Transition Metals - Redox

Question 1

Fe2+ -> Fe3+ reaction

Answer 1

Fe2+ can be oxidised to Fe3+ using potassium manganate, KMnO4.

Fe2+ -> Fe3+
MnO4- -> Mn2+

MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O

Question 2

Fe2+ -> Fe3+ colour change

Answer 2

Purple -> colourless

MnO4- ions are purple in solution, whilst Mn2+ ions are colourless.

Question 3

Fe3+ -> Fe2+ reaction

Answer 3

Fe3+ can be reduced to Fe2+ using I- ions.

Fe3+ + e- -> Fe2+
2I- -> I2 + 2e-

2Fe3+ + 2I- -> 2Fe2+ + I2

Question 4

Fe3+ -> Fe2+ colour change

Answer 4

orange-brown -> brown

Question 5

Cr6+ -> Cr3+ reaction

Answer 5

Cr2O7 2- is reduced to Cr3+ using alcohol or zinc.

(incomplete half equations) =

Cr2O7 2- + 6e- -> 2Cr3+
Zn -> Zn2+ + 2e-

Question 6

Cr6+ -> Cr3+ colour change

Answer 6

orange -> green

Question 7

Cr3+ -> Cr6+ reaction

Answer 7

Cr3+ can be oxidised to Cr6+ by hot, alkaline H2O2.

Cr3+ + 8OH- -> CrO4 2- + 4H2O + 3e-
H2O2 + 2e- -> 2OH-

Question 8

Cu2+ -> Cu+ reaction

Answer 8

Cu2+ reduced to Cu+ using I-.

Cu2+ + e- -> Cu+
2I- -> I2 + 2e-

2Cu2+ + 2I- -> 2Cu+ + I2

Question 9

Cu2+ -> Cu+ colour change

Answer 9

pale blue -> brown (due to iodine) and white precipitate (due to some of the CuI that forms)

Question 10

Cu+ -> Cu2+ reaction

Answer 10

Cu+ ions in Cu2O.
Cu2O is oxidised to Cu2+ with H2SO4.

Cu2O + H2SO4 -> CuSO4 + Cu (s) + H2O

Disproportionation reaction as copper is both oxidised and reduced.

Question 11

Cu+ -> Cu2+ colour change

Answer 11

Blue solution forms (copper (II) sulfate) and brown precipitate forms (Cu (s))