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General Chemistry IIGeneral Chemistry IICHEM 152 Unit 4CHEM 152 Unit 4

Week 13

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A 3.41 x 10-6 g sample of a compound is known to

contain 4.67 x 1016 molecules. This compound

is:

1. CO2

2. CH4

3. NH3

4. H2O

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Solution

A 3.41 x 10-6 g sample of a compound is known to contain 4.67 x 1016 molecules. This compound is?

Gameplan: Find the molecular weight (g/mole), then assess what compounds could have that MW

4.67 x 1016 molecules 1 mole = 7.65 10-8

6.02 1023 molecules moles

3.41 x 10-6 g = 44.0 g/mole

7.65 10-8 moles What molecule has this MW?

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A 3.41 x 10-6 g sample of a compound is known to contain 4.67 x 1016 molecules.

This compound is? 44.0 g/mole

1. CO2 44 g/mole

2. CH4 16 g/mole

3. NH3 17 g/mole

4. H2O 18 g/mole

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In a neutral COMPOUND some elements often have a fixed OXIDATION NUMBER

H almost always is +1O almost always is -2F always is -1

3) In a NEUTRAL COMPOUND the SUM of all the atomic oxidation numbers is ZERO

CH4

CH3COOH

? +1CH4 1C + 4H = 0

C + 4(+1) = -4

-4 +1 CH4

? +1 ? -2 -2 +1 CH3C O O H 2C + 4H + 2O = 0

2C + 4(+1) +2(-2) = 02C + 4 – 4 = 0

0 +1 0 -2 -2 +1 CH3 C O O H

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4) In a COMPLEX ION the SUM of all the atomic oxidation numbers is equal to the TOTAL CHARGE on the ion.

ClO3−

MnO4−

PO33-

? -2 ClO3

1Cl + 3O = -1Cl + 3(-2) = -1

+5 -2 ClO3

? -2 MnO4

1Mn + 4O = -1Mn + 4(-2) = -1

+7 -2 MnO4

? -2 PO3

3 1P + 3O = -3P + 3(-2) = -3

+3 -2 PO3

3

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What is the oxidation number of P in H3PO4?

P = +5What is the oxidation number

of Cr in Cr2O72-?

Cr = +6What is the oxidation number

of C in (NH4)2CO3?C = +4

What is the oxidation number of C in C3H8(propane)?

C = -2 2/3 or -8/3

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When the carbon-carbon bond lengths in ethane (C2H6),

ethene (C2H4), and benzene (C6H6) are arranged in order of

increasing bond length (shortest first), which is the

correct order?

1. C2H6 < C2H4 < C6H6

2. C2H4 < C2H6 < C6H6

3. C6H6 < C2H4 < C2H6

4. C2H4 < C6H6 < C2H6

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Propane again…

Using the structure method, what is the oxidation number of each carbon in propane (C3H8)?

C

C

C

H H

H H

HHH H

12

3

1 4 7 = -32 4 6 = -23 4 7 = -3

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What is the oxidation number of the oxygen-bonded C in

CH3COOH (acetic acid)?

CC

HH

O

HO—H

=

:

:

:

:

1

22 is the O-bonded CO.N. 4-1 = +3

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What is the oxidation number of the C in NH2CONH2? In

ClCH2CH2Cl?

H—N—C—N—H

H H

O=

: :

: :

H—C — C—H

H H

Cl Cl: : : : : :

O.N. 4-0 = +4both carbons are equivalentO.N. 4-5 = -1

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S8 + 12 O2 8 SO3

Is it redox?

If so,What is oxidized?What is the reducing agent?

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0 0 +6 -2S8 + 12 O2 8 SO3

Is it redox?

If so,What is oxidized?

0 +6Sulfur is oxidized0 -2Oxygen is reduced

What is the reducing agent? S8

What is the reducing agent? O2

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2 AgNO3 + Na2S Ag2S + 2 NaNO3

Is it redox?

5 As4O6 + 8 MnO4- + 18 H2O

20 AsO43- + 8 Mn2+ + 36 H+

What is oxidized and reduced?What is the oxidizing agent (OA) and reducing agent (RA)?

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NOT REDOX

+1 +5-2 +1 -2 +1 -2 +1 +5-22 AgNO3 + Na2S Ag2S + 2 NaNO3

Is it redox?

Na will always be +1 in ionic compoundsAg is a transition element, but because we know nitrate and sulfide, it is +1 in both

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Is it redox?

+3 -2 +7 -2 +1 -25 As4O6 + 8 MnO4

- + 18 H2O +5 -2 +2

+1 20 AsO4

3- + 8 Mn2+ + 36 H+

As is oxidizedMn is reducedAs4O6 is the reducing agentMnO4

- is the oxidizing agent

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Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored.

