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Published bySara Reynolds Modified over 8 years ago
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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
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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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Anode Cathode Ve--+
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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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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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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.
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Anode: Pb → Pb 2+ Cathode: PbO 2 → Pb 2+ + 2H 2 0 + 4H + + 2e -
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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+
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Cu → Cu 2+ Cu → Cu 2+ NO 3 - → NO 2 NO 3 - → NO 2 + H 2 0 + 2H + + e - + 2e -
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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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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-
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-2+1 +6 -2 +4 -2 +7
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A disproportionation reaction is a redox reaction where the same substance is both oxidised and reduced
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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.
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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
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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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X 3
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Co Co 2+ C Cr 3+ Cr 2 O 7 2- e- K+K+ NO 3 - CATHODE ANODE
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e- -+
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X 2 X 3 k u
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