Question

Cr2O72- + 14H+ + 6Isup>- → 2Cr3++ ____ +7H2O

Answer 1

Answer:

Explanation:

6 (I)- + Cr2O72- + 14 H+ → 3 I2 + 2 (Cr)3+ + 7 H2O

This is an oxidation-reduction (redox) reaction:

6 I-I - 6 e- → 6 I0 (oxidation)

2 CrVI + 6 e- → 2 CrIII (reduction)

I- is a reducing agent, Cr2O72- is an oxidizing agent.

Reactants:

I-

Cr2O72-

H+

Products:

I2

Names: Iodine source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Jod source: ICSC, accessed: 2019-09-04, Iode source: ICSC, accessed: 2019-09-04

Appearance: Bluish black or dark purple crystals with pungent odour source: ICSC, accessed: 2019-09-04; Violet solid with a sharp, characteristic odor. source: NIOSH NPG, accessed: 2019-09-02

Cr3+

H2O – Water, oxidane source: wikipedia, accessed: 2019-09-27

Other names: Water (H2O) source: wikipedia, accessed: 2019-09-27, Hydrogen hydroxide (HH or HOH) source: wikipedia, accessed: 2019-09-27, Hydrogen oxide source: wikipedia, accessed: 2019-09-27

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas

Answer 2

Answer:

3 I2 is the right answer