Y13 Redox

Question 1

What are the 2 exceptions for oxygen’s oxidation number?

Answer 1

1. Hydrogen peroxide (H2O2)
2. F2O - fluorine is always -1

Question 2

What are the 4 steps to constructing a half equation in ACIDIC conditions?

Answer 2

e.g. VO2(+) –> VO(2+)

1. Balance V (already balanced)
2. Balance oxygen: VO2(+) –> VO(2+) + H2O
3. Balance hydrogen: VO2(+) + 2H(+) –> VO(2+) + H2O
4. Balance the charges: VO2(+) + 2H(+) + 2e(-) –> VO(2+) + H2O

Question 3

What must you check after constructing a half equation in acidic conditions?

Answer 3

That V’s change in oxidation number matches the number of electrons added/subtracted

Question 4

How do you construct a half equation in ALKALINE conditions?

Answer 4

1. Construct the half equation as if it is in acidic conditions (balance O2, H+, charge)
2. Cancel any H+ ions by adding OH- ions (put them on both sides of the equation)

Question 5

What are the 4 steps to predicting products in redox reactions?

Answer 5

Each
Total
Total
Each

Question 6

What is the final species containing silver for this reaction: Pb + 2Ag(+) –> Pb(2+) + x?

Answer 6

Each: Pb state change 0 –> +2 (increase 2)

Total: 2 x 1(mole of Pb) = 2

Total: -2

Each: -2 / 2 moles of Ag(+) = -1

So 2Ag(+) –> Ag

Question 7

How do you combine half equations?

Answer 7

1. Multiply to ensure the number of electrons of both equations match
2. Cancel out the e- and combine all products and reactants together

Question 8

Construct the fully balanced redox ionic equation from these 2 half equations:
MnO4(-) + 8H(+) + 5e(-) –> Mn(2+) + 4H2O
Fe(2+) –> Fe(3+) + e(-)

Answer 8

Fe(2+) –> Fe(3+) + e(-) becomes 5Fe(2+) –> 5Fe(3+) + 5e(-) to balance electrons

Combined: MnO4(-) + 8H(+) + 5Fe(2+) –> Mn(2+) + 4H2O + 5Fe(3+)

Question 9

What is the colour change in the titration of a known volume of iron(II) against potassium manganate(VII)?

Answer 9

Colourless to pale pink

Question 10

What is the exam technique for manganate(VII) titration calculations (e.g. % purity, calculating x, conc. of iron)?

Answer 10

1. Equation (molar ratio)
2. n (MnO4-)
3. Deduce n(Fe2+) from ratio
4. Scale up e.g. moles x (250/25)
5. Depends on the question

Question 11

Why does the iron(II) compound need to be acidified in a titration?

Answer 11

To obtain the correct products

Question 12

What are equations 1 and 2 of redox titrations using thiosulfate?

Answer 12

1. Oxidising agent + 2I(-) –> I2 + ____
2. 2S2O3(2-) + I2 –> S4O6(2-) + 2I(-)

Question 13

What are the colour changes as thiosulfate is added, what extra substance is added, and why?

Answer 13

Brown –> orange –> yellow –> colourless
Starch is added when the solution is yellow so the colour change from yellow to colourless is easier to distinguish. Instead the colour change is blue-black to colourless

Question 14

How are the moles of oxidising agent calculated in a thiosulfate titration?

Answer 14

1. Divide n(S2O3(2-)) by 2 to get n(I2)
2. Carries over to the first equation because I2 is the same in both equations
3. Use molar ratios to calculate n(oxidising agent) from n(I2)