Question

Does anyone know how to do these

Answer 1

First acidify all the 3 samples. Add H2O2 to all three samples and chromate , manganate , and vanadate wii oxidize h2o2 to O2 while reducing themselves.

cr2o7(2-) +6e + 14H(+) >2cr3+ +7 H2O.

H2O2 > O2 + 2H(+) + 2e

MnO4(-) + 8H(+) +5e > Mn(2+) + 4H2O

H2O2 > O2 + 2H(+) + 2e

VO3(-) + 6H(+) + 3e. > V(2+) +3H2O

H2O2 +> O2 + 2H(+) + 2e

Step-by-step explanation:

As aresult chromate will form cr3+ ion(blue violet) and manganate will form mn2+ ion (colour less or pale pink). vanadate to V2+ (purple)

Answer 2

How about this?

Step-by-step explanation:

The following ideas might get you started. This is a practical, so I leave the experimental details to you.

1) Reduction of dichromate ion

Slow addition of 5 % hydrogen peroxide to an acidic solution of potassium dichromate.

The skeleton equation is

1×[Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⟶ 2Cr³⁺ + 7H₂O]

3×[H₂O₂ ⟶ O₂ + 2H⁺ + 2e⁻]

Cr₂O₇²⁻ + 3H₂O₂ + 8H⁺ ⟶ 2Cr³⁺ + 3O₂ + 7H₂O

orange colourless blue-green colourless gas

2) Reduction of permanganate ion

Slow addition of an acidic potassium permanganate solution to a solution of iron(II) chloride.

The equations are

2×[MnO₄⁻ + 8H⁺ + 5e⁻ ⟶ Mn²⁺ + 4H₂O]

5×[Fe²⁺⟶ Fe³⁺ + e⁻]

2MnO₄⁻ + 5Fe²⁺ + 16H⁺ ⟶ 2Mn²⁺ + 5Fe³⁺ + 8H₂O

deep purple pale green pale orange-pink yellow-brown

3) Reduction of vanadium(V) to vanadium(II)

Addition of an acidic solution of ammonium metavanadate to granular zinc. There are various colour changes as the vanadium is reduced by the zinc from V(V) ⟶ V(IV) ⟶ V(III) ⟶ V(II).

The equations are

2×[VO₂⁺ + 2H⁺ + e⁻ ⟶ VO²⁺ + H₂O]

1×[Zn ⟶ Zn²⁺ + 2e⁻]

2VO₂⁺ + Zn + 4H⁺ ⟶ 2VO²⁺ + Zn²⁺ + 2H₂O

yellow blue

2×[VO²⁺ + 2H⁺ + e⁻ ⟶ V³⁺ + H₂O]

1×[Zn ⟶ Zn²⁺ + 2e⁻]

2VO²⁺ + Zn + 4H⁺ ⟶ 2V³⁺ + Zn²⁺ + 2H₂O

blue green

2×[V³⁺ + e⁻ ⟶ V²⁺]

1×[Zn ⟶ Zn²⁺ + 2e⁻]

2V³⁺ + Zn ⟶ 2V²⁺ + Zn²⁺

green lilac