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^- → SO₄^2- + 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 SO₃^2-. It can reduce other species and change to SO₄^2-: SO₃^2- + H₂O → SO₄^2- + 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
-Electrolysis

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