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