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Lect w13 152_electrochemistry_key

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  • 1. General Chemistry II CHEM 152 Unit 4 Week 13
  • 2. A 3.41 x 10 -6 g sample of a compound is known to contain 4.67 x 10 16 molecules. This compound is:
    • CO 2
    • CH 4
    • NH 3
    • H 2 O
  • 3. Solution
    • A 3.41 x 10 -6 g sample of a compound is known to contain 4.67 x 10 16 molecules. This compound is?
    • Gameplan: Find the molecular weight (g/mole), then assess what compounds could have that MW
    • 4.67 x 10 16 molecules  1 mole = 7.65  10 -8
    • 6.02  10 23 molecules moles
    • 3.41 x 10 -6 g = 44.0 g/mole
    • 7.65  10 -8 moles What molecule has this MW?
  • 4. A 3.41 x 10 -6 g sample of a compound is known to contain 4.67 x 10 16 molecules. This compound is? 44.0 g/mole
    • CO 2 44 g/mole
    • CH 4 16 g/mole
    • NH 3 17 g/mole
    • H 2 O 18 g/mole
  • 5. In a neutral COMPOUND some elements often have a fixed OXIDATION NUMBER H almost always is +1 O almost always is -2 F always is -1 3) In a NEUTRAL COMPOUND the SUM of all the atomic oxidation numbers is ZERO CH 4 CH 3 COOH ? +1 CH 4 1C + 4H = 0 C + 4(+1) = -4 -4 +1 CH 4 ? +1 ? -2 -2 +1 CH 3 C O O H 2C + 4H + 2O = 0 2C + 4(+1) +2(-2) = 0 2C + 4 – 4 = 0 0 +1 0 -2 -2 +1 CH 3 C O O H
  • 6. 4) In a COMPLEX ION the SUM of all the atomic oxidation numbers is equal to the TOTAL CHARGE on the ion. ClO 3 − MnO 4 − PO 3 3- ? -2 ClO 3  1Cl + 3O = -1 Cl + 3(-2) = -1 +5 -2 ClO 3  ? -2 MnO 4  1Mn + 4O = -1 Mn + 4(-2) = -1 +7 -2 MnO 4  ? -2 PO 3 3  1P + 3O = -3 P + 3(-2) = -3 +3 -2 PO 3 3 
  • 7. What is the oxidation number of P in H 3 PO 4 ? P = +5 What is the oxidation number of Cr in Cr 2 O 7 2- ? Cr = +6 What is the oxidation number of C in (NH 4 ) 2 CO 3 ? C = +4 What is the oxidation number of C in C 3 H 8 (propane)? C = -2 2/3 or -8/3
  • 8. When the carbon-carbon bond lengths in ethane (C 2 H 6 ), ethene (C 2 H 4 ), and benzene (C 6 H 6 ) are arranged in order of increasing bond length (shortest first), which is the correct order?
    • C 2 H 6 < C 2 H 4 < C 6 H 6
    • C 2 H 4 < C 2 H 6 < C 6 H 6
    • C 6 H 6 < C 2 H 4 < C 2 H 6
    • C 2 H 4 < C 6 H 6 < C 2 H 6
  • 9. Propane again… Using the structure method, what is the oxidation number of each carbon in propane (C 3 H 8 )? 1 4  7 = -3 2 4  6 = -2 3 4  7 = -3 1 2 3
  • 10. What is the oxidation number of the oxygen-bonded C in CH 3 COOH (acetic acid)? 2 is the O-bonded C O.N. 4-1 = +3 C C H H O H O—H = : : : : 1 2
  • 11. What is the oxidation number of the C in NH 2 CONH 2 ? In ClCH 2 CH 2 Cl? O.N. 4-0 = +4 both carbons are equivalent O.N. 4-5 = -1 H—N—C—N—H H H O = : : : : H—C — C—H H H Cl Cl : : : : : :
  • 12. S 8 + 12 O 2  8 SO 3 Is it redox? If so, What is oxidized? What is the reducing agent?
  • 13. 0 0 +6 -2 S 8 + 12 O 2  8 SO 3 Is it redox? If so, What is oxidized? 0  +6 Sulfur is oxidized 0  -2 Oxygen is reduced What is the reducing agent? S 8 What is the reducing agent? O 2
  • 14. 2 AgNO 3 + Na 2 S  Ag 2 S + 2 NaNO 3 Is it redox? 5 As 4 O 6 + 8 MnO 4 - + 18 H 2 O  20 AsO 4 3- + 8 Mn 2+ + 36 H + What is oxidized and reduced? What is the oxidizing agent (OA) and reducing agent (RA)?
