Redox reactions

Question 1

How to create an ionic half equation?

Answer 1

1. Write reagents and products
2. Balance atoms present
3. Add hydrogen ions to combine with oxygen to form water
4. Find charges on both sides to add the electrons needed

Question 2

What is the dichromate half equation?

Answer 2

Cr2O7^2- + 14H+ +6e- —> 2Cr^3+ + 7H2O

Question 3

What is the permanganate half equation?

Answer 3

MnO4- +8H+ +5e- —> Mn2+ +4H2O

Question 4

How to combine half equations?

Answer 4

1. Multiply equations to match the number of electrons
2. Combine the equations and cancel out the electrons

Question 5

Describe a redox titration

Answer 5

25cm3 of a solution is measured using a volumetric pipette and placed in a conical flask. A second solution is added a bit at a time from the burette, swirling through mixture during addition until a colour change is seen. This procedure is repeated until readings are taken sufficiently close together.

Question 6

Why do many redox titrations not need indicator?

Answer 6

As the colours of the reactants frequently allow end-points to be seen.

Question 7

Describe a titration of acidified manganate (VII) ions with iron (II) ions

Answer 7

Manganate goes from purple to colourless when reduced. The colour change of the solution as Fe2+ (pale green) to Fe3+ (yellow) is hard to see which is why manganate is used. The iron ions are then oxidised and H2SO4 is added to get the overall equation:
MnO4- + 8H+ 5Fe2+ —> Mn2+ + 4H2O + 5Fe3+

Question 8

Describe a titration of acidified dichromate (VI) ions with iron (II) ions

Answer 8

Potassium dichromate in acid solution will oxidise Fe2+ to Fe3+ with a colour change from dark orange Cr2O7^2- to green Cr3+ and H2SO4 being added with the overall equation being
Cr2O7^2- + 14H+ + 6Fe2+ —> 2Cr3+ + 7H2O + 6Fe3+

Question 9

Describe a redox titration for copper (II) ions

Answer 9

Addition of a colourless solution containing iodide ions to a blue solution containing copper (II) ions leads to a cloudy brown solution. Cu2+ ions in solution react with iodide ions to generate a brown solution of iodine, and are reduced to Cu+ in a precipitate of CuI. We can titrate this iodine with sodium thiosulphate to work out the amount of copper present originally, a brown to colourless change is seen as thiosulphate reduces I2 to I- in the overall equation 2S2O3^2- + I2 —> S4O6^2- + 2I-