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Appearance & state of reductants & oxidants Reduced form Oxidised Form FormulaAppearanceFormulaAppearance Cu Brown solid Cu 2+ Blue ion SO 2.

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Presentation on theme: "Appearance & state of reductants & oxidants Reduced form Oxidised Form FormulaAppearanceFormulaAppearance Cu Brown solid Cu 2+ Blue ion SO 2."— Presentation transcript:

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7 Appearance & state of reductants & oxidants Reduced form Oxidised Form FormulaAppearanceFormulaAppearance Cu Brown solid Cu 2+ Blue ion SO 2 SO 4 2- Mn 2+ MnO 4 - Purple ion H2O2H2O2H2O2H2O2 O2O2O2O2 H2OH2OH2OH2O H2O2H2O2H2O2H2O2 Cr 3+ Green ion Cr 2 O 7 2- Orange ion Fe 2+ Green ion Fe 3+ Orange ion Cl - Cl 2 Pale green gas I-I-I-I- I 2 (aq) Brown solution H2H2H2H2 H+H+H+H+ Zn Grey solid Zn 2+ Br - Br 2 (aq) Orange solution

8 Appearance & state of reductants & oxidants Reduced form Oxidised Form FormulaAppearanceFormulaAppearance MnO 2 Brown ppt H 2 O/MnO 4 - Purple ion MnO 4 2- OH - /MnO 4 - I2I2I2I2 IO 3 - Colourless ion H2SH2SH2SH2SS Pb 2+ PbO 2 NO 2 Brown gas NO 3 - Colourless ion C 2 O 4 2- Colourless ion CO 2 Colourless gas S 2 O 3 2- S 4 O 6 2- Colourless ion Br - BrO 3 - Colourless ion SO 3 2- SO 4 2-

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10 Anode Cathode Ve--+

11 Cu(s)/ Cu 2+ (aq)// MnO 4 – (aq), Mn 2+ (aq)/ C(s) Pt(s)/ Cl - (aq)/ Cl 2 (g)// BrO 3 - (aq),Br 2 / C(s)

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13 Strongest & weakest Standard reduction potentials (E  Cell) are written with reductants on the right Standard reduction potentials (E  Cell) are written with reductants on the right Strongest reductant is the species on the right with the most negative E  Strongest reductant is the species on the right with the most negative E  Strongest oxidant is the species on the left with the most positive E  Strongest oxidant is the species on the left with the most positive E  Weakest oxidant = strongest reductant = most –ve Weakest oxidant = strongest reductant = most –ve Weakest reductant = strongest oxidant = most +ve Weakest reductant = strongest oxidant = most +ve

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15 0 -2 +4 -2 +2 0 +4 +2 ON of Pb has increased from 0 to +2 so Pb is being oxidised and is acting as the reductant. ON of Pb in PbO 2 has decreased from +4 to +2 so PbO 2 is being reduced and is acting as the oxidant.

16 Anode: Pb → Pb 2+ Cathode: PbO 2 → Pb 2+ + 2H 2 0 + 4H + + 2e -

17 Brown metal is placed in colourless solution and slowly disappears, producing heat. A brown gas (NO 2 ) is produced The liquid turns increasingly blue due to Cu 2+

18 Cu → Cu 2+ Cu → Cu 2+ NO 3 - → NO 2 NO 3 - → NO 2 + H 2 0 + 2H + + e - + 2e -

19 Cu 2+ → Cu + Cu 2+ → Cu + 2I - → I 2 2I - → I 2 + 2e - + e - + e - I - is oxidised to I 2 while the Cu 2+ is reduced to Cu + I - is oxidised to I 2 while the Cu 2+ is reduced to Cu + I - combines with Cu + to form white solid CuI I - combines with Cu + to form white solid CuI 2Cu 2+ + 2I - → 2Cu + + I 2 2Cu 2+ + 2I - → 2Cu + + I 2 222 I 2 → 2I - I 2 → 2I - + 2e - S 2 O 3 2- → S 4 O 6 2- S 2 O 3 2- → S 4 O 6 2-2 + 2e - brown solution of I 2 is reduced by colourless S 2 O 3 2- to form colourless I -. S 4 O 6 2- also formed is colourless too brown solution of I 2 is reduced by colourless S 2 O 3 2- to form colourless I -. S 4 O 6 2- also formed is colourless too I 2 + 2S 2 O 3 2- → 2I - + S 4 O 6 2- I 2 + 2S 2 O 3 2- → 2I - + S 4 O 6 2-

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22 MnO 4 - → Mn 2+ MnO 4 - → Mn 2+ SO 3 2- → SO 4 2- SO 3 2- → SO 4 2- + 4H 2 0 + 8H + + 2e - + 5e - + H 2 0 + 2H + X 2 X 5 2MnO 4 - + 16H + +5SO 3 2- + 5H 2 O → 2Mn 2+ + 8H 2 O + 5SO 4 2- + 10H + 2MnO 4 - + 6H + +5SO 3 2- → 2Mn 2+ + 3H 2 O + 5SO 4 2-

23 -2+1 +6 -2 +4 -2 +7

24 A disproportionation reaction is a redox reaction where the same substance is both oxidised and reduced

25 Solution is intially green due to the presence of MnO 4 2-. It then turns purple due to the presence of MnO 4 - and a dark brown/black ppt of MnO 2 forms.

26 MnO 4 2- → MnO 4 - MnO 4 2- → MnO 4 - + e - MnO 4 2- → MnO 2 MnO 4 2- → MnO 2 + 2H 2 0 + 4H + + 2e - X 2 3MnO 4 2- + 4H + → 2MnO 4 - + MnO 2 + 2H 2 O 3MnO 4 2- + 4H + → 2MnO 4 - + MnO 2 + 2H 2 O

27 M(K 2 MnO 4 ) = 197.1 gmol -1 3MnO 4 2- + 4H + → 2MnO 4 - + MnO 2 + 2H 2 O 3MnO 4 2- + 4H + → 2MnO 4 - + MnO 2 + 2H 2 O n(K 2 MnO 4 ) = m/M = 1.05/197.1 = 1.05/197.1 = 0.00533 mol = 0.00533 mol n(MnO 2 ) = n(K 2 MnO 4 )/3 = 0.00178 mol = 0.00178 mol m(MnO 2 ) = n X M = 0.00178 X 86.9 = 0.00178 X 86.9 = 0.154 g = 0.154 g

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30 Co Co 2+ C Cr 3+ Cr 2 O 7 2- e- K+K+ NO 3 - CATHODE ANODE

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36 e- -+

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38 X 2 X 3 k u

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