Sunday 28 November 2010

Chemistry : Redox II

Balancing two ionic-half equation
The reaction occurs when a ionic-half equation at a higher position in E.C.S. goes as reverse, and another ionic-half equation at a lower position in E.C.S. is added together.
e.g. I2 + 2e- → 2I- is higher than MnO4- + 8H+ + 5e- → Mn2+ + 4H2O, so we write the first equation in reverse form: 2I- → I2 + 2e-, then we balance them by eliminating the electrons:
5x     2I- → I2 + 2e-  and  2x       MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Overall: 10I- + 2MnO4- + 16H+ → 5I2 + 2Mn2+ + 8H2O
Note that there shouldn’t be any electron in the balanced equation and the charge conserves.
Mentionable oxidizing and reducing agent:
1)       Chlorine as a strong oxidizing agent: Cl2 + 2e- → 2Cl-
i)                     Displacement on KBr and KI: Cl2 + 2X- → 2Cl- + X2, when Cl2 gas (pale green) is bubbled into the solution, Br2 (yellow-brown) / I2 (orange) solution is formed. When organic solvent (hexane) is added, Br and I become orange and violet respectively.
ii)                   Disproportionation against NaOH:
When it’s added to cold, dilute NaCl: Cl2 + NaOH → NaCl + NaOCl + H2O, O.N. of Cl change from 0 to +1 (in NaOCl) and -1(NaCl) respectively.
When it’s added to hot, conc., NaOH: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O, where O.N. of Cl in NaClO3 changed from 0 to +5.
The reaction where the same species is simultaneously reduced and oxidized is called disproportionation.

2)       Reactions with hot and conc. HNO3 (about 16M) gives brown and toxic gas NO2:
NO3- + 2H+ + e- → NO2 + H2O. e.g. Cu + 2NO3- + 4H+ → Cu2+ + 2NO2 + 2H2O
Reactions between non-metals and conc. HNO3:
C + 4HNO3 → CO2 + 4NO2 + 2H2O and S + 4HNO3 → SO2 + 4NO2 + 2H2O
Dilute HNO3 (about 2M) reacts to give NO: NO3- + 4H+ + 3e- → NO + 2H2O, but NO released will mix with oxygen in the air to give out brown NO2 (2NO + O2 → 2NO2), so when it’s mixed in a test tube, brown gas is only observed at the opening of the tube.
Very dilute HNO3 (less than 1M) only perform like a dilute acid which gives CO2.
3)       Reaction with dilute H2SO4: like a normal dilute acid, reacts with Zn but not Cu.
Hot and conc. H2SO4 oxidizes most metals: 2H2SO4 + 2e- → SO42- + SO2 + 2H2O
Reaction with hot and conc. H2SO4 with non-metals:
C + 4H2SO4 → CO2 + 4SO2 + 2H2O and S + 2H2SO4 → 3SO2 + 2H2O
Reaction with halides: for F and Cl, NaX + H2SO4 → NaHSO4 + HX
Bromides: NaBr + H2SO4 → NaHSO4 + HBr and 2HBr + H2SO4 → SO2 + Br2 + 2H2O, the overall reaction is : 2NaBr + 3H2SO4 → 2NaHSO4 + SO2 + Br2 + 2H2O
Iodides: NaI + H2SO4 → NaHSO4 + HI and 8HI + H2SO4 → H2S + 4I2 + 4H2O, the overall reaction is 8NaBr + 9H2SO4 → 8NaHSO4 + H2S + 4I2 + 4H2O
NaBr and NaI reacts in different behavior since HBr and HI is further oxidized by H2SO4.
4)       Aqueous sulphur dioxide (sulphurous acid, H2SO3) as a reducing agent is equivalent to SO32-. It can reduce other species and change to SO42-: SO32- + H2O → SO42- + 2H+ + 2e-
Test for reducing agent (especially SO2): filter paper soaked with KCr2O7/H+, if reducing agent presents, it reduces the filter paper from orange to green.
Test for oxidizing agent: filter paper soaked with (colourless) starch solution and KI, if oxidizing agent presents, it oxidizes 2I- to I2, and turn the starch solution to dark blue.
Test for HCl and HBr: white fumes formed with NH3(aq) is added (fumes from HCl is denser)
Test for H2S: turn lead(II) ethanoate paper black.

Electrochemistry is an extremely long topic and this is only one of the five chapters, doc. version won't be released until it's finished under the following sequences:
-Redox (completed)
-Chemical Cell
-Daily used cell

And I'm sorry that there's a little AL and non-DSE stuffs in the note.

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