Redox

The ion/electron half equations should be written out when oxidation states are changing, showing a redox reaction.

Oxidation is loss of electrons and Reduction is gaining of electrons

e.g Mg(s) --> Mg2+ + e-  or  Fe3+(aq) + e- --> Fe2+(aq)

More complex half equations do exist. Ions such as dichromate (VI) and manganate (VII) include oxygen which require 2H+ ions to form water.

e.g MnO4- + 8H+ + 5e- --> Mn2+ + 4H20     Cr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O

worked example where bromate (V) ion BrO3- can be reduced to Br2

1. Write the reagents and products:   BrO3- --> Br2

2. Balance the atoms:   2BrO3- --> Br2

3. Add two H+ ions to combine with each oxygen to form water:  2BrO3- + 12H+ --> Br2 + 6H2O

4. Find the total changes on both sides to find the number of electrons needed:

charge at the start = 2- + 12+ = 10+

charge at the end = 0 + 0 = 0

so 10e- are needed, which would give us the ion/electron half equation as:

2BrO3- + 12H+ + 10e- --> Br2 + 6H2O

2BrO3- + 12H+ + 10e- --> Br2 + 6H2O

2Br- --> Br2 + 2e-

Combining these half quations above we need to multiply the second by five to give the same amount of electrons as the first:  10Br- --> 5Br2 + 10e-

2BrO3- + 12H+ +10e- --> Br2 + 6H2O

10Br- --> 5Br2 + 10e-

These two half equations are now added together:

2BrO3- + 12H+ + 10e- + 10Br- --> Br2 + 6H2O +5Br2 + 10e-

Cancel anything out and then as they are even numbers we'd need to divide by two:

BrO3- + 6H+ + 5Br- --> 3Br2 + 3H2O

Carried out in the same way as acid-base titrations. This involves 25.0cm3 of a solution being measured using a volumetric pipette and placed in a conical flask. A second solution is added a little at a time from a burette, swirling the mixture during addition. This is continued until the desired colour change is seen. Many redox titrations do not require an in indicator as the colours of the reactants frequently allow the end point to be seen. The volume of solution added is measured using initial and final burette readings. 

Example:

Acidified manganate (VII) ions with iron (II) ions (H2SO4 normally used)

1. MnO4-(aq) + 8H+(aq) + 5e- --> Mn2+(aq) + 4H2O(l)  colour change = purple --> pale pink (this is caused because a small amount of purlpe remains, Mn2+ is almost colourless)

2. Fe2+(aq) --> Fe3+(aq) + e-   colour change = pale green (Fe2+) --> yellow (Fe3+)

3. MnO4-(aq) + 8H+(aq) +5Fe2+ --> Mn2+(aq) + 4H20(l) + 5Fe3+(aq)

Calculations: number of moles of Fe2+/number of moles of MnO4- = 5/1

A sample of 25.00cm3 of a solution of acidified iron(II) sulfate was titrated using a solution of potassium manganate (VII) of concentration 0.0200moldm-3. The results obtained are shown in the table below. Calculate the concentration of FeSO4. 

initial reading (cm3) : 0.00   0.55   0.20   0.75

final reading (cm3) : 30.04  30.35  30.10  30.60

volume used (cm3) : 30.04  29.80  29.90  29.85

nKMnO4 = c x v / 1000  (0.0200 x 29.85)/1000 = 5.97x10-4mol

ratio 1:5  so, 5.97x10-4 x 5 = 2.985x10-3 is the number of moles for FeSO4.

c = n x 1000/v  (2.985x10-3 x 1000)/25.00 = 0.119 moldm-3

Write an ionic equation for the oxidation of oxalic acid (ethanedioic acid), C2O4H2 to CO2 by acidified KMnO4.

C2O42- --> 2CO2 + 2H+ + 2e-   x5

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O   x2

Make sure there's an even number of electrons in both equations to allow cancellation to take place: 

Full equation : 2MnO4- + 16H+ + 5C2O42- --> 2Mn2+ + 10CO2 + 8H2O

Potassium dichromate (VI) with iron(II) ions half equations (H2SO4 used):

Cr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O   [2- + 14+ = 12+ - 6+ = 6+]

Fe2+ + e- --> Fe3+ 

Full equation: Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) --> 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

Colour change from dark orange to green

An indirect method is used to measure Cu(II) ions. An addition of a colourless solution containing iodide ions such as potassium iodide, to a blue solution containing copper (II) ions leads to the formation of a cloudy brown solution. Cu2+ ions in solution react with iodide ions to generate a brown solution of iodine and are reduced to copper (I) in a precipitate of CuI. Equation:

2Cu2+(aq) + 4I-(aq) --> 2CuI(s) + I2(aq) Iodine then reacts with sodium thiosulfate (reducing agent)

2S2O32-(aq) --> S4O62-(aq) + 2e-

I2(aq) + 2e- --> 2I-(aq)  colour change: brown --> colourless

overall equation: 2S2O32-(aq) + I2(aq) --> S4O62-(aq) + 2I-(aq) 

Cu2+ (blue solutions0 --> (cloudy brown forms after mixing with KI)

Titrate the brown Iodine against sodium thiosulfate (Na2S2O3), until the mixture is starw-coloured.

Add starch indicator which goes blue-black, and continue until the colour vanishes. The mixture is often described as flesh-coloured.