:

2K1+ Cr2O72- + H1+ Cl1- + C2H5OH Cr3+ + 3Cl1- + CO2 + K1+ + Cl1- + H2O +1

+6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2

Only those where oxidation number changes : Cr2O72-

+ C2H5OH Cr3+ + CO2 +6 -2 +3 +4

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Balancing Redox Equations pp

K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O Only those where oxidation number changes :

Cr2O72- + C2H5OH Cr3+ + CO2

+6 -2 +3 +4

For reactions in acidic solution: 8-step procedure.

Write separate half-reactions

Cr2O72- Cr3+

C2H5OH CO2

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Balancing Redox Equations pp

Cr2O72- Cr3+ - C2H5OH

CO2

For each half-reaction balance all reactants except H and O

Cr2O72- Cr3+ C2H5OH

CO2

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Balancing Redox Equations pp

Cr2O72- Cr3+ - C2H5OH

CO2

Balance O using H2O

Cr2O72- Cr3+ + 7H2O

3H2O + C2H5OH CO2

Balance H using H1+

14H1+ + Cr2O72- Cr3+ + 7H2O 3H2O

+ C2H5OH CO2 + 12H1+

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Example pp

14H1+ + Cr2O72- Cr3+ + 7H2O 3H2O +

C2H5OH CO2 + 12H1+

Balance the charge using e-

6e- + 14H1+ + Cr2O72- Cr3+ + 7H2O 3H2O +

C2H5OH CO2 + 12H1+ + 12e- Multiply equations to equalize the charges

2(6e- + 14H1+ + Cr2O72- Cr3+ + 7H2O)

3H2O + C2H5OH CO2 + 12H1+ + 12e-

Get: 12e- + 28H1+ + 2Cr2O7

2- Cr3+ + 14H2O 3H2O + C2H5OH CO2 + 12H1+ + 12e-

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Example pp

12e- + 28H1+ + 2Cr2O72- Cr3+ + 14H2O

3H2O + C2H5OH CO2 + 12H1+ + 12e-

Add equations and cancel identical species (cancel coefficients if needed)

12e- + 1628H1+ + 2Cr2O72- Cr3+ + 1114H2O

3H2O + C2H5OH CO2 + 12H1+ + 12e-

•16H1+ + 2Cr2O72- + C2H5OH CO2 + Cr3+ + 11H2O

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Acidic Solution pp

16H1+ + 2Cr2O72- + C2H5OH CO2 + Cr3+ + 11H2O

Combine the net ions to form the original compounds and check that everything is balanced.

The original compounds were:K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O

So the balanced equation is . . .(write it out)2K2Cr2O7 + 16HCl + C2H5OH CrCl3 + 2CO2 + 4KCl + 11H2O

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Balance the following in acidic solution

MnO41- + Fe+2 Mn+2 + Fe+3 Ans. . .

8H1+ + MnO4- + 5Fe+2 Mn+2 + 5Fe+3 + 4H2O

Cu + NO31- Cu+2 + NO(g) Ans. . .

3Cu + 2NO31- + 8H1+ 3Cu+2 + 2NO(g) + 4H2O

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More Practice if time

The following reactions occur in aqueous solution. Balance them

Pb + PbO2 + SO4-2 PbSO4 (not done)

Mn+2 + NaBiO3 Bi+3 + MnO4- (not done)

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Now for a tough one Fe(CN)6

-4 + MnO4- Mn+2 + Fe+3 + CO2 + NO3

- . . .

188 H+ + 5 Fe(CN)64- + 61 MnO4

1- 94 H2O + 61 Mn2+ + 5 Fe3+ + 30 CO2 + 30 NO3

1-

85

Basic Solution pp

Do everything you would with acid, but we have to add a step because . . .

Because there is no H1+ in basic solution.So, add OH1- sufficient to convert the H1+

to water (OH1- + H1+ H2O)Be sure to add the OH1- to both sides of

the reaction (conservation of mass law) Cr(OH)3 + OCl- + OH- CrO4

-2 + Cl- + H2Owe’ll do this on the board to get . . .

2Cr(OH)3 + 3OCl- + 4OH- CrO4-2 + 3Cl- + 5H2O

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Basic Solution

After balancing like it was in acidic solution, add OH1- (to both sides) to cancel all the H1+ in basic solution.

You try CrO21- + ClO1- CrO4

1- + Cl1- in basic solution to get . . .

CrO21- + 2ClO1- CrO4

1- + 2Cl1-

Try NO21- + Al NH3 + AlO2

1- in basic solution to get . . .

OH1- + H2O + NO21- + 2Al NH3 + 2AlO2

1-

87

Basic Solution - more problems if time

Do everything you would with acid, but add one more step.

Add enough OH- to both sides to neutralize the H+ (by forming water)

CrI3 + Cl2 CrO4- + IO4

- + Cl-

Fe(OH)2 + H2O2 Fe(OH)-

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Redox Titrations Same as any other titration. The permanganate ion is used often

because it is its own indicator. MnO4- is

purple, Mn+2 is colorless. When reaction solution remains clear, MnO4

- is gone.

(Know for AP lab portion) Chromate ion (CrO4

1-) is also useful, but color change, orangish yellow to green, is harder to detect.

89

Redox Titration Example pp

The iron content of iron ore can be determined by titration with standard KMnO4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMnO4 solution, producing Fe+3 and Mn+2 ions in acidic solution.

If it requires 41.95 mL of 0.205 M KMnO4 to titrate a solution made with 6.128 g of iron ore, what percent of the ore was iron? Steps. . .

90

Example pp

Iron ore + xs HCl iron reduced to Fe+2. Then titrated w/ KMnO4

solution Fe+3 + Mn+2 ions in acidic solution. 41.95 mL of 0.205 M KMnO4 titrates 6.128 g of iron ore, what percent is

iron?

Write NR Fe2+ reacting with MnO41-

8H1+ + 5Fe2+ + MnO41- 5Fe3+ + Mn2+ +

4H20

Do stoich (calc moles MnO41- 1st) to get

0.0086 moles MnO41- and 0.043 moles Fe2+

Find g of Fe2+ = 0.043 mol x 55.847 g/mol = 2.40 g (2.40 g Fe/6.128 g Fe ore) x 100% = 39.2%