  • 15. NOT REDOX +1 +5 -2 +1 -2 +1 -2 +1 +5-2 2 AgNO 3 + Na 2 S  Ag 2 S + 2 NaNO 3 Is it redox? Na will always be +1 in ionic compounds Ag is a transition element, but because we know nitrate and sulfide, it is +1 in both
  • 16. Is it redox? +3 -2 +7 -2 +1 -2 5 As 4 O 6 + 8 MnO 4 - + 18 H 2 O +5 -2 +2 +1  20 AsO 4 3- + 8 Mn 2+ + 36 H + As is oxidized Mn is reduced As 4 O 6 is the reducing agent MnO 4 - is the oxidizing agent
  • 17. Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored. Fe(s) + Sn 2+ (aq)  O 2 (g) + Au(s)  Na(s) + H 2 O(l)  Al 3+ (aq) + Hg(l) 
  • 18. Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored. Fe(s) + Sn 2+ (aq)  O 2 (g) + Au(s)  Na(s) + H 2 O(l)  Al 3+ (aq) + Hg(l)  product favored
  • 19. Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored. Fe(s) + Sn 2+ (aq)  O 2 (g) + Au(s)  Na(s) + H 2 O(l)  Al 3+ (aq) + Hg(l)  reactant favored
  • 20. Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored. Fe(s) + Sn 2+ (aq)  O 2 (g) + Au(s)  Na(s) + H 2 O(l)  Al 3+ (aq) + Hg(l)  product favored
  • 21. Predict the outcome of the following reaction. Classify each of the reactions as processes as product-favored or reactant-favored. Fe(s) + Sn 2+ (aq)  O 2 (g) + Au(s)  Na(s) + H 2 O(l)  Al 3+ (aq) + Hg(l)  reactant favored
  • 22. Calculate the E  cell for these reactions. Fe(s) + Sn 2+ (aq)  O 2 (g) + Au(s)  Na(s) + H 2 O(l)  Al 3+ (aq) + Hg(l) 
  • 23. Fe(s)  Fe 2+ + 2 e - E  ox = +0.44 V (the negative of the E  red ) Sn 2+ + 2e -  Sn(s) E  red = -0.14 V Overall reaction: Fe(s) + Sn 2+  Fe 2+ + Sn(s) E  cell = +0.30 V positive E  cell , therefore spontaneous (product-favored) You could choose either oxygen reaction, I just chose this one to do the calculation. 4[ Au(s)  Au 3+ + 3 e - ] E  ox = -1.50 V (the negative of the E  red ) 3[ O 2 (g) + 2 H 2 O(l)+ 4e -  4 OH - ] E  red = +0.40 V multiply the reactions to get charge balance, but NOT the E  ox and E  red Overall reaction: 4 Au(s) + O 2 (g) + 6 H 2 O(l)  4 Au 3+ + 2 OH - E  cell = -1.10 V negative E  cell , therefore non-spontaneous (reactant-favored)
  • 24. 2[ Na(s)  Na + + 1 e - ] E  ox = +2.714 V (the negative of the E  red ) 2 H 2 O(l)+ 2e -  H 2 + 2 OH - E  red = -0.8277 V multiply the reactions to get charge balance, but NOT the E  ox and E  red Overall reaction: 2 Na(s) + 2 H 2 O(l)  2 Na + + H 2 + 2 OH - E  cell = +1.886 V positive E  cell , therefore spontaneous (product-favored) 3[ Hg(l)  Hg 2+ + 2 e - ] E  ox = -0.855 V (the negative of the E  red ) 2[ Al 3+ + 3e -  Al(s) ] E  red = -1.66 V multiply the reactions to get charge balance, but NOT the E  ox and E  red Overall reaction: 3 Hg(l) + 2 Al 3+  3 Hg 2+ + 2 Al(s) E  cell = -2.52 V negative E  cell , therefore non-spontaneous (reactant-favored)
  • 25. Summary Activity A lead acid battery uses the following two reactions to provide a voltage: Pb + SO 4 2-  PbSO 4 + 2e -    +0.36V PbSO 4 + 2H 2 O  PbO 2 + 4H + + SO 4 2- + 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?
  • 26. A lead acid battery uses the following two reactions to provide a voltage: Pb + SO 4 2-  PbSO 4 + 2e -    E  1 = +0.36V PbSO 4 + 2H 2 O  PbO 2 + 4H + + SO 4 2- + 2e -   E  2 = -1.70V What is the overall reaction? (reversing the 2 nd reaction produces a spontaneous E  ; electrons are equal in both reactions, so no multiplication) Pb + PbO 2 + 4H + + 2SO 4 2-  2PbSO 4 + 2H 2 O What is the overall standard cell potential? (since the second reaction was reversed, the E  2 = +1.70) E  cell = 0.36 + 1.70 E  cell = 2.06 V Which reaction is the oxidation and which reaction is the reduction? The 1 st reaction proceeds as-is, with the electrons being lost. Therefore the 1 st reaction is an oxidation. The 2 nd reaction is written as an oxidation, but proceeds in the reverse direction, therefore it is a reduction.
  • 27. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) If 0.10 mol of methane is reacted with 0.10 mol oxygen gas, how many moles of CO 2 can form?
    • 0.050 mol
    • 0.10 mol
    • 0.20 mol
    • 0.15 mol
  • 28. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) If 0.10 mol of methane is reacted with 0.10 mol oxygen gas, how many moles of CO 2 can form?
    • Limiting reagent problem: calculate the moles of CO2 can form from both 0.10 mol CH 4 and 0.10 mole O 2 . The lower number is the correct answer.
    • 0.10 mol CH 4  1 mol CO 2 = 0.10 moles CO 2
    • 1 mol CH 4
    • 0.10 mol O 2  1 mol CO 2 = 0.050 moles CO 2
    • 2 mol O 2