Fe(s) + Sn2+(aq)

O2(g) + Au(s)

Na(s) + H2O(l)

Al3+ (aq) + Hg(l)

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Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored.

Fe(s) + Sn2+(aq)

O2(g) + Au(s)

Na(s) + H2O(l)

Al3+ (aq) + Hg(l)

product favored

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Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored.

Fe(s) + Sn2+(aq)

O2(g) + Au(s)

Na(s) + H2O(l)

Al3+ (aq) + Hg(l)

reactant favored

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Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored.

Fe(s) + Sn2+(aq)

O2(g) + Au(s)

Na(s) + H2O(l)

Al3+ (aq) + Hg(l)

product favored

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Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored.

Fe(s) + Sn2+(aq)

O2(g) + Au(s)

Na(s) + H2O(l)

Al3+ (aq) + Hg(l)

reactant favored

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Calculate the Ecell for these reactions.

Fe(s) + Sn2+(aq)

O2(g) + Au(s)

Na(s) + H2O(l)

Al3+ (aq) + Hg(l)

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Fe(s) Fe2+ + 2 e- Eox = +0.44 V (the negative of the Ered)Sn2+ + 2e- Sn(s) Ered = -0.14 V

Overall reaction: Fe(s) + Sn2+ Fe2+ + Sn(s) Ecell = +0.30 Vpositive Ecell , therefore spontaneous (product-favored)

You could choose either oxygen reaction, I just chose this one to do the calculation.

4[Au(s) Au3+ + 3 e-] Eox = -1.50 V (the negative of the

Ered)3[O2(g) + 2 H2O(l)+ 4e- 4 OH-] Ered = +0.40 V

multiply the reactions to get charge balance, but NOT the Eox and Ered Overall reaction: 4 Au(s) + O2(g) + 6 H2O(l) 4 Au3+ + 2 OH-

Ecell = -1.10 Vnegative Ecell , therefore non-spontaneous (reactant-favored)

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2[Na(s) Na+ + 1 e-] Eox = +2.714 V (the negative of the Ered)

2 H2O(l)+ 2e- H2 + 2 OH- Ered = -0.8277 Vmultiply the reactions to get charge balance, but NOT the Eox and Ered

Overall reaction: 2 Na(s) + 2 H2O(l) 2 Na+ + H2 + 2 OH- Ecell = +1.886 V

positive Ecell , therefore spontaneous (product-favored)

3[Hg(l) Hg2+ + 2 e-] Eox = -0.855 V (the negative of the

Ered)2[Al3+ + 3e- Al(s)] Ered = -1.66 V

multiply the reactions to get charge balance, but NOT the Eox and Ered

Overall reaction: 3 Hg(l) + 2 Al3+ 3 Hg2+ + 2 Al(s) Ecell = -2.52 V

negative Ecell , therefore non-spontaneous (reactant-favored)

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Summary Activity

A lead acid battery uses the following two reactions to provide a voltage:

Pb + SO42- PbSO4 + 2e-    +0.36V

PbSO4 + 2H2O PbO2 + 4H+ + SO42- + 2e-  -

1.70V

What is the overall reaction?

What is the overall standard cell potential?

Which reaction is the oxidation and which reaction is the reduction?

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A lead acid battery uses the following two reactions to provide a voltage:

Pb + SO42- PbSO4 + 2e-    E1 = +0.36V

PbSO4 + 2H2O PbO2 + 4H+ + SO42- + 2e-  E2 = -1.70V

What is the overall reaction?

(reversing the 2nd reaction produces a spontaneous E; electrons are equal in both reactions, so no multiplication)

Pb + PbO2 + 4H+ + 2SO42- 2PbSO4 + 2H2O

What is the overall standard cell potential?

(since the second reaction was reversed, the E2 = +1.70)

Ecell = 0.36 + 1.70

Ecell = 2.06 V

Which reaction is the oxidation and which reaction is the reduction?

The 1st reaction proceeds as-is, with the electrons being lost. Therefore the 1st reaction is an oxidation.

The 2nd reaction is written as an oxidation, but proceeds in the reverse direction, therefore it is a reduction.

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CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

If 0.10 mol of methane is reacted with 0.10 mol oxygen gas, how many moles of CO2

can form?

1. 0.050 mol2. 0.10 mol3. 0.20 mol4. 0.15 mol

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CH4(g) + 2O2(g) CO2(g) + 2H2O(g)If 0.10 mol of methane is reacted with 0.10

mol oxygen gas, how many moles of CO2 can form?

Limiting reagent problem: calculate the moles of CO2 can form from both 0.10 mol CH4 and 0.10 mole O2. The lower number is the correct answer.

0.10 mol CH4 1 mol CO2 = 0.10 moles CO2

1 mol CH4

0.10 mol O2 1 mol CO2 = 0.050 moles CO2

2 mol O2