balance this equation by ion electron method:?

Question

Cr(OH)3+H2O2_______>CrO4(it has _2 charge on it)+OH-


why hydroxyl ions possess negative charge?

Answer 1

(Cr3+ + 4 H2O>> CrO42- + 8H+ + 3e-) 2

(H2O2 + 2e- >> 2OH- )3
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2 Cr3+ + 8H2O >>2 CrO42- + 16 H+ +6e-

3H2O2 + 6e- >> 6 OH-


2Cr(OH)3 + 8H2O >> 2CrO42- +6OH- + 16H+ + 6e-

3H2O2 +6e- >> 6OH-


2Cr(OH)3 + 2H2O >> 2CrO42- +10 H+ +6e-
3 H2O2 + 6e- >> 6 OH-


2Cr(OH)3 + 2H2O + 3 H2O2 >>2CrO42- + 10H+ + 6 OH-

2Cr(OH)3 + 3H2O2 >> 2CrO42- + 4H2O +4 H+

2Cr(OH)3 + 3H2O2 + 4OH- >> 2CrO42- + 8H2O

I hope this help you

Answer 2

Answering question 2 first, OH- is the part of water that indicates you are in a basic solution, and not an acid solution. If the situation were reversed, you would see H+.

The reaction is
Cr(OH)3 + H2O2 -> CrO4= + OH-
First, you identify the species being oxidized and the species being reduced.

The one being oxidized is Cr+3, which goes to Cr+6 in the chromate ion. The one being reduced is the O-1 (peroxide is H-O-O-H), which goes to O=.

Balance the oxidation and reduction reactions. Supply H2O as needed to balance the O and H requirements.

Cr(OH)3 + 5OH- -> CrO4 = + 4 H2O +3e
H2O2 + 2e -> 2 OH-

We multiply the Cr(OH)3 eqtn by 2 and the H2O2 equation by 3 to get
2 Cr(OH)3 +10 OH- -> 2 CrO4= + 8 H2O + 6e
3 H2O2 + 6e -> 6 OH-
We add equations and move common species to the side where they are positive-valued.
2 Cr(OH)3 + 4OH- + 3 H2O2 -> 2 CrO4= + 8H